diff options
| author | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
| commit | 7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch) | |
| tree | 449d555969bfd7befe906877abab098c6e63a0e8 /3886 | |
| parent | d1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff) | |
| download | Scilab-TBC-Uploads-master.tar.gz Scilab-TBC-Uploads-master.tar.bz2 Scilab-TBC-Uploads-master.zip | |
Diffstat (limited to '3886')
554 files changed, 5515 insertions, 0 deletions
diff --git a/3886/CH10/EX10.13/10_13.sce b/3886/CH10/EX10.13/10_13.sce new file mode 100644 index 000000000..117f8007c --- /dev/null +++ b/3886/CH10/EX10.13/10_13.sce @@ -0,0 +1,12 @@ +//Determine radius of gyration
+//refer fig.10.23
+//composite body may be divided into
+//1.a solid block of size (80*120*100 mm) and 2.two semicircular grooves each of 40 mm radius and 80 mm length
+//Let's assign random value to rho
+rho=1
+//Ig=1.029*10^8*rho
+//Ix2=Ig+M2*d'^2
+Ixx=10.0816*(10^8)*rho
+M=557876.14*rho //units
+k=sqrt(Ixx/M) //mm
+printf("\nk=%.2f mm",k)
diff --git a/3886/CH10/EX10.13/10_13.txt b/3886/CH10/EX10.13/10_13.txt new file mode 100644 index 000000000..c706d22d0 --- /dev/null +++ b/3886/CH10/EX10.13/10_13.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\10. Centre of gravity and mass moment of inertia\10.13.sce', -1)
+
+k=42.51 mm
\ No newline at end of file diff --git a/3886/CH10/EX10.14/10_14.sce b/3886/CH10/EX10.14/10_14.sce new file mode 100644 index 000000000..0caf2b0fa --- /dev/null +++ b/3886/CH10/EX10.14/10_14.sce @@ -0,0 +1,26 @@ +//Moment of Inertia of flywheel
+//refer fig. 10.24
+//Moment of inertia of rim
+aRo=1.5/2
+aRi=1.4/2
+at=0.30
+rho=7200 //kg/m^3
+I1=((%pi)*0.3*7200*(0.75^4-0.7^4))/(2) //units
+//Moment of inertia of hub
+bRo=0.25/2 //m
+bRi=0.1/2 //m
+bt=0.2 //m
+I2=(%pi)*(0.2*7200)*(0.125^4-0.05^4)/(2) //units
+//Moment of inertia of Arms
+A=8000*(10^-9) //m^2
+l=0.575 //m
+d=(0.575/2)+0.125 //m
+M=l*A*rho //kg
+//there are six such arms
+I3=6*0.03312*((0.575)^2/(12))*(0.4125^2) //units
+//moment of inertia of flywheel
+I=I1+I2+I3 //units
+printf("\nmoment of inertia of flywheel=%.2f units",I)
+
+
+
diff --git a/3886/CH10/EX10.14/10_14.txt b/3886/CH10/EX10.14/10_14.txt new file mode 100644 index 000000000..98023a3c2 --- /dev/null +++ b/3886/CH10/EX10.14/10_14.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\10. Centre of gravity and mass moment of inertia\10.14.sce', -1)
+
+moment of inertia of flywheel=259.44 units
\ No newline at end of file diff --git a/3886/CH10/EX10.4/10_4.sce b/3886/CH10/EX10.4/10_4.sce new file mode 100644 index 000000000..527e1339f --- /dev/null +++ b/3886/CH10/EX10.4/10_4.sce @@ -0,0 +1,12 @@ +//Locate centre of gravity
+//refer fig. 10.5
+W1=0.6*0.75*0.5*25000 //N
+W2=(%pi*(0.2^2)*0.3*25000)/(4) //N
+sumWi=7889.38
+sumWixi=3241.57
+sumWiyi=2593.25
+sumWizi=1745.91
+xbar=(sumWixi)/(sumWi)
+ybar=(sumWiyi)/(sumWi)
+zbar=(sumWizi)/(sumWi)
+printf("\nxbar=%.3f m\nybar=%.3f m\nzbar=%.3f m",xbar,ybar,zbar)
diff --git a/3886/CH10/EX10.4/10_4.txt b/3886/CH10/EX10.4/10_4.txt new file mode 100644 index 000000000..a0ec67862 --- /dev/null +++ b/3886/CH10/EX10.4/10_4.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\10. Centre of gravity and mass moment of inertia\10.4.sce', -1)
+
+xbar=0.411 m
+ybar=0.329 m
+zbar=0.221 m
\ No newline at end of file diff --git a/3886/CH10/EX10.5/10_5.sce b/3886/CH10/EX10.5/10_5.sce new file mode 100644 index 000000000..443b4c898 --- /dev/null +++ b/3886/CH10/EX10.5/10_5.sce @@ -0,0 +1,13 @@ +//Locate centroid
+//Refer fig.10.6
+//lets assign random value to w
+w=1
+sumWi=1053.98*w
+sumWixi=95055.54*w
+sumWiyi=125214.83*w
+sumWizi=59201.4*w
+xbar=(sumWixi)/(sumWi)
+ybar=(sumWiyi)/(sumWi)
+zbar=(sumWizi)/(sumWi)
+printf("\nxbar=%.2f mm\nybar=%.2f mm\nzbar=%.2f mm",xbar,ybar,zbar)
+
diff --git a/3886/CH10/EX10.5/10_5.txt b/3886/CH10/EX10.5/10_5.txt new file mode 100644 index 000000000..4c265f11d --- /dev/null +++ b/3886/CH10/EX10.5/10_5.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\10. Centre of gravity and mass moment of inertia\10.5.sce', -1)
+
+xbar=90.19 mm
+ybar=118.80 mm
+zbar=56.17 mm
\ No newline at end of file diff --git a/3886/CH12/EX12.10/12_10.sce b/3886/CH12/EX12.10/12_10.sce new file mode 100644 index 000000000..73c10b698 --- /dev/null +++ b/3886/CH12/EX12.10/12_10.sce @@ -0,0 +1,14 @@ +//cage and mine shaft
+//t is time during which stone is in motion
+//s=(9.81*t^2)/2
+//consider motion of cage
+//t1 be the time taken to travel first 30 m
+a=0.6 //m/sec^2
+t1=10 //sec
+//When the stone strikes
+//s=(0.6*(t+10)^2)/2
+//solving
+t=3.286 //sec
+s=(9.81*3.286^2)/2 //m
+printf("\nt=%.2f sec\ns=%.2f m",t,s)
+
diff --git a/3886/CH12/EX12.10/12_10.txt b/3886/CH12/EX12.10/12_10.txt new file mode 100644 index 000000000..067c14e7a --- /dev/null +++ b/3886/CH12/EX12.10/12_10.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.10.sce', -1)
+
+t=3.29 sec
+s=52.96 m
\ No newline at end of file diff --git a/3886/CH12/EX12.11/12_11.sce b/3886/CH12/EX12.11/12_11.sce new file mode 100644 index 000000000..2344998d6 --- /dev/null +++ b/3886/CH12/EX12.11/12_11.sce @@ -0,0 +1,20 @@ +//Train B overtakes train A
+//refer fig.12.12
+//speed of A
+v1=7.5 //m/sec
+//speed of B
+v2=15 //m/sec
+//motion of train A
+//using equation of motion
+t1=7.5/0.15 //sec
+//distance travelled in time t
+//s=7.5*t-187.5
+//Motion of train B
+//using equation of motion
+t2=15/0.3 //sec
+//distance travelled t seconds after train A started
+//s=15*t-975
+//solving
+t=(975-187.5)/(15-7.5) //sec
+s=15*t-975 //m
+printf("\nTrain B overtakes train A %.2d sec\ns=%.2d m",t,s)
diff --git a/3886/CH12/EX12.11/12_11.txt b/3886/CH12/EX12.11/12_11.txt new file mode 100644 index 000000000..5a9698117 --- /dev/null +++ b/3886/CH12/EX12.11/12_11.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.11.sce', -1)
+
+Train B overtakes train A 105 sec
+s=600 m
\ No newline at end of file diff --git a/3886/CH12/EX12.12/12_12.sce b/3886/CH12/EX12.12/12_12.sce new file mode 100644 index 000000000..07c7e6818 --- /dev/null +++ b/3886/CH12/EX12.12/12_12.sce @@ -0,0 +1,29 @@ +//Two cars
+//refer fig.12.13
+//Let A and B be the positions of cars when the drivers see each other and apply brakes
+//Let they meet at C
+//1.car A
+au=12 //m/sec
+av=0
+//s=x
+//a1 be acceleration
+//using equation of motion
+//a1=(-12)/t
+//x=6*t
+//2.car B
+bu=9 //m/sec
+bv=0
+//a=a2
+//time=t
+//s=100-x
+//using equation of motion
+//a2=-9/t
+//100-x=4.5*t
+//solving
+t=100/10.5 //sec
+a1=-12/t //m/sec^2
+a2=-9/t //m/sec^2
+x=57.14 //m
+//distance traveled by second car
+bx=100-x //m
+printf("\nt=%.2f sec\na1=%.2f m/sec^2\na2=%.2f m/sec^2\nDistance travelled by first car=%.2f m\nDistance travelled by second car=%.2f m",t,a1,a2,x,bx)
diff --git a/3886/CH12/EX12.12/12_12.txt b/3886/CH12/EX12.12/12_12.txt new file mode 100644 index 000000000..e96634eef --- /dev/null +++ b/3886/CH12/EX12.12/12_12.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.12.sce', -1)
+
+t=9.52 sec
+a1=-1.26 m/sec^2
+a2=-0.95 m/sec^2
+Distance travelled by first car=57.14 m
+Distance travelled by second car=42.86 m
\ No newline at end of file diff --git a/3886/CH12/EX12.13/12_13.sce b/3886/CH12/EX12.13/12_13.sce new file mode 100644 index 000000000..f245fc9fb --- /dev/null +++ b/3886/CH12/EX12.13/12_13.sce @@ -0,0 +1,14 @@ +//Car and truck
+//refer fig.12.14 and 12.15
+u=12.5 //m/sec
+//sT=10+12.5*t+(aT*t^2)/2
+aT=-2 //m/sec^2
+//t is the time at any instant after the brakes are applied
+//sT=10+12.5*t-t^2
+//distance moved by car
+//sC=u*2+u*(t-2)+(aC*(t-2)^2)/2
+//sT=sC
+//Apply equations of motion
+//we get quadratic equation whose solution gives
+aC=-10/3 //m/sec^2
+printf("the deceleration of the car is=%.2f m/sec^2",aC)
diff --git a/3886/CH12/EX12.13/12_13.txt b/3886/CH12/EX12.13/12_13.txt new file mode 100644 index 000000000..06606ec3b --- /dev/null +++ b/3886/CH12/EX12.13/12_13.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.13.sce', -1)
+the deceleration of the car is=-3.33 m/sec^2
\ No newline at end of file diff --git a/3886/CH12/EX12.14/12_14.sce b/3886/CH12/EX12.14/12_14.sce new file mode 100644 index 000000000..5076245d2 --- /dev/null +++ b/3886/CH12/EX12.14/12_14.sce @@ -0,0 +1,13 @@ +//motion of particle
+//s=t^3-3*t^2+2*t+5
+//v=ds/dt=3*t^2-6*t+2
+//a=6*t-6
+//after 4 seconds
+v=3*4*4-6*4+2 //m/sec
+a=6*4-6 //m/sec^2
+//minimum velocity Vmin by using maxima and minima principle
+Vmin=-1 //m/sec
+//let at time t the velocity be zero,then
+t1=1.577 //sec
+t2=0.423 //sec
+printf("\nv=%.2f m/sec\na=%.2f m/sec^2\nMinimum velocity=%.2f m/sec\nVelocity is zero at t=%.2f sec and %.2f sec",v,a,Vmin,t1,t2)
diff --git a/3886/CH12/EX12.14/12_14.txt b/3886/CH12/EX12.14/12_14.txt new file mode 100644 index 000000000..2d8bf6721 --- /dev/null +++ b/3886/CH12/EX12.14/12_14.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.14.sce', -1)
+
+v=26.00 m/sec
+a=18.00 m/sec^2
+Minimum velocity=-1.00 m/sec
+Velocity is zero at t=1.58 sec and 0.42 sec
\ No newline at end of file diff --git a/3886/CH12/EX12.15/12_15.sce b/3886/CH12/EX12.15/12_15.sce new file mode 100644 index 000000000..96c055e8a --- /dev/null +++ b/3886/CH12/EX12.15/12_15.sce @@ -0,0 +1,16 @@ +//particle in motion
+//v=t^3-t^2-2*t+2
+//a=3*t^2-2*t-2
+//acceleration after 4 seconds
+a=3*4^2-2*4-2 //m/sec^2
+//s=(t^4)/4-(t^3)/3-(t^2)+2*t+C
+//c is constant of acceleration
+//applying given condition
+C=4/3
+s=(4^4)/4-(4^3)/3-(4^2)+2*4+(4/3) //m
+//using maxima and minima principle
+//minimum value of acceleration (amin)
+amin=3*((1/3)^2)-2*(1/3)-2 //m/sec^2
+printf("\nMinimum value of acceleration=%.2f m/sec^2",amin)
+
+
diff --git a/3886/CH12/EX12.15/12_15.txt b/3886/CH12/EX12.15/12_15.txt new file mode 100644 index 000000000..d76f99291 --- /dev/null +++ b/3886/CH12/EX12.15/12_15.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.15.sce', -1)
+
+Minimum value of acceleration=-2.33 m/sec^2
\ No newline at end of file diff --git a/3886/CH12/EX12.16/12_16.sce b/3886/CH12/EX12.16/12_16.sce new file mode 100644 index 000000000..2c4775d4b --- /dev/null +++ b/3886/CH12/EX12.16/12_16.sce @@ -0,0 +1,22 @@ +//body moving along straight line
+//refer fig. 12.16
+//a=2-3*t
+//v=2*t-(3/2)*(t^2)+C1
+//C1 is constant of integration
+//v=20 //m/sec
+//t=5 //sec
+//thus
+C1=47.5
+//s=47.5*t+t^2-0.5*t^3+C2
+//s=85 m when t=10 sec thus
+C2=10
+a=2-3*0 //m/sec^2
+v=47.5 //m/sec
+as=10 //m
+//let the time when velocity becomes zero be t, thus
+t=6.33 //sec
+//Corresponding distance from origin
+s=10+47.5*6.33+6.33^2-0.5*6.33^3
+printf("\na=%.2f m/sec^2\nv=%.2f m/sec\ns=%.2f m\nt=%.2f sec\nDistance from origin\ns=%.3f",a,v,as,t,s)
+
+
diff --git a/3886/CH12/EX12.16/12_16.txt b/3886/CH12/EX12.16/12_16.txt new file mode 100644 index 000000000..60d5088f9 --- /dev/null +++ b/3886/CH12/EX12.16/12_16.txt @@ -0,0 +1,9 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.16.sce', -1)
+
+a=2.00 m/sec^2
+v=47.50 m/sec
+s=10.00 m
+t=6.33 sec
+Distance from origin
+s=223.926
\ No newline at end of file diff --git a/3886/CH12/EX12.18/12_18.sce b/3886/CH12/EX12.18/12_18.sce new file mode 100644 index 000000000..bc53c3a12 --- /dev/null +++ b/3886/CH12/EX12.18/12_18.sce @@ -0,0 +1,14 @@ +//Car moving
+//let the expression for retardation be
+//a=-k*s ...k=constant
+//v^2/2=(-k*(s^2)/2)+C1
+//When brakes are applied
+//s=0 and v=72 kmph
+v=20 //m/sec
+C1=200
+//when vehicle stops
+//v=0 s=15 m
+k=400/225
+//expression for retardation is
+//a=-1.778*s ...theory approach
+printf("The expression for retardation is a=-1.778*s")
\ No newline at end of file diff --git a/3886/CH12/EX12.18/12_18.txt b/3886/CH12/EX12.18/12_18.txt new file mode 100644 index 000000000..95c1bcb9d --- /dev/null +++ b/3886/CH12/EX12.18/12_18.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.18.sce', -1)
+The expression for retardation is a=-1.778*s
\ No newline at end of file diff --git a/3886/CH12/EX12.2/12_2.sce b/3886/CH12/EX12.2/12_2.sce new file mode 100644 index 000000000..9e1744a76 --- /dev/null +++ b/3886/CH12/EX12.2/12_2.sce @@ -0,0 +1,21 @@ +//Steel ball shot vertically up
+//refer fig.12.6
+//For upward motion
+au=18 //m/sec
+av=0
+aa=-9.81 //m/sec^2
+//s=h
+//let t1 be the time required to reach maximum height
+t1=1.83 //sec
+h=(18^2)/(2*9.81) //m
+//total height from the ground
+ah=25+h //m
+//Downward motion
+bu=0
+bs=41.51 //m
+ba=9.81 //m/sec^2
+v2=sqrt(2*9.81*41.51) //m/sec
+t2=28.54/9.81 //m/sec
+//total time during which the body is in motion
+t=t1+t2 //sec
+printf("\nt1=%.2f sec\nh=%.2f m\nv2=%.2f m/sec\nt2=%.2f sec\nt=%.2f sec",t1,h,v2,t2,t)
diff --git a/3886/CH12/EX12.2/12_2.txt b/3886/CH12/EX12.2/12_2.txt new file mode 100644 index 000000000..95719b60c --- /dev/null +++ b/3886/CH12/EX12.2/12_2.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.2.sce', -1)
+
+t1=1.83 sec
+h=16.51 m
+v2=28.54 m/sec
+t2=2.91 sec
+t=4.74 sec
\ No newline at end of file diff --git a/3886/CH12/EX12.3/12_3.sce b/3886/CH12/EX12.3/12_3.sce new file mode 100644 index 000000000..2bcb11b87 --- /dev/null +++ b/3886/CH12/EX12.3/12_3.sce @@ -0,0 +1,13 @@ +//Height from which stone fell
+//refer fig.12.7
+//Let the stone be dropped from A at a height h above window
+//h=(g*t^2)/2 ...(1)
+//h+2.45=((g)*(t+0.5)^2)/2 ...(2)
+//from (1) and (2)
+t=0.2495 //sec
+g=9.81 //m/sec^2
+h=(g*t^2)/2 //m
+printf("\nh=%.3f m",h)
+
+
+
diff --git a/3886/CH12/EX12.3/12_3.txt b/3886/CH12/EX12.3/12_3.txt new file mode 100644 index 000000000..d7144aab7 --- /dev/null +++ b/3886/CH12/EX12.3/12_3.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.3.sce', -1)
+
+h=0.305 m
\ No newline at end of file diff --git a/3886/CH12/EX12.4/12_4.sce b/3886/CH12/EX12.4/12_4.sce new file mode 100644 index 000000000..cc04c1e0f --- /dev/null +++ b/3886/CH12/EX12.4/12_4.sce @@ -0,0 +1,24 @@ +//Crossing of balls
+//refer fig. 12.8
+//1.for motion of first ball
+au=0
+//1s=30-h
+aa=9.81 //m/sec^2
+//2.for motion of second ball
+bu=15 //m/sec
+//s=h
+ba=-9.81 //m/sec^2
+//30-h=0*t+(9.81*t^2)/2 ...(1)
+//h=15*t-(9.81*t^2)/2 ...(2)
+//solving (1) and (2)
+t=30/15
+h=15*2-(9.81*2^2)/2 //m
+//at t=2
+//downward velocity of first ball
+v1=0+9.81*2 //m/sec
+//Upward velocity of second ball
+v2=15-9.81*2 //m/sec
+//relative velocity vr
+vr=v1-(-v2) //m/sec
+printf("\nt=%.2f sec\nh=%.2f m\nvr=%.2f m/sec",t,h,vr)
+
diff --git a/3886/CH12/EX12.4/12_4.txt b/3886/CH12/EX12.4/12_4.txt new file mode 100644 index 000000000..1d8a08fce --- /dev/null +++ b/3886/CH12/EX12.4/12_4.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.4.sce', -1)
+
+t=2.00 sec
+h=10.38 m
+vr=15.00 m/sec
\ No newline at end of file diff --git a/3886/CH12/EX12.5/12_5.sce b/3886/CH12/EX12.5/12_5.sce new file mode 100644 index 000000000..9a2ca3f70 --- /dev/null +++ b/3886/CH12/EX12.5/12_5.sce @@ -0,0 +1,13 @@ +//Stone dropped into well
+//let
+//h=depth of well
+//t1=time tataken by stone to strike water
+//t2=time taken by sound to travel h
+//t1+t2=4
+//h=(g*t1^2)/2
+//h=335*t2
+//solving
+//t1^2+68.30*t1-273.19=0
+t1=3.79 //sec
+h=(9.81*t1^2)/2 //m
+printf("h=%.2f m",h)
diff --git a/3886/CH12/EX12.5/12_5.txt b/3886/CH12/EX12.5/12_5.txt new file mode 100644 index 000000000..b50a1c27c --- /dev/null +++ b/3886/CH12/EX12.5/12_5.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.5.sce', -1)
+h=70.46 m
\ No newline at end of file diff --git a/3886/CH12/EX12.6/12_6.sce b/3886/CH12/EX12.6/12_6.sce new file mode 100644 index 000000000..bf9583733 --- /dev/null +++ b/3886/CH12/EX12.6/12_6.sce @@ -0,0 +1,19 @@ +//Distance covered
+//refer fig.12.9
+//Let the particle start from A and come to halt at E
+//Let initial velocity be u m/sec
+//consider motion between A and B
+//u+a=10
+//consider motion between A and C
+//70=7*u+7*a
+//solving
+a=-10/17.5 //m/sec^2
+u=10-(a) //m/sec
+//Let distance AD be s1
+s1=10.571*10+(-0.571*10^2)/2 //m
+//Distance covered in the interval 7 sec to 10 sec
+CD=77.16-60 //m
+//Let AE=s
+s=(10.571^2)/(2*0.571) //m
+printf("\nCD=%.2f m\ns=%.2f m",CD,s)
+
diff --git a/3886/CH12/EX12.6/12_6.txt b/3886/CH12/EX12.6/12_6.txt new file mode 100644 index 000000000..3115d1422 --- /dev/null +++ b/3886/CH12/EX12.6/12_6.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.6.sce', -1)
+
+CD=17.16 m
+s=97.85 m
\ No newline at end of file diff --git a/3886/CH12/EX12.7/12_7.sce b/3886/CH12/EX12.7/12_7.sce new file mode 100644 index 000000000..caca3452b --- /dev/null +++ b/3886/CH12/EX12.7/12_7.sce @@ -0,0 +1,13 @@ +//motorist and traffic light
+//initial velocity
+u=(80*1000)/(60*60) //m/sec
+t=10 //sec
+s=200 //m
+//a be acceleration
+//using equation of motion
+a=(200-22.22*10)*2/10^2 //m/sec^2
+//final velocity
+v=(22.22-0.444*10)*(3600/1000) //kmph
+printf("\na=%.2f m/sec^2\nv=%.2f kmph",a,v)
+
+
diff --git a/3886/CH12/EX12.7/12_7.txt b/3886/CH12/EX12.7/12_7.txt new file mode 100644 index 000000000..c1888fa21 --- /dev/null +++ b/3886/CH12/EX12.7/12_7.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.7.sce', -1)
+
+a=-0.44 m/sec^2
+v=64.01 kmph
\ No newline at end of file diff --git a/3886/CH12/EX12.9/12_9.sce b/3886/CH12/EX12.9/12_9.sce new file mode 100644 index 000000000..ea7da1ee6 --- /dev/null +++ b/3886/CH12/EX12.9/12_9.sce @@ -0,0 +1,13 @@ +//time required to cover the distance between two stations
+//refer fig.12.11
+v=(48*1000)/(60*60) //m/sec
+t1=30 //sec
+//after application of brakes the vehicle retards from 13.33 m/sec to 0 in t3 sec
+t3=13.33 //sec
+//Let t2 be the time during which the automobile travels with uniform velocity
+//s=s1+s2+s3
+s=5200 //m
+t2=((5200)-(13.33*30/2)-(13.33*13.33/2))/13.33 //sec
+//total time
+t=t1+t2+t3 //sec
+printf("Total time taken=%.2f sec",t)
diff --git a/3886/CH12/EX12.9/12_9.txt b/3886/CH12/EX12.9/12_9.txt new file mode 100644 index 000000000..ea732ee2b --- /dev/null +++ b/3886/CH12/EX12.9/12_9.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.9.sce', -1)
+Total time taken=411.76 sec
\ No newline at end of file diff --git a/3886/CH13/EX13.1/13_1.sce b/3886/CH13/EX13.1/13_1.sce new file mode 100644 index 000000000..43f13dc2a --- /dev/null +++ b/3886/CH13/EX13.1/13_1.sce @@ -0,0 +1,13 @@ +//Pilot and his bomber
+//refer fig.13.3
+h=2000 //m
+u=(600*1000)/(60*60) //m/sec
+//initial velocity in vertical direction
+//gravitational acceleration=9.81 m/sec^2
+//if t is the time of flight
+t=sqrt((2000*2)/(9.81)) //sec
+//during this period horizontal distance travelled by the bomb must be (d)
+d=u*t //m
+printf("Bomb should be released at %.2f m from the target",d)
+
+
diff --git a/3886/CH13/EX13.1/13_1.txt b/3886/CH13/EX13.1/13_1.txt new file mode 100644 index 000000000..283fd2ebd --- /dev/null +++ b/3886/CH13/EX13.1/13_1.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.1.sce', -1)
+Bomb should be released at 3365.46 m from the target
\ No newline at end of file diff --git a/3886/CH13/EX13.10/13_10.sce b/3886/CH13/EX13.10/13_10.sce new file mode 100644 index 000000000..f279c5ec8 --- /dev/null +++ b/3886/CH13/EX13.10/13_10.sce @@ -0,0 +1,9 @@ +//cricket ball
+//refer fig. 13.15
+u=20 //m/sec
+alpha=30 //degree
+Y0=-1.5 //m
+t=2.179 //sec
+//Distance of the fielder from the wickets
+Range=u*t*cosd(alpha) //m
+printf("The distance of the fielder from the wickets=%.3f m",Range)
diff --git a/3886/CH13/EX13.10/13_10.txt b/3886/CH13/EX13.10/13_10.txt new file mode 100644 index 000000000..340d1d676 --- /dev/null +++ b/3886/CH13/EX13.10/13_10.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.10.sce', -1)
+The distance of the fielder from the wickets=37.741 m
\ No newline at end of file diff --git a/3886/CH13/EX13.11/13_11.sce b/3886/CH13/EX13.11/13_11.sce new file mode 100644 index 000000000..79b734ed7 --- /dev/null +++ b/3886/CH13/EX13.11/13_11.sce @@ -0,0 +1,22 @@ +//Gravel is thrown in bin
+//refer fig. 13.16 and 13.17
+//taking O as origin
+u=5 //m/sec
+alpha=50 //degree
+//for point B
+y=-10 //m
+t=1.871 //sec
+//Horizontal distance travelled in this time=6.012 m
+//Vertical component of velocity of gravel at the time of striking the bin is=14.524 m/sec (downwards)
+//Horizontal component of velocity=5*cosd(50) m/sec
+//Velocity of strike
+v=sqrt((14.524^2)+(3.214^2)) //m/sec
+theta=atand(14.524/3.214) //degree to the horizontal
+printf("\nt=%.2f sec\nHorizontal distance travelled=6.012 m\nv=%.2f m/sec\ntheta=%.2f degree to horizontal",t,v,theta)
+
+
+
+
+
+
+
diff --git a/3886/CH13/EX13.11/13_11.txt b/3886/CH13/EX13.11/13_11.txt new file mode 100644 index 000000000..32955cd47 --- /dev/null +++ b/3886/CH13/EX13.11/13_11.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.11.sce', -1)
+
+t=1.87 sec
+Horizontal distance travelled=6.012 m
+v=14.88 m/sec
+theta=77.52 degree to horizontal
\ No newline at end of file diff --git a/3886/CH13/EX13.12/13_12.sce b/3886/CH13/EX13.12/13_12.sce new file mode 100644 index 000000000..8e7171a05 --- /dev/null +++ b/3886/CH13/EX13.12/13_12.sce @@ -0,0 +1,26 @@ +//soldier fires a bullet
+//refer fig 13.18
+//equation of trajectory of bullet is known thus
+//For the point on ground where bullet strikes
+y=-50 //m
+x=100 //m
+u=31.32 //m/sec
+//alpha=0 or
+alpha=atand(2) //degree
+//when alpha =0
+//Horizontal component of velocity
+vx=31.32 //m/sec
+//Vertical component of velocity
+vy=sqrt(2*9.81*50) //m/sec
+//Velocity of strike
+v=sqrt((31.32^2)+(31.32^2)) //m/sec
+theta=atand(1) //degree
+//when alpha=63.435 degree vx=14.007 m/sec
+//vy=42.02 m/sec
+bv=sqrt((14.007^2)+(42.02^2)) //m/sec
+btheta= atand(42.02/14.007) //degree to horizontal
+printf("\nalpha=%.2f degree\nv=%.2f m/sec\ntheta=%.2f degree\nv=%.2f m/sec\ntheta=%.2f degree to horizontal",alpha,v,theta,bv,btheta)
+
+
+
+
diff --git a/3886/CH13/EX13.12/13_12.txt b/3886/CH13/EX13.12/13_12.txt new file mode 100644 index 000000000..85d20f29d --- /dev/null +++ b/3886/CH13/EX13.12/13_12.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.12.sce', -1)
+
+alpha=63.43 degree
+v=44.29 m/sec
+theta=45.00 degree
+v=44.29 m/sec
+theta=71.56 degree to horizontal
\ No newline at end of file diff --git a/3886/CH13/EX13.13/13_13.sce b/3886/CH13/EX13.13/13_13.sce new file mode 100644 index 000000000..e02bea7cd --- /dev/null +++ b/3886/CH13/EX13.13/13_13.sce @@ -0,0 +1,12 @@ +//a rebounding ball
+//refer fig.13.19
+//at A the vertical component of velocity =u*asind(alpha)
+//when h=19 m Vertical component of velocity =0
+//y-coordinate of B=-24.033m
+//considering the motion in vertically upward direction
+t=4.93 //sec
+//x-coordinate of B=72.1 m
+//considering the horizontal motion of the ball
+alpha=atand(19.308/14.625) //degree
+u=14.625/cosd(52.86) //m/sec
+printf("\nalpha=%.2f degree\nu=%.2f m/sec",alpha,u)
diff --git a/3886/CH13/EX13.13/13_13.txt b/3886/CH13/EX13.13/13_13.txt new file mode 100644 index 000000000..713d65210 --- /dev/null +++ b/3886/CH13/EX13.13/13_13.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.13.sce', -1)
+
+alpha=52.86 degree
+u=24.22 m/sec
\ No newline at end of file diff --git a/3886/CH13/EX13.14/13_14.sce b/3886/CH13/EX13.14/13_14.sce new file mode 100644 index 000000000..80f611478 --- /dev/null +++ b/3886/CH13/EX13.14/13_14.sce @@ -0,0 +1,28 @@ +//Flying bomber
+//refer fig. 13.20
+h=2400 //m
+//Let the time required for bomb to reach ground be t seconds
+//then
+t=sqrt((2400*2)/(9.81)) //sec
+u=(1000*1000)/(60*60) //m/sec
+//Horizontal distance moved by bomb d
+d=u*t //m
+//muzzle velocity=600 m/sec
+//velocity of projection u=600 m/sec
+//alpha=60 degree
+//shell has to hit the bomber at height h=2400 m
+//let time required for the shell to rise to this height be t1
+//then
+//t1=110.370 sec or 4.433 sec
+//when t1=110.370 sec
+//horizontal distance moved by the shell=600*cosd(60)*110.370 m
+//distance moved by plane during this period=30658.58 m
+//the gun must fire the shell when the bomber is at a distance=33111+30658.58 m
+//when t1=4.839 sec
+//horizontal distance moved by the shell=1331.70 m
+//distance moved by plane during this period=1233.07 m
+//the gun must fire the shell when the bomber is at a distance=2564.77 m
+printf("\nThe bomb should be released when the bomber is %.2f m away from the target",d)
+printf("\nWhen the shell is fired at a distance of 63769.58 m, it will hit the plane in its downward motion.")
+printf("\nIf the shell is fired when the bomber is at a distance of 2795.87 m, then it will hit the bomber during its upward motion")
+
diff --git a/3886/CH13/EX13.14/13_14.txt b/3886/CH13/EX13.14/13_14.txt new file mode 100644 index 000000000..1f5f654da --- /dev/null +++ b/3886/CH13/EX13.14/13_14.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.14.sce', -1)
+
+The bomb should be released when the bomber is 6144.46 m away from the target
+When the shell is fired at a distance of 63769.58 m, it will hit the plane in its downward motion.
+If the shell is fired when the bomber is at a distance of 2795.87 m, then it will hit the bomber during its upward motion
\ No newline at end of file diff --git a/3886/CH13/EX13.15/13_15.sce b/3886/CH13/EX13.15/13_15.sce new file mode 100644 index 000000000..30800ed82 --- /dev/null +++ b/3886/CH13/EX13.15/13_15.sce @@ -0,0 +1,13 @@ +//A plane
+//initial velocity
+u=200 //m/sec
+//angle of projection
+alpha=30 //degree
+//Inclination of the plane=atand(5/12) degree
+//(a) When the shot is fired up the plane
+abeta=22.62 //degree
+aRange=((200^2)/(9.81*(cosd(22.62))^2))*((sind(2*30-22.62))-(sind(22.62))) //m
+//(b) When the shot is fired down the plane
+bbeta=-22.62 //degree
+bRange=((200*200)/(9.81*(cosd(22.62))^2))*(sind(82.62)+sind(22.62)) //m
+printf("\nWhen the shot is fired up the plane\nRange=%.2f m\nWhen the shot is fired down the plane\nRange=%.2f m",aRange,bRange)
diff --git a/3886/CH13/EX13.15/13_15.txt b/3886/CH13/EX13.15/13_15.txt new file mode 100644 index 000000000..984206ce0 --- /dev/null +++ b/3886/CH13/EX13.15/13_15.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.15.sce', -1)
+
+When the shot is fired up the plane
+Range=1064.65 m
+When the shot is fired down the plane
+Range=6586.27 m
\ No newline at end of file diff --git a/3886/CH13/EX13.16/13_16.sce b/3886/CH13/EX13.16/13_16.sce new file mode 100644 index 000000000..635c1dee2 --- /dev/null +++ b/3886/CH13/EX13.16/13_16.sce @@ -0,0 +1,15 @@ +//person throws a ball
+//refer fig. 13.23
+//(a) Up the plane
+atheta=35 //degree
+aalpha=atheta+20 //degree
+//maximum range
+aRangemax=((30*30)/(9.81*(cosd(20))^2))*(sind(2*55-20)-sind(20)) //m
+//(b) Down the plane
+//refer fig. 13.24
+btheta=(90+20)/2 //degree
+balpha=55-20 //degree
+//maximum range
+bRangemax=((30*30)/(9.81*(cosd(-20))^2))*(sind(2*35+20)-sind(-20)) //m
+printf("\nUp the plane\nMax Range=%.3f m",aRangemax)
+printf("\nDown the plane\nMax Range=%.3f m",bRangemax)
diff --git a/3886/CH13/EX13.16/13_16.txt b/3886/CH13/EX13.16/13_16.txt new file mode 100644 index 000000000..8e74c42e4 --- /dev/null +++ b/3886/CH13/EX13.16/13_16.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.16.sce', -1)
+
+Up the plane
+Max Range=68.362 m
+Down the plane
+Max Range=139.432 m
\ No newline at end of file diff --git a/3886/CH13/EX13.2/13_2.sce b/3886/CH13/EX13.2/13_2.sce new file mode 100644 index 000000000..bee681c1a --- /dev/null +++ b/3886/CH13/EX13.2/13_2.sce @@ -0,0 +1,10 @@ +//Person jumping over ditch
+//refer fig. 13.4
+h=2 //m
+Range=3 //m
+//let t be the time of flight and u the minimum horizontal velocity required
+//consider vertical motion
+t=sqrt((2*2)/(9.81)) //sec
+//consider horizontal motion of uniform velocity
+u=3/0.6386 //m/sec
+printf("Person should jump with u=%.2f m/sec",u)
diff --git a/3886/CH13/EX13.2/13_2.txt b/3886/CH13/EX13.2/13_2.txt new file mode 100644 index 000000000..118faf399 --- /dev/null +++ b/3886/CH13/EX13.2/13_2.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.2.sce', -1)
+Person should jump with u=4.70 m/sec
\ No newline at end of file diff --git a/3886/CH13/EX13.3/13_3.sce b/3886/CH13/EX13.3/13_3.sce new file mode 100644 index 000000000..aef50ecb8 --- /dev/null +++ b/3886/CH13/EX13.3/13_3.sce @@ -0,0 +1,21 @@ +//Pressure tank
+//refer fig. 13.5
+//Required velocity to enter at B
+h=1 //m
+//If t1 is the time of flight , considering vertical motion
+t1=sqrt(2/9.81) //sec
+//Considering horizontal motion
+u1=3/t1 //m/sec
+//Required velocity to enter at C
+//let t2 be the time required for flight from A to C
+bh=2.5 //m
+Range=3 //m
+//Considering Vertical motion
+t2=sqrt((2*2.5)/9.81) //sec
+//Considering horizontal motion
+u2=3/t2 //m/sec
+printf("The range of velocity for which the jet can enter the opening BC is %.2f m/sec to %.2f m/sec",u2,u1)
+
+
+
+
diff --git a/3886/CH13/EX13.3/13_3.txt b/3886/CH13/EX13.3/13_3.txt new file mode 100644 index 000000000..f2bd18a76 --- /dev/null +++ b/3886/CH13/EX13.3/13_3.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.3.sce', -1)
+The range of velocity for which the jet can enter the opening BC is 4.20 m/sec to 6.64 m/sec
\ No newline at end of file diff --git a/3886/CH13/EX13.4/13_4.sce b/3886/CH13/EX13.4/13_4.sce new file mode 100644 index 000000000..f361c7c11 --- /dev/null +++ b/3886/CH13/EX13.4/13_4.sce @@ -0,0 +1,13 @@ +//Rocket released from fighter jet
+//refer fig. 13.6
+h=3000 //m
+//If t is time of flight then
+//using equations of motion
+t=sqrt((2*3000)/(9.81)) //sec
+u=(1200*1000)/(60*60) //m/sec
+//Horizontal distance covered during the time of flight=range
+a=6 //m/sec^2
+Range=u*t+(1/2)*a*(t^2) //m
+//Angle theta below the horizontal at which the pilot must see the target while releasing the rocket is
+theta=atand(3000/10078.5) //degree
+printf("Angle theta below the horizontal at which the pilot must see the target while releasing the rocket is=%.3f degree",theta)
diff --git a/3886/CH13/EX13.4/13_4.txt b/3886/CH13/EX13.4/13_4.txt new file mode 100644 index 000000000..8b9c2dcb0 --- /dev/null +++ b/3886/CH13/EX13.4/13_4.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.4.sce', -1)
+Angle theta below the horizontal at which the pilot must see the target while releasing the rocket is=16.576 degree
\ No newline at end of file diff --git a/3886/CH13/EX13.5/13_5.sce b/3886/CH13/EX13.5/13_5.sce new file mode 100644 index 000000000..5d85db0b0 --- /dev/null +++ b/3886/CH13/EX13.5/13_5.sce @@ -0,0 +1,9 @@ +//Body is projected
+//u be the velocity of projection and alpha the angle of projection
+//then maximum height reached=((u^2)*(sind(alpha))^2)/(2*g)
+//range=((u^2)*sind(2*alpha))/(g)
+//in this case
+//Range=3*maximum height reached
+//thus
+alpha=atand(4/3) //degree
+printf("\n alpha=%.2f degree",alpha)
diff --git a/3886/CH13/EX13.5/13_5.txt b/3886/CH13/EX13.5/13_5.txt new file mode 100644 index 000000000..c04edd3bc --- /dev/null +++ b/3886/CH13/EX13.5/13_5.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.5.sce', -1)
+
+ alpha=53.13 degree
\ No newline at end of file diff --git a/3886/CH13/EX13.6/13_6.sce b/3886/CH13/EX13.6/13_6.sce new file mode 100644 index 000000000..a7bd5829e --- /dev/null +++ b/3886/CH13/EX13.6/13_6.sce @@ -0,0 +1,19 @@ +//Projectile aimed at target
+//refer fig. 13.9
+//let s be the distance of the target from the point of projection
+//u be the velocity of projection
+//range
+//R=((u^2)*sind(2*alpha))/(g)
+//applying it to first case
+//s-12=(u^2)/(2*g)
+//from second case
+//s+24=(u^2)/(g)
+//solving we get
+s=24+24 //m
+//let the correct angle of projection be alpha, then
+//sind(2*alpha)=48/72
+alpha=41.81/2 //degree
+printf("Angle of projection=%.3f degree",alpha)
+
+
+
diff --git a/3886/CH13/EX13.6/13_6.txt b/3886/CH13/EX13.6/13_6.txt new file mode 100644 index 000000000..dceeda60a --- /dev/null +++ b/3886/CH13/EX13.6/13_6.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.6.sce', -1)
+Angle of projection=20.905 degree
+-->
diff --git a/3886/CH13/EX13.7/13_7.sce b/3886/CH13/EX13.7/13_7.sce new file mode 100644 index 000000000..dbb03fadd --- /dev/null +++ b/3886/CH13/EX13.7/13_7.sce @@ -0,0 +1,13 @@ +//Projectile
+//let u be the initial velocity and alpha its angle of projection
+//Vertical component of velocity=u*sind(alpha)
+//Horizontal component of velocity=u*cosd(alpha)
+//thus according to given condition
+alpha=atand(1/2) //degree
+//when x=18 m y=3 m
+//using equation of trajectory
+u=sqrt((9.81*(18^2))/(6*2*(cosd(26.565))^2)) //m/sec
+g=9.81 //m/sec
+//range on the horizontal plane (range)
+range=((u^2)*sind(2*alpha))/(g) //m
+printf("Range on the horizontal plane=%.2f m",range)
diff --git a/3886/CH13/EX13.7/13_7.txt b/3886/CH13/EX13.7/13_7.txt new file mode 100644 index 000000000..2936a3822 --- /dev/null +++ b/3886/CH13/EX13.7/13_7.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.7.sce', -1)
+Range on the horizontal plane=27.00 m
\ No newline at end of file diff --git a/3886/CH13/EX13.8/13_8.sce b/3886/CH13/EX13.8/13_8.sce new file mode 100644 index 000000000..5d900a1a7 --- /dev/null +++ b/3886/CH13/EX13.8/13_8.sce @@ -0,0 +1,15 @@ +//Find the least initial velocity
+//refer fig. 13.10
+//Let u the initial velocity required and alpha the angle of projection
+//here
+range=9 //m
+//at P x=5m and y=4m
+//u^2=(9*g)/(sind(2*alpha))
+//from the equation of trajectory
+alpha=atand(1.8) //degree
+//thus
+u=sqrt((9*9.81)/(sind(2*60.95))) //m/sec
+printf("u=%.2f m/sec",u)
+
+
+
diff --git a/3886/CH13/EX13.8/13_8.txt b/3886/CH13/EX13.8/13_8.txt new file mode 100644 index 000000000..c162baaf8 --- /dev/null +++ b/3886/CH13/EX13.8/13_8.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.8.sce', -1)
+u=10.20 m/sec
\ No newline at end of file diff --git a/3886/CH13/EX13.9/13_9.sce b/3886/CH13/EX13.9/13_9.sce new file mode 100644 index 000000000..bbde6602b --- /dev/null +++ b/3886/CH13/EX13.9/13_9.sce @@ -0,0 +1,36 @@ +//Bullet fired
+//refer fig. 13.13
+//velocity of projection
+u=(360*1000)/(60*60) //m/sec
+//(a) total time of flight
+//method 1
+y0=-120 //m
+//considering vertical motion and solving quadratic equation
+t=12.20 //sec
+//method 2
+//t1=(100*sind(30))/(9.81) //sec
+//maximum height reached in this time
+//h=((100^2)*(sind(30))^2)/(2*9.81) //m
+//during downward motion
+//t2=7.1 //sec
+//t=t1+t2 //sec
+//method 3
+//time required to travel from A to D
+//t1=10.19 //sec
+//g=9.81 //m/sec^2
+//distance travelled=120 m
+t=12.20 //sec
+//(b) Maximum height reached by the bullet
+//h=((100^2)*(sind(30))^2)/(2*9.81) m above point A
+h=127.42+120 //m above the ground
+//(c)Horizontal range
+Hrange=100*12.2*cosd(30) //m
+//(d)Velocity of the bullet just before striking the ground
+//vertical component of velocity=69.682 m/sec
+//horizontal component of velocity=86.603 m/sec
+//velocity at strike
+v=sqrt((69.682^2)+(86.603^2)) //m/sec
+theta=atand(69.682/86.603) //degree
+printf("\nt=%.2f sec\nh=%.2f m above the ground\nHorizontal range=%.2f m\nv=%.2f m/sec\ntheta=%.2f degree",t,h,Hrange,v,theta)
+
+
diff --git a/3886/CH13/EX13.9/13_9.txt b/3886/CH13/EX13.9/13_9.txt new file mode 100644 index 000000000..5683f2c41 --- /dev/null +++ b/3886/CH13/EX13.9/13_9.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\13. Projectiles\13.9.sce', -1)
+
+t=12.20 sec
+h=247.42 m above the ground
+Horizontal range=1056.55 m
+v=111.16 m/sec
+theta=38.82 degree
\ No newline at end of file diff --git a/3886/CH14/EX14.1/14_1.sce b/3886/CH14/EX14.1/14_1.sce new file mode 100644 index 000000000..c8cad45f6 --- /dev/null +++ b/3886/CH14/EX14.1/14_1.sce @@ -0,0 +1,10 @@ +//Passenger train
+//refer fig. 14.7
+//let vb= velocity of goods train
+vA=(72*1000)/(60*60) //m/sec
+//relative velocity of B w.r.t A
+//vB/A=20-vB
+//relative distance moved to overtake the goods train=250+200 m
+//45 seconds are required to cover this relative distance
+vB=(10*60*60)/(1000) //kmph
+printf("\nvB=%.2f kmph",vB)
diff --git a/3886/CH14/EX14.1/14_1.txt b/3886/CH14/EX14.1/14_1.txt new file mode 100644 index 000000000..abfe1c8a8 --- /dev/null +++ b/3886/CH14/EX14.1/14_1.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\14.Relative velocity\14.1.sce', -1)
+
+vB=36.00 kmph
\ No newline at end of file diff --git a/3886/CH14/EX14.10/14_10.sce b/3886/CH14/EX14.10/14_10.sce new file mode 100644 index 000000000..94f37567f --- /dev/null +++ b/3886/CH14/EX14.10/14_10.sce @@ -0,0 +1,22 @@ +//Jet of water
+//refer fig. 14.18 and 14.19
+//time taken to move a horizontal distance of 5m
+t=5/20 //sec
+//During this period vertical downward velocity gained by water (Vw)
+Vw=0+(9.81/4)
+//Horizontal component of velocity of plate (HCp)
+HCp=0
+//Vertical component of velocity of plate (VCp)
+VCp=1 //m/sec
+//relative velocity of water w.r.t. plate
+vry=Vw-VCp //m/sec
+vrx=20 //m/sec
+vr=sqrt((20)^2+(1.453)^2) //m/sec
+alpha=atand(1.453/20) //degree
+printf("\nvr=%.2f m/sec\nalpha=%.2f degree",vr,alpha)
+
+
+
+
+
+
diff --git a/3886/CH14/EX14.10/14_10.txt b/3886/CH14/EX14.10/14_10.txt new file mode 100644 index 000000000..5757095bd --- /dev/null +++ b/3886/CH14/EX14.10/14_10.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\14.Relative velocity\14.10.sce', -1)
+
+vr=20.05 m/sec
+alpha=4.16 degree
\ No newline at end of file diff --git a/3886/CH14/EX14.11/14_11.sce b/3886/CH14/EX14.11/14_11.sce new file mode 100644 index 000000000..f902e8589 --- /dev/null +++ b/3886/CH14/EX14.11/14_11.sce @@ -0,0 +1,23 @@ +//railway carriage
+//refer fig. 14.20
+//Velocity of components of train are
+v1x=96 //kmph
+v1y=0
+//Velocity components of bullet are
+//v2x=0.9848*v
+//v2y=0.1736*v
+//Component of relative velocity of B w.r.t. A are
+//vrx=0.9848*v-96
+//vry=0.1736*v
+//Direction of relative velocity alpha with x-axis
+//tand(alpha)=((vry)/(vrx))
+//thus
+v=183.96 //kmph
+//consider motion in y direction
+vry=0.1736*51.1 //m/sec
+//Time period (t)
+t=(1.8)/(0.1736*51.1) //sec
+printf("\nv=%.2f kmph\nt=%.3f sec",v,t)
+
+
+
diff --git a/3886/CH14/EX14.11/14_11.txt b/3886/CH14/EX14.11/14_11.txt new file mode 100644 index 000000000..618f69e04 --- /dev/null +++ b/3886/CH14/EX14.11/14_11.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\14.Relative velocity\14.11.sce', -1)
+
+v=183.96 kmph
+t=0.203 sec
\ No newline at end of file diff --git a/3886/CH14/EX14.2/14_2.sce b/3886/CH14/EX14.2/14_2.sce new file mode 100644 index 000000000..ea302b69e --- /dev/null +++ b/3886/CH14/EX14.2/14_2.sce @@ -0,0 +1,13 @@ +//passenger train
+//velocity
+vA=20 //m/sec
+//let velocity of goods train be vB m/sec
+//relative velocity=20-vB m/sec
+//when t=25 relative distance moved is x metres
+//(20-vB)*25=x
+//In the next t=30 seconds
+//relative distance moved=length of passenger train=240 m
+vB=(20-(240/30))*((60*60)/(1000)) //km/h
+x=(20-12)*25 //m
+printf("\nLength is %.2f m\nSpeed is %.2f km/h",x,vB)
+
diff --git a/3886/CH14/EX14.2/14_2.txt b/3886/CH14/EX14.2/14_2.txt new file mode 100644 index 000000000..db648b8c2 --- /dev/null +++ b/3886/CH14/EX14.2/14_2.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\14.Relative velocity\14.2.sce', -1)
+
+Length is 200.00 m
+Speed is 43.20 km/h
\ No newline at end of file diff --git a/3886/CH14/EX14.3/14_3.sce b/3886/CH14/EX14.3/14_3.sce new file mode 100644 index 000000000..125117821 --- /dev/null +++ b/3886/CH14/EX14.3/14_3.sce @@ -0,0 +1,8 @@ +//Two trains
+//Taking the direction of motion of train A as positive
+//let velocity of A be v m/sec
+//Relative velocity=1.5*v
+vA=20*((60*60)/(1000)) //kmph
+//velocity of B
+vB=-10*((60*60)/(1000)) //kmph
+printf("\nVelocity of A=%.2f kmph\nVelocity of B=%.2f kmph",vA,-vB)
diff --git a/3886/CH14/EX14.3/14_3.txt b/3886/CH14/EX14.3/14_3.txt new file mode 100644 index 000000000..fbdc6bc01 --- /dev/null +++ b/3886/CH14/EX14.3/14_3.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\14.Relative velocity\14.3.sce', -1)
+
+Velocity of A=72.00 kmph
+Velocity of B=36.00 kmph
\ No newline at end of file diff --git a/3886/CH14/EX14.4/14_4.sce b/3886/CH14/EX14.4/14_4.sce new file mode 100644 index 000000000..f8bb4e9d4 --- /dev/null +++ b/3886/CH14/EX14.4/14_4.sce @@ -0,0 +1,18 @@ +//Two ships
+//refer fig.14.8,14.9 and 14.10
+//Taking west direction as x-axis and north direction as y-axis
+//velocities in kmph are
+vAx=30*sind(30)
+vAy=30*cosd(30)
+vBx=40*sind(45)
+vBy=-40*sind(45)
+vrx=15-28.284
+vry=25.98-(-28.284)
+vr=sqrt((13.284^2)+(54.264^2))
+theta=atand((13.284)/(54.264)) //degree
+printf("\nFrom B, ship A appears to move with a velocity of %.2f kmph in N %.2f degree E direction",vr,theta)
+//relative distance after half an hour (drel)
+drel=55.866*(1/2) //km
+printf("\nRelative distance after half an hour=%.2f km",drel)
+
+
diff --git a/3886/CH14/EX14.4/14_4.txt b/3886/CH14/EX14.4/14_4.txt new file mode 100644 index 000000000..cc391eaf6 --- /dev/null +++ b/3886/CH14/EX14.4/14_4.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\14.Relative velocity\14.4.sce', -1)
+
+From B, ship A appears to move with a velocity of 55.87 kmph in N 13.76 degree E direction
+Relative distance after half an hour=27.93 km
\ No newline at end of file diff --git a/3886/CH14/EX14.5/14_5.sce b/3886/CH14/EX14.5/14_5.sce new file mode 100644 index 000000000..4bb4f42db --- /dev/null +++ b/3886/CH14/EX14.5/14_5.sce @@ -0,0 +1,26 @@ +//Enemy ship location
+//refer fig. 14.10
+//taking north as y direction and west as x direction
+//vAy=36*cosd(theta)
+//vAx=36*sind(theta)
+//Components of velocity of enemy ship
+vBy=18*cosd(30) //kmph
+vBx=-18*sind(30) //kmph
+//then
+//vrx=36*sind(theta)+9
+//vry=36*cosd(theta)-15.588
+//solving
+x=0.2777
+theta=16.12 //degree
+printf("\nWar ship must move in N %.2f W direction",theta)
+vrx=36*sind(theta)+9
+vry=36*cosd(theta)-15.588
+vr=sqrt((19^2)+(19^2)) //kmph
+//relative distance moved
+dr=25-5 //km
+//time interval
+t=20/26.870 //hour
+printf("\n%.2f hour after sighting the enemy ship the shell is to be fired",t)
+
+
+
diff --git a/3886/CH14/EX14.5/14_5.txt b/3886/CH14/EX14.5/14_5.txt new file mode 100644 index 000000000..b54896698 --- /dev/null +++ b/3886/CH14/EX14.5/14_5.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\14.Relative velocity\14.5.sce', -1)
+
+War ship must move in N 16.12 W direction
+0.74 hour after sighting the enemy ship the shell is to be fired
\ No newline at end of file diff --git a/3886/CH14/EX14.6/14_6.sce b/3886/CH14/EX14.6/14_6.sce new file mode 100644 index 000000000..fed361c26 --- /dev/null +++ b/3886/CH14/EX14.6/14_6.sce @@ -0,0 +1,16 @@ +//Motor boat crossing river
+//refer fig. 14.11 and 14.12
+//let the motor boat start from A and reaches C
+vrx=15 //kmph
+//distance to be moved in x direction=1 km
+//time required t is
+t=4 //min
+//boat will move down the stream
+vy=5 //kmph
+//Distance moved in downstream direction (d)
+d=333.33 //m
+//Let the direction of boat be set at theta to x-direction
+theta=asind(1/3) //degree
+printf("The boat should be set in the direction theta=%.2f degree and time required is t=%.2f min\nDistance moved in downstream direction=%.2f m",theta,t,d)
+
+
diff --git a/3886/CH14/EX14.6/14_6.txt b/3886/CH14/EX14.6/14_6.txt new file mode 100644 index 000000000..12c71bd1f --- /dev/null +++ b/3886/CH14/EX14.6/14_6.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\14.Relative velocity\14.6.sce', -1)
+The boat should be set in the direction theta=19.47 degree and time required is t=4.00 min
+Distance moved in downstream direction=333.33 m
\ No newline at end of file diff --git a/3886/CH14/EX14.7/14_7.sce b/3886/CH14/EX14.7/14_7.sce new file mode 100644 index 000000000..152de87a0 --- /dev/null +++ b/3886/CH14/EX14.7/14_7.sce @@ -0,0 +1,23 @@ +//ship approaching port
+//refer fig. 14.13 and 14.14
+//let west be x and north be y axes
+//speed in kmph is
+vBx=25*sind(45)
+vBy=25*cosd(45)
+vAx=-15
+vAy=0
+//Let vr be the relative velocity of B w.r.t. A
+vrx=17.678-(-15) //kmph
+vry=17.678 //kmph
+vr=sqrt((32.678^2)+(17.678^2)) //kmph
+alpha=atand(17.678/32.678) //degree
+t=(50*cosd(alpha))/(vr) //hours
+//during this time A has moved in east by (da)
+da=15*1.1837 //km
+//and B has moved in N 45 degree W a distance (db)
+db=25*1.1837 //km
+printf("\nt=%.2f hours\nA has moved in east by da=%.2f km\nB has moved in N 45 degree W a distance db=%.2f km",t,da,db)
+
+
+
+
diff --git a/3886/CH14/EX14.7/14_7.txt b/3886/CH14/EX14.7/14_7.txt new file mode 100644 index 000000000..44e21c7b4 --- /dev/null +++ b/3886/CH14/EX14.7/14_7.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\14.Relative velocity\14.7.sce', -1)
+
+t=1.18 hours
+A has moved in east by da=17.76 km
+B has moved in N 45 degree W a distance db=29.59 km
\ No newline at end of file diff --git a/3886/CH14/EX14.8/14_8.sce b/3886/CH14/EX14.8/14_8.sce new file mode 100644 index 000000000..2d76b882b --- /dev/null +++ b/3886/CH14/EX14.8/14_8.sce @@ -0,0 +1,31 @@ +//ship B approaching port
+//refer fig. 14.15 and 14.16
+//Considering west as x-axis and south as y-axis
+vAx=24*cosd(30)
+vAy=24*sind(30)
+vBx=-18
+vBy=0
+//Let relative velocity of A w.r.t. B be vr at an angle alpha to western direction
+vrx=vAx-vBx //kmph
+vry=vAy-vBy
+v=sqrt((vrx)^2+(vry)^2) //kmph
+alpha=atand(vry/vrx) //degree
+//Holding B stationary and allowing A to move with relative velovity,BC is given by
+BC=60*sind(alpha) //km
+//from triangle BCD
+DC=sqrt((25^2)+(17.735)^2) //km
+CE=DC //km
+AC=60*cosd(alpha) //km
+AD=AC-DC //km
+AE=AC+CE //km
+//Time taken to reach D
+t1=39699*60/40899 //min
+//time taken to reach E
+t2=74939*60/40599 //min
+printf("\nThe two ships can start exchanging signals %.2f min after ship A leaves the port and continue to do so for %.2f min",t1,t2-t1)
+
+
+
+
+
+
diff --git a/3886/CH14/EX14.8/14_8.txt b/3886/CH14/EX14.8/14_8.txt new file mode 100644 index 000000000..b6ba50eaa --- /dev/null +++ b/3886/CH14/EX14.8/14_8.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\14.Relative velocity\14.8.sce', -1)
+
+The two ships can start exchanging signals 58.24 min after ship A leaves the port and continue to do so for 52.51 min
\ No newline at end of file diff --git a/3886/CH14/EX14.9/14_9.sce b/3886/CH14/EX14.9/14_9.sce new file mode 100644 index 000000000..58ff6af69 --- /dev/null +++ b/3886/CH14/EX14.9/14_9.sce @@ -0,0 +1,18 @@ +//passenger observing rain drops
+//refer fig. 14.17
+//Let the true velocity of rain be v kmph at a true angle theta with vertical
+//Taking the direction of train as x and that of vertical downward as y
+//Velocity components of train are
+//v1x=v*sind(theta)
+//v1y=v*cosd(theta)
+//when the velocity of train was 36 kmph
+v2x=36
+v2y=0
+//alpha is the direction of relative velocity and is given as 30 degree and when the velocity of train is 54 kmph alpha=45 degree thus
+//v*cosd(theta)=v*sind(theta)-54
+//v*sind(theta)=-11.402
+//solving we get
+v=sqrt(4407.43) //kmph
+theta=asind(-(11.402)/(66.388))
+printf("\nv=%.3f kmph\ntheta=%.2f degree",v,theta)
+
diff --git a/3886/CH14/EX14.9/14_9.txt b/3886/CH14/EX14.9/14_9.txt new file mode 100644 index 000000000..13e7c8ac8 --- /dev/null +++ b/3886/CH14/EX14.9/14_9.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\14.Relative velocity\14.9.sce', -1)
+
+v=66.388 kmph
+theta=-9.89 degree
\ No newline at end of file diff --git a/3886/CH15/EX15.10/15_10.sce b/3886/CH15/EX15.10/15_10.sce new file mode 100644 index 000000000..9ee0f376a --- /dev/null +++ b/3886/CH15/EX15.10/15_10.sce @@ -0,0 +1,11 @@ +//Two bodies hung to the rope ends
+//refer fig. 15.12 (a),(b) and (c)
+//Let a be the acceleration with which the system moves and T be the tension in the string
+//Considering 300 N body
+//T-(300*a)/(9.81)=300
+//Considering 450 N body
+//T+(450*a)/(9.81)=450
+//solving we get
+a=(450-300)*9.81/(450+300) //m/sec^2
+T=300+((300*1.962)/(9.81)) //N
+printf("\na=%.4f m/sec^2\nT=%.0f N",a,T)
diff --git a/3886/CH15/EX15.10/15_10.txt b/3886/CH15/EX15.10/15_10.txt new file mode 100644 index 000000000..21e9f7e21 --- /dev/null +++ b/3886/CH15/EX15.10/15_10.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\15. DAlemberts principle\15.10.sce', -1)
+
+a=1.9620 m/sec^2
+T=360 N
\ No newline at end of file diff --git a/3886/CH15/EX15.11/15_11.sce b/3886/CH15/EX15.11/15_11.sce new file mode 100644 index 000000000..4b94490ab --- /dev/null +++ b/3886/CH15/EX15.11/15_11.sce @@ -0,0 +1,11 @@ +//Tension in the string and accelerations of blocks
+//refer fig. 15.13 (a),(b) and (c)
+//Considering 1500 N block
+//2*T+(1500*a)/(9.81)=1500
+//Considering 500N block
+//T-(2*500*a)/(9.81)=500
+//Solving this we get
+a=(500*9.81)/(1500+2000) //m/sec^2
+T=(1500-((1500*1.401)/(9.81)))/2 //N
+printf("\na=%.3f m/sec^2\nT=%.2f N",a,T)
+
diff --git a/3886/CH15/EX15.11/15_11.txt b/3886/CH15/EX15.11/15_11.txt new file mode 100644 index 000000000..d88e58ad3 --- /dev/null +++ b/3886/CH15/EX15.11/15_11.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\15. DAlemberts principle\15.11.sce', -1)
+
+a=1.401 m/sec^2
+T=642.89 N
\ No newline at end of file diff --git a/3886/CH15/EX15.12/15_12.sce b/3886/CH15/EX15.12/15_12.sce new file mode 100644 index 000000000..f4ccaa9cf --- /dev/null +++ b/3886/CH15/EX15.12/15_12.sce @@ -0,0 +1,33 @@ +//Train along an inclined plane
+//refer fig. 15.14 and 15.15
+u=0
+v=(36*1000)/(60*60) //m/sec^2
+s=1000 //m
+//From kinematic equation
+a=100/2000 //m/sec^2
+//Tractive resistance (Tr)
+Tr=5*1500 //N
+//Component of weight of train (Wt1)
+Wt=1500/100 //kN
+//Inertia force (I1)
+I=(1500*0.05)/(9.81) //kN (Down the plane)
+//Dynamic equilibrium equation gives
+T=7.5+15+7.645 //kN
+//Consider dynamic equilibrium of train
+//Total tractive resistance (Rt)
+Rt=5*2000 //N
+//Inertia force (I2)
+I2=(2000*0.05)/(9.81) //kN (Down the plane)
+//Component of weight down the plane (Wt2)
+Wt2=(2000)/(100) //kN
+//Dynamic equilibrium equation gives
+P=10+10.194+20 //kN
+printf("\nT=%.3f kN\nP=%.3f kN",T,P)
+
+
+
+
+
+
+
+
diff --git a/3886/CH15/EX15.12/15_12.txt b/3886/CH15/EX15.12/15_12.txt new file mode 100644 index 000000000..acb1e29f0 --- /dev/null +++ b/3886/CH15/EX15.12/15_12.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\15. DAlemberts principle\15.12.sce', -1)
+
+T=30.145 kN
+P=40.194 kN
\ No newline at end of file diff --git a/3886/CH15/EX15.2/15_2.sce b/3886/CH15/EX15.2/15_2.sce new file mode 100644 index 000000000..15eb61df6 --- /dev/null +++ b/3886/CH15/EX15.2/15_2.sce @@ -0,0 +1,13 @@ +//Elevator cage
+//refer fig. 15.4
+u=0
+v=25 //m/sec
+s=187.5 //m
+//using equations of motion
+a=(25^2)/(2*187.5) //m/sec^2
+//summing up the forces in vertical direction
+T=8600-((8600*1.667)/(9.81)) //N
+//Equilibrium condition gives
+R=600-((600*1.667)/(9.81)) //N
+printf("\nT=%.2f N\nR=%.2f N",T,R)
+
diff --git a/3886/CH15/EX15.2/15_2.txt b/3886/CH15/EX15.2/15_2.txt new file mode 100644 index 000000000..600a3c32c --- /dev/null +++ b/3886/CH15/EX15.2/15_2.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\15. DAlemberts principle\15.2.sce', -1)
+
+T=7138.61 N
+R=498.04 N
\ No newline at end of file diff --git a/3886/CH15/EX15.3/15_3.sce b/3886/CH15/EX15.3/15_3.sce new file mode 100644 index 000000000..477278c0e --- /dev/null +++ b/3886/CH15/EX15.3/15_3.sce @@ -0,0 +1,13 @@ +//Motorist travelling
+//refer fig. 15.5
+u=(70*1000)/(60*60) //m/sec
+v=0
+s=50 //m
+//Using equation of linear motion
+a=-(19.44^2)/(2*50) //m/sec^2
+//again
+t=19.44/3.78 //sec
+//Applying equilibrium equationswe get
+mu=(3.78)/(9.81)
+printf("\nt=%.2f sec\nmu=%.3f ",t,mu)
+
diff --git a/3886/CH15/EX15.3/15_3.txt b/3886/CH15/EX15.3/15_3.txt new file mode 100644 index 000000000..6daba659b --- /dev/null +++ b/3886/CH15/EX15.3/15_3.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\15. DAlemberts principle\15.3.sce', -1)
+
+t=5.14 sec
+mu=0.385
\ No newline at end of file diff --git a/3886/CH15/EX15.4/15_4.sce b/3886/CH15/EX15.4/15_4.sce new file mode 100644 index 000000000..bb62eb174 --- /dev/null +++ b/3886/CH15/EX15.4/15_4.sce @@ -0,0 +1,12 @@ +//block on horizontal plane
+//refer fig. 15.6 (a) and (b)
+//Inertia force of block m*a=3/9.81 kN
+//applying equilibrium conditions
+//N=1+P/2
+//P*cosd(30)-F-3/9.81
+//From law of friction
+//F=mu*N
+//Solving above equations
+P=((3/9.81)+(0.25))/(cosd(30)-(0.125)) //kN
+printf("\nP=%.3f kN",P)
+//The answers vary due to round off error
diff --git a/3886/CH15/EX15.4/15_4.txt b/3886/CH15/EX15.4/15_4.txt new file mode 100644 index 000000000..22c02cafd --- /dev/null +++ b/3886/CH15/EX15.4/15_4.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\15. DAlemberts principle\15.4.sce', -1)
+
+P=0.750 kN
\ No newline at end of file diff --git a/3886/CH15/EX15.5/15_5.sce b/3886/CH15/EX15.5/15_5.sce new file mode 100644 index 000000000..b4bd55496 --- /dev/null +++ b/3886/CH15/EX15.5/15_5.sce @@ -0,0 +1,12 @@ +//Crate resting on cart
+//refer fig. 15.7 (a),(b) and (c)
+//Applying equilibrium condition
+//N=W=750 N
+//Frictional force
+mu=0.3
+N=750
+F=mu*N
+a=(225*9.81)/(750) //m/sec^2
+//Consider dynamic equilibrium of the system
+P=250+((1250*2.943)/(9.81)) //N
+printf("\nMaximum allowable P=%.2f N and a=%.3f m/sec^2",P,a)
diff --git a/3886/CH15/EX15.5/15_5.txt b/3886/CH15/EX15.5/15_5.txt new file mode 100644 index 000000000..7799e21b8 --- /dev/null +++ b/3886/CH15/EX15.5/15_5.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\15. DAlemberts principle\15.5.sce', -1)
+
+Maximum allowable P=625.00 N and a=2.943 m/sec^2
\ No newline at end of file diff --git a/3886/CH15/EX15.6/15_6.sce b/3886/CH15/EX15.6/15_6.sce new file mode 100644 index 000000000..8cdc10100 --- /dev/null +++ b/3886/CH15/EX15.6/15_6.sce @@ -0,0 +1,20 @@ +//Body on an inclined plane
+//Refer fig. 15.8 (a),(b) and (c)
+//Consider 1200 N block
+//applying equilibrium condition
+N=1200*cosd(12) //N
+mu=0.2
+//From Law of friction
+F=mu*N
+//applying equilibrium condition
+a=(800-484.25)/(122.32+((800)/(9.81))) //m/sec^2
+//solving for T
+T=800-((800*1.549)/(9.81)) //N
+//initial velocity=0
+t=3 //sec
+//distance moved in 3 sec
+s=0*3+((1.549*3^2)/(2)) //m
+printf("\na=%.3f m/sec^2\nT=%.2f N\ns=%.3f m",a,T,s)
+
+
+
diff --git a/3886/CH15/EX15.6/15_6.txt b/3886/CH15/EX15.6/15_6.txt new file mode 100644 index 000000000..1c8d54c21 --- /dev/null +++ b/3886/CH15/EX15.6/15_6.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\15. DAlemberts principle\15.6.sce', -1)
+
+a=1.549 m/sec^2
+T=673.68 N
+s=6.970 m
\ No newline at end of file diff --git a/3886/CH15/EX15.7/15_7.sce b/3886/CH15/EX15.7/15_7.sce new file mode 100644 index 000000000..1914ea74f --- /dev/null +++ b/3886/CH15/EX15.7/15_7.sce @@ -0,0 +1,21 @@ +//Two weights connected by weight
+//refer fig.15.9 (a) and (b)
+//Consider dynamic equilibrium of 200 N Weight
+N1=200 //N
+mu=0.3
+//From law of friction
+F1=mu*N1 //N
+//applying equilibrium condition
+//T1-(200*a)/9.81=60
+//Consider 800N body
+N2=800 //N
+//From Law of friction
+F2=mu*N2 //N
+//applying equilibrium condition
+//T+(800*a)/9.81=160 N
+//Solving
+a=((160-60)*9.81)/(200+800) //m/sec^2
+T=160-((800*a)/(9.81)) //N
+printf("\na=%.3f m/sec^2\nT=%.2f N",a,T)
+
+
diff --git a/3886/CH15/EX15.7/15_7.txt b/3886/CH15/EX15.7/15_7.txt new file mode 100644 index 000000000..c3c5f57c3 --- /dev/null +++ b/3886/CH15/EX15.7/15_7.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\15. DAlemberts principle\15.7.sce', -1)
+
+a=0.981 m/sec^2
+T=80.00 N
\ No newline at end of file diff --git a/3886/CH15/EX15.8/15_8.sce b/3886/CH15/EX15.8/15_8.sce new file mode 100644 index 000000000..8d8a838f4 --- /dev/null +++ b/3886/CH15/EX15.8/15_8.sce @@ -0,0 +1,23 @@ +//Two incloned planes
+//refer fig.15.10 (a),(b) and (c)
+//Let the assembly move down the 60 degree plane by an acceleration a m/sec^2
+//Consider the block weighing 100 N
+//Applying equilibrium conditions
+N1=50 //N
+mu=1/3
+//From law of friction
+F1=mu*N1 //N
+//T+((100*a)/(9.81))=69.93
+//Now consider 50 N block
+N2=50*cosd(30) //N
+//From the law of friction
+F2=mu*N2
+//((50*a)/(9.81))-T=-39.43
+//Solving we get
+a=(69.93-39.43)*9.81/(100+50) //m/sec^2
+T=69.93-(100*1.9947/9.81) //N
+printf("\na=%.4f m/sec^2\nT=%.2f N",a,T)
+
+
+
+
diff --git a/3886/CH15/EX15.8/15_8.txt b/3886/CH15/EX15.8/15_8.txt new file mode 100644 index 000000000..c64f7042f --- /dev/null +++ b/3886/CH15/EX15.8/15_8.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\15. DAlemberts principle\15.8.sce', -1)
+
+a=1.9947 m/sec^2
+T=49.60 N
\ No newline at end of file diff --git a/3886/CH15/EX15.9/15_9.sce b/3886/CH15/EX15.9/15_9.sce new file mode 100644 index 000000000..59b595c49 --- /dev/null +++ b/3886/CH15/EX15.9/15_9.sce @@ -0,0 +1,28 @@ +//Two blocks on an inclined plane
+//refer fig. 15.11 (a,b,c) and (d)
+//Let block A move with an acceleration a1 and block B with an acceleration a2
+//Consider block A
+//Using equilibrium conditions
+//NA=WA*cosd(30)
+mu1=0.2
+WA=100 //N
+//From the law of friction
+FA=mu1*WA*cosd(30) //
+a1=3.2058 //m/sec^2
+//Consider block B
+//NB=WB*cosd(30)
+mu2=0.4
+//From law of friction
+//FB=mu2*WB*cosd(30)
+a2=1.5067 //m/sec^2
+//Let t be the time elapsed until the blocks touch each other
+//displacement of block A in this period be s1
+//displacement of block B in this period be s2
+//when the two blocks touch each other
+//s1=s2+18
+//thus
+t=4.60 //sec
+//After the blocks touch each other
+a=2.45 //m/sec^2
+P=100*sind(30)-(0.2*100*cosd(30))-((100*2.45)/(9.81)) //N
+printf("\nt=%.2f sec\nP=%.1f N",t,P)
diff --git a/3886/CH15/EX15.9/15_9.txt b/3886/CH15/EX15.9/15_9.txt new file mode 100644 index 000000000..7254c9480 --- /dev/null +++ b/3886/CH15/EX15.9/15_9.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\15. DAlemberts principle\15.9.sce', -1)
+
+t=4.60 sec
+P=7.7 N
\ No newline at end of file diff --git a/3886/CH16/EX16.1/16_1.sce b/3886/CH16/EX16.1/16_1.sce new file mode 100644 index 000000000..658726feb --- /dev/null +++ b/3886/CH16/EX16.1/16_1.sce @@ -0,0 +1,15 @@ +//Pump
+//Work done in lifting 40 m^3 of water to a height of 50 m (W1)
+W1=40*9810*50 //N-m
+//Kinetic energy at delivery KE1
+KE1=(40*9810*25)/(2*9.81) //N-m
+//Total energy spent (TE)
+TE=19620000+500000 //N-m
+//This energy is spent by the pump in half an hour
+//Pump output power (PO)
+PO=(20120000)/(1800*1000) //kW
+//Input power (Ip)
+Ip=PO/0.7 //kW
+printf("\Energy spent=%.2f N-m\nInput power=%.4f kW",TE,Ip)
+
+
diff --git a/3886/CH16/EX16.1/16_1.txt b/3886/CH16/EX16.1/16_1.txt new file mode 100644 index 000000000..84c50c4cb --- /dev/null +++ b/3886/CH16/EX16.1/16_1.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.1.sce', -1)
+Energy spent=20120000.00 N-m
+Input power=15.9683 kW
\ No newline at end of file diff --git a/3886/CH16/EX16.10/16_10.sce b/3886/CH16/EX16.10/16_10.sce new file mode 100644 index 000000000..0a8eaf629 --- /dev/null +++ b/3886/CH16/EX16.10/16_10.sce @@ -0,0 +1,16 @@ +//Body A
+//refer fig. 16.14
+mu=0.2
+//let theta1 and theta2 be the slopes of the inclined planes
+//sind(thets1)=4/5 cosd(theta1)=0.6
+//sind(theta2)=3/5 cosd(theta2)=0.8
+//1500*sind(theta1)=1200 N down the plane
+F1=mu*1500*0.6 //N up the plane
+F2=0.2*2000*0.8 //N down the plane
+//Equating work done to change in kinetic energy
+//v=3 m/sec
+s=((1500*3*3+2000*1.5*1.5)/(2*9.81*260)) //m
+printf("\nThus s=%.3f m",s)
+
+
+
diff --git a/3886/CH16/EX16.10/16_10.txt b/3886/CH16/EX16.10/16_10.txt new file mode 100644 index 000000000..ec68dd228 --- /dev/null +++ b/3886/CH16/EX16.10/16_10.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.10.sce', -1)
+
+Thus s=3.529 m
\ No newline at end of file diff --git a/3886/CH16/EX16.11/16_11.sce b/3886/CH16/EX16.11/16_11.sce new file mode 100644 index 000000000..4c37ae884 --- /dev/null +++ b/3886/CH16/EX16.11/16_11.sce @@ -0,0 +1,10 @@ +//Two bodies hung to rope
+//refer fig. 16.15 (a) and (b)
+s=(450+300)*(4*4-2*2)/(2*9.81*150) //m
+//Let T be the tension in the string
+//apply work energy principle
+T=((450*3.058)-((450*12)/(2*9.81)))/3.058 //N
+printf("\nT=%.0f N\ns=%.3f m",T,s)
+
+
+
diff --git a/3886/CH16/EX16.11/16_11.txt b/3886/CH16/EX16.11/16_11.txt new file mode 100644 index 000000000..b897d5e10 --- /dev/null +++ b/3886/CH16/EX16.11/16_11.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.11.sce', -1)
+
+T=360 N
+s=3.058 m
\ No newline at end of file diff --git a/3886/CH16/EX16.12/16_12.sce b/3886/CH16/EX16.12/16_12.sce new file mode 100644 index 000000000..d758c5937 --- /dev/null +++ b/3886/CH16/EX16.12/16_12.sce @@ -0,0 +1,16 @@ +//Block slides down a plane
+//refer fig. 16.17 and 16.18
+N=3000*cosd(50) //N
+mu=0.2
+F=mu*N //N
+//let the maximum deformation of the spring be s mm
+//then
+s=721.43 //mm
+//Velocity will be maximum when the acceleration is zero
+//Let x be the deformation when net force on the body in the direction of motion is zero
+x=(3000*sind(50)-385.67)/(20) //mm
+//applying work energy principle
+v=5.061 //m/sec
+printf("\nv=%.3f m/sec\ns=%.3f mm",v,s)
+
+
diff --git a/3886/CH16/EX16.12/16_12.txt b/3886/CH16/EX16.12/16_12.txt new file mode 100644 index 000000000..60a948d58 --- /dev/null +++ b/3886/CH16/EX16.12/16_12.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.12.sce', -1)
+
+v=5.061 m/sec
+s=721.430 mm
\ No newline at end of file diff --git a/3886/CH16/EX16.13/16_13.sce b/3886/CH16/EX16.13/16_13.sce new file mode 100644 index 000000000..541f1bcdf --- /dev/null +++ b/3886/CH16/EX16.13/16_13.sce @@ -0,0 +1,23 @@ +//Wagon strikes bumper post
+//refer fig. 16.19
+W=500
+//Component of weight down the plane Wd
+Wd=W/100 //kN
+//Track resistance Rt
+Rt=2.5 //kN
+//u=0
+s=30 //m
+//Let the velocity of wagon while striking be v m/sec
+//Applying work energy equation
+v=1.716 //m/sec
+//Let spring compresion be x
+k=15000 //kN/m
+//Applying work energy equation and solving quadratic equation
+x=100.2 //mm
+printf("\nThe spring will be compressed by %.1f mm",x)
+
+
+
+
+
+
diff --git a/3886/CH16/EX16.13/16_13.txt b/3886/CH16/EX16.13/16_13.txt new file mode 100644 index 000000000..2375bf47b --- /dev/null +++ b/3886/CH16/EX16.13/16_13.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.13.sce', -1)
+
+The spring will be compressed by 100.2 mm
\ No newline at end of file diff --git a/3886/CH16/EX16.2/16_2.sce b/3886/CH16/EX16.2/16_2.sce new file mode 100644 index 000000000..fac3e4695 --- /dev/null +++ b/3886/CH16/EX16.2/16_2.sce @@ -0,0 +1,28 @@ +//Man and his wish
+//refer fig. 16.4 (a),(b)
+//Work done in sliding
+N=1 //kN
+W=N //kN
+mu=0.3
+F=mu*N //kN
+//Applied force
+P=F //kN
+//Work to be done in sliding to a distance of 5 m (W1)
+W1=0.3*5 //kJ
+//Work to be done in tipping
+//Height (h)
+h=(1/sqrt(2))-0.5 //m
+//Work done in one tipping (W2)
+W2=W*h //kJ
+//To move a distance of 5m, Five tippings are required
+//Hence
+W3=5*W2 //kJ
+printf("\nThe man needs to spend only %.2f kJ while tipping and it is less than %.2f kJ spent in sliding\nHe should move the box by tipping",W3,W1)
+
+
+
+
+
+
+
+
diff --git a/3886/CH16/EX16.2/16_2.txt b/3886/CH16/EX16.2/16_2.txt new file mode 100644 index 000000000..54554555b --- /dev/null +++ b/3886/CH16/EX16.2/16_2.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.2.sce', -1)
+
+The man needs to spend only 1.04 kJ while tipping and it is less than 1.50 kJ spent in sliding
+He should move the box by tipping
\ No newline at end of file diff --git a/3886/CH16/EX16.3/16_3.sce b/3886/CH16/EX16.3/16_3.sce new file mode 100644 index 000000000..037cb8404 --- /dev/null +++ b/3886/CH16/EX16.3/16_3.sce @@ -0,0 +1,19 @@ +//body pushed up the plane
+//refer fig. 16.6
+//applying equilibrium condition
+N=300*cosd(30) //N
+mu=0.2
+//Frictional force
+F=mu*N //N
+//initial velocity
+u=1.5 //m/sec
+//displacement
+s=6 //m
+//let final velocity be v m/sec
+//Equating work done by forces along the plane to change in K.E
+v=sqrt(77.71+2.25) //m/sec
+printf("After moving 6 m the body will have velocity v=%.4f m/sec",v)
+
+
+
+
diff --git a/3886/CH16/EX16.3/16_3.txt b/3886/CH16/EX16.3/16_3.txt new file mode 100644 index 000000000..cfd2b040a --- /dev/null +++ b/3886/CH16/EX16.3/16_3.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.3.sce', -1)
+After moving 6 m the body will have velocity v=8.9420 m/sec
\ No newline at end of file diff --git a/3886/CH16/EX16.4/16_4.sce b/3886/CH16/EX16.4/16_4.sce new file mode 100644 index 000000000..6c6ef94d5 --- /dev/null +++ b/3886/CH16/EX16.4/16_4.sce @@ -0,0 +1,21 @@ +//Power of a locomotive
+//refer fig.16.7 (a) and (b)
+v=(56*1000)/(60*60) //m/sec
+F=5*420/1000 //kN
+W=420 //kN
+P=F+W*(1/120) //kN
+//Power of Locomotive Pw
+Pw=P*v //kW (mistake in book)
+u=15.556 //m/sec
+//Resultant force parallel to the plane R
+R=F+W*sind(theta) //kN (Down the plane)
+s=((420*(15.556^2))/(2*9.81*5.6)) //m
+printf("Power of locomotive=%.4f kW\ns=%.4f m",Pw,s)
+//The answers vary due to round off error
+
+
+
+
+
+
+
diff --git a/3886/CH16/EX16.4/16_4.txt b/3886/CH16/EX16.4/16_4.txt new file mode 100644 index 000000000..1fecd1e8a --- /dev/null +++ b/3886/CH16/EX16.4/16_4.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.4.sce', -1)
+Power of locomotive=87.1111 kW
+s=925.0349 m
\ No newline at end of file diff --git a/3886/CH16/EX16.5/16_5.sce b/3886/CH16/EX16.5/16_5.sce new file mode 100644 index 000000000..aba7c080d --- /dev/null +++ b/3886/CH16/EX16.5/16_5.sce @@ -0,0 +1,26 @@ +//A tram car
+//refer fig.16.8 (a),(b) and (c)
+//frictional resistance
+W=120 //kN
+F=5*120/1000 //kN
+v=(20*1000)/(60*60) //m/sec
+// (1) on level track
+P1=F //kN
+//output power Pw1
+Pw1=P1*v //kW
+eta1=0.8
+//input power Ip1
+Ip1=Pw1/0.8 //kW
+// (2) Up the plane
+P2=F+W*(1/300) //kN
+//output power required Pw2
+Pw2=P2*v //kW
+//Input power of engine Ip2
+Ip2=Pw2/0.8 //kW
+// (3) Down the incline plane
+Pd=F-W*(1/300)
+Pwd=0.2*5.5556 //kW
+//Input power
+Ipd=1.1111/0.8 //kW
+printf("\On level track Input Power=%.3f kW\nUp the plane Input Power=%.3f kW\nDown the incline plane Input Power=%.3f kW",Ip1,Ip2,Ipd)
+
diff --git a/3886/CH16/EX16.5/16_5.txt b/3886/CH16/EX16.5/16_5.txt new file mode 100644 index 000000000..2eba45c68 --- /dev/null +++ b/3886/CH16/EX16.5/16_5.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.5.sce', -1)
+On level track Input Power=4.167 kW
+Up the plane Input Power=6.944 kW
+Down the incline plane Input Power=1.389 kW
\ No newline at end of file diff --git a/3886/CH16/EX16.6/16_6.sce b/3886/CH16/EX16.6/16_6.sce new file mode 100644 index 000000000..564987fc4 --- /dev/null +++ b/3886/CH16/EX16.6/16_6.sce @@ -0,0 +1,10 @@ +//Police investigation
+//refer fig. 16.9
+//Let the probable speed of the car just before brakes are applied be u m/sec
+//F=0.5*W
+//Final velocity=0
+s=60 //m
+//applying work energy equation
+u=((sqrt(0.5*60*2*9.81))*60*60)/1000 //m/sec
+printf("\nThe probable speed of the car just before brakes are applied is %.2f kmph",u)
+
diff --git a/3886/CH16/EX16.6/16_6.txt b/3886/CH16/EX16.6/16_6.txt new file mode 100644 index 000000000..a73233fe0 --- /dev/null +++ b/3886/CH16/EX16.6/16_6.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.6.sce', -1)
+
+The probable speed of the car just before brakes are applied is 87.34 kmph
\ No newline at end of file diff --git a/3886/CH16/EX16.7/16_7.sce b/3886/CH16/EX16.7/16_7.sce new file mode 100644 index 000000000..8e351d1bc --- /dev/null +++ b/3886/CH16/EX16.7/16_7.sce @@ -0,0 +1,20 @@ +//Block being pulled
+//refer fig. 16.10 (a) and (b)
+//when pull P is acting
+W=2500 //N
+P=1000 //N
+N=W-P*sind(30)
+mu=0.2
+F=mu*N //N
+//Initial velocity=0
+//Let final velocity be v
+s=30 //m
+//Applying work energy equation for the horizontal motion
+v=sqrt((0.866*1000-400)*30*2*9.81/2500)
+printf("\nv=%.3f m/sec",v)
+//Now if the 1000 N force is removed,let the distance moved before rest be s
+//Initial velocity=10.4745 //m/sec
+//Final velocity=0
+s=(2500*(10.4745^2))/(400*2*9.81) //m
+printf("\ns=%.3f m",s)
+//The answer provided in the textbook is wrong
diff --git a/3886/CH16/EX16.7/16_7.txt b/3886/CH16/EX16.7/16_7.txt new file mode 100644 index 000000000..5f5014663 --- /dev/null +++ b/3886/CH16/EX16.7/16_7.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.7.sce', -1)
+
+v=10.474 m/sec
+s=34.950 m
\ No newline at end of file diff --git a/3886/CH16/EX16.8/16_8.sce b/3886/CH16/EX16.8/16_8.sce new file mode 100644 index 000000000..1c5e9f496 --- /dev/null +++ b/3886/CH16/EX16.8/16_8.sce @@ -0,0 +1,18 @@ +//Small block sliding down the plane
+//refer fig. 16.11 (a),(b) and (c)
+//Length AB
+AB=sqrt((3^2)+(4^2))
+//Consider FBD of the block on inclined plane A
+//It moves down the plane, hence
+//N1=W*0.8
+mu=0.3
+//F1=0.3*W
+//Applying work energy equation for the motion from A to B
+vB=sqrt((0.6-0.24)*5*2*9.81) //m/sec
+//For the motion on horizontal plane
+//final velocity=0
+//Writing work energy equation for the motion along BC
+s=(5.943^2)/(2*9.81*0.3) //m
+printf("\ns=%.2f m",s)
+
+
diff --git a/3886/CH16/EX16.8/16_8.txt b/3886/CH16/EX16.8/16_8.txt new file mode 100644 index 000000000..ee1b89366 --- /dev/null +++ b/3886/CH16/EX16.8/16_8.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.8.sce', -1)
+
+s=6.00 m
\ No newline at end of file diff --git a/3886/CH16/EX16.9/16_9.sce b/3886/CH16/EX16.9/16_9.sce new file mode 100644 index 000000000..9dcdf4232 --- /dev/null +++ b/3886/CH16/EX16.9/16_9.sce @@ -0,0 +1,16 @@ +//Force P required
+//refer fig. 16.13 (a),(b)
+//The system of forces acting on connecting bodies is shown in figure
+N1=250 //N
+mu=0.3
+F1=mu*N1 //N
+N2=(1000*3)/(5) //N
+F2=0.3*N2 //N
+N3=500 //N
+F3=mu*N3 //N
+//Let the constant force be P
+//writing work energy equation
+P=((250+1000+500)*3*3/(2*9.81*4.5))+75+180+1000*0.8+150 //N
+printf("\nThus P=%.3f N",P)
+
+
diff --git a/3886/CH16/EX16.9/16_9.txt b/3886/CH16/EX16.9/16_9.txt new file mode 100644 index 000000000..4b2cf51c1 --- /dev/null +++ b/3886/CH16/EX16.9/16_9.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\16. Work energy method\16.9.sce', -1)
+
+Thus P=1383.389 N
\ No newline at end of file diff --git a/3886/CH17/EX17.1/17_1.sce b/3886/CH17/EX17.1/17_1.sce new file mode 100644 index 000000000..3dcb67a12 --- /dev/null +++ b/3886/CH17/EX17.1/17_1.sce @@ -0,0 +1,17 @@ +//glass marble
+//refer fig. 17.1
+g=9.81 //m/sec^2
+//applying kinematic equations
+//velocity with which marble strikes the floor Vm
+Vm=sqrt(2*g*10) //m/sec (downward)
+//applying kinematic equations
+//Velocity of rebound Vr
+Vr=sqrt(2*g*8) //m/sec (upward)
+//Taking upward direction as positive and applying impulse momentum equation
+//Impulse I
+I=(0.2*(12.52+14.007))/9.81 //N-sec
+//average force F
+F=0.541*10 //N
+printf("\nImpulse=%.3f N-sec\nF=%.2f N",I,F)
+
+
diff --git a/3886/CH17/EX17.1/17_1.txt b/3886/CH17/EX17.1/17_1.txt new file mode 100644 index 000000000..7b8a655cc --- /dev/null +++ b/3886/CH17/EX17.1/17_1.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.1.sce', -1)
+
+Impulse=0.541 N-sec
+F=5.41 N
\ No newline at end of file diff --git a/3886/CH17/EX17.10/17_10.sce b/3886/CH17/EX17.10/17_10.sce new file mode 100644 index 000000000..adb4afc01 --- /dev/null +++ b/3886/CH17/EX17.10/17_10.sce @@ -0,0 +1,15 @@ +//Nozzle issuing jet of water
+//refer fig. 17.10 and 17.11 (a) and (b)
+//Weight of water whose momentum is changed in t second is (W)
+//W=(%pi*(0.05^2)*30*9810*t)/4 N
+Px=236.75 //N
+Py=883.58 //N
+P=sqrt((Px^2)+(Py^2)) //N
+//Inclination with horizontal
+theta=atand(Py/Px) //degree
+printf("The force exerted by jet is P=%.3f N\nInclination to horizontal=%.3f degree",P,theta)
+
+
+
+
+
diff --git a/3886/CH17/EX17.10/17_10.txt b/3886/CH17/EX17.10/17_10.txt new file mode 100644 index 000000000..c4a6c5d36 --- /dev/null +++ b/3886/CH17/EX17.10/17_10.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.10.sce', -1)
+The force exerted by jet is P=914.748 N
+Inclination to horizontal=75.000 degree
\ No newline at end of file diff --git a/3886/CH17/EX17.11/17_11.sce b/3886/CH17/EX17.11/17_11.sce new file mode 100644 index 000000000..2b839ab0f --- /dev/null +++ b/3886/CH17/EX17.11/17_11.sce @@ -0,0 +1,17 @@ +//Vane is moving
+//refer fig. 17.12 (a) and (b)
+//Velocity of approach Va
+Va=20 //m/sec
+//Weight of water impinging in t seconds=385.24*t
+//Velocity of departure Vd
+Vd=30-10 //m/sec
+//Writing impulse momentum equation in x direction
+Px=105.22 //N
+Py=392.70 //N
+P=sqrt((Px^2)+(Py^2)) //N
+//inclination
+theta=atand(Py/Px) //degree
+printf("\nPressure exerted P=%.3f N",P)
+
+
+
diff --git a/3886/CH17/EX17.11/17_11.txt b/3886/CH17/EX17.11/17_11.txt new file mode 100644 index 000000000..28fc05948 --- /dev/null +++ b/3886/CH17/EX17.11/17_11.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.11.sce', -1)
+
+Pressure exerted P=406.552 N
\ No newline at end of file diff --git a/3886/CH17/EX17.12/17_12.sce b/3886/CH17/EX17.12/17_12.sce new file mode 100644 index 000000000..9f6e8a9da --- /dev/null +++ b/3886/CH17/EX17.12/17_12.sce @@ -0,0 +1,13 @@ +//Man moving
+//Weight of man
+W1=800 //N
+v=3 //m/sec
+//Weight of system after man jumps into boat
+W2=800+3200 //N
+//(a) Initial velocity of boat
+//using principle of conservation of momentum
+v=0.6 //m/sec
+//(b) Initial velocity of boat =0.9 m/sec towards the pier
+//Applying principle of conservation of momentum
+v1=-0.12 //m/sec
+printf("\nVelocity of boat and man will be %.3f m/sec towards the pier",-v1)
diff --git a/3886/CH17/EX17.12/17_12.txt b/3886/CH17/EX17.12/17_12.txt new file mode 100644 index 000000000..b21f6deb4 --- /dev/null +++ b/3886/CH17/EX17.12/17_12.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.12.sce', -1)
+
+Velocity of boat and man will be 0.120 m/sec towards the pier
\ No newline at end of file diff --git a/3886/CH17/EX17.13/17_13.sce b/3886/CH17/EX17.13/17_13.sce new file mode 100644 index 000000000..6a9b657e5 --- /dev/null +++ b/3886/CH17/EX17.13/17_13.sce @@ -0,0 +1,9 @@ +//Car running
+//(1) When three men jump off in succession
+u=0
+v1=10+((700*5)/(11000+3*700))
+v2=v1+((700*5)/(11000+2*700))
+v3=v2+((700*5)/(11000+700)) //m/sec
+//(2) When three men jump together
+v=10+((3*5*700)/(11000+3*700)) //m/sec
+printf("\nWhen three men jump off in succession v=%.3f m/sec\nWhen three men jump together v=%.3f m/sec",v3,v)
diff --git a/3886/CH17/EX17.13/17_13.txt b/3886/CH17/EX17.13/17_13.txt new file mode 100644 index 000000000..33977cb88 --- /dev/null +++ b/3886/CH17/EX17.13/17_13.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.13.sce', -1)
+
+When three men jump off in succession v=10.849 m/sec
+When three men jump together v=10.802 m/sec
\ No newline at end of file diff --git a/3886/CH17/EX17.14/17_14.sce b/3886/CH17/EX17.14/17_14.sce new file mode 100644 index 000000000..0c1e94235 --- /dev/null +++ b/3886/CH17/EX17.14/17_14.sce @@ -0,0 +1,11 @@ +//car and lorry
+//refer fig. 17.13 (a) and (b)
+//Let the velocity of vehicle after collision be vx in x direction and vy in y direction
+vx=18 //kmph
+//applying impulse momentum equation in y-direction
+vy=12 //kmph
+//Resultant velocity
+v=sqrt((vx^2)+(vy^2)) //kmph
+//its inclination to main road
+theta=atand(vy/vx) //degree
+printf("\nv=%.3f kmph\ntheta=%.3f degree",v,theta)
diff --git a/3886/CH17/EX17.14/17_14.txt b/3886/CH17/EX17.14/17_14.txt new file mode 100644 index 000000000..50f934b3e --- /dev/null +++ b/3886/CH17/EX17.14/17_14.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.14.sce', -1)
+
+v=21.633 kmph
+theta=33.690 degree
\ No newline at end of file diff --git a/3886/CH17/EX17.15/17_15.sce b/3886/CH17/EX17.15/17_15.sce new file mode 100644 index 000000000..7ecdfd4e6 --- /dev/null +++ b/3886/CH17/EX17.15/17_15.sce @@ -0,0 +1,11 @@ +//A gun
+//applying principle of conservation of momentum
+v=-5 //m/sec
+printf("\nGun will have a velocity of %.2d m/sec in the direction opposite to that of bullet",-v)
+//Let the gun recoil for a distance s
+//Using work energy equation
+s=(300*25)/(2*9.81*600) //m
+//Applying impulse momentum equation to gun
+t=(300*5)/(600*9.81) //sec
+printf("\ns=%.3f m\nt=%.3f sec",s,t)
+
diff --git a/3886/CH17/EX17.15/17_15.txt b/3886/CH17/EX17.15/17_15.txt new file mode 100644 index 000000000..1a1ab9296 --- /dev/null +++ b/3886/CH17/EX17.15/17_15.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.15.sce', -1)
+
+Gun will have a velocity of 05 m/sec in the direction opposite to that of bullet
+s=0.637 m
+t=0.255 sec
\ No newline at end of file diff --git a/3886/CH17/EX17.16/17_16.sce b/3886/CH17/EX17.16/17_16.sce new file mode 100644 index 000000000..5f8f1226e --- /dev/null +++ b/3886/CH17/EX17.16/17_16.sce @@ -0,0 +1,18 @@ +//Bullet fired horizontally
+//refer fig. 17.14
+//Let the velocity of the block be u immediately after bullet strikes it
+//Applying work energy principle
+u=1.025 //m/sec
+//Let v be the velocity of the bullet before striking the block
+//Principle of conservation of momentum gives
+v=342.69 //m/sec
+//Initial energy of bullet Ei
+Ei=(0.3*342.69^2)/(2*9.81) //J
+//Energy of the block and the bullet system E
+E=((100+0.3)*1.025^2)/(9.81*2) //J
+//Loss of energy El
+El=1795.68-5.37 //J
+printf("\nv=%.3f m/sec\nLoss of energy=%.3f J",v,El)
+
+
+
diff --git a/3886/CH17/EX17.16/17_16.txt b/3886/CH17/EX17.16/17_16.txt new file mode 100644 index 000000000..b2488a637 --- /dev/null +++ b/3886/CH17/EX17.16/17_16.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.16.sce', -1)
+
+v=342.690 m/sec
+Loss of energy=1790.310 J
\ No newline at end of file diff --git a/3886/CH17/EX17.17/17_17.sce b/3886/CH17/EX17.17/17_17.sce new file mode 100644 index 000000000..858265a7b --- /dev/null +++ b/3886/CH17/EX17.17/17_17.sce @@ -0,0 +1,7 @@ +//Bullet
+//refer fig. 17.15
+//Initial momentum of the system=Final momentum
+v=21.31 //m/sec
+//Kinetic Energy lost= Initial K.E- Final K.E
+loss=((0.5*400^2)/(9.81*2)+(30*15^2)/(2*9.81))-((30.5*21.31^2)/(2*9.81)) //J
+printf("\nLoss of energy=%.3f J",loss)
diff --git a/3886/CH17/EX17.17/17_17.txt b/3886/CH17/EX17.17/17_17.txt new file mode 100644 index 000000000..c9dbeb7a5 --- /dev/null +++ b/3886/CH17/EX17.17/17_17.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.17.sce', -1)
+
+Loss of energy=3715.569 J
\ No newline at end of file diff --git a/3886/CH17/EX17.19/17_19.sce b/3886/CH17/EX17.19/17_19.sce new file mode 100644 index 000000000..b41aa5d48 --- /dev/null +++ b/3886/CH17/EX17.19/17_19.sce @@ -0,0 +1,12 @@ +//A pile hammer
+u=0
+h=0.75 //m
+g=9.81 //m/sec^2
+//at the time of strike
+v=sqrt(2*g*h) //m/sec
+//Applying principle of conservation of momentum of pile and hammer
+V=(20*3.836)/(30) //m/sec
+//Applying work energy equation
+R=130000/1000 //kN
+printf("\nResistance to penetration of the ground=%.3f kN",R)
+
diff --git a/3886/CH17/EX17.19/17_19.txt b/3886/CH17/EX17.19/17_19.txt new file mode 100644 index 000000000..d8aa93ee9 --- /dev/null +++ b/3886/CH17/EX17.19/17_19.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.19.sce', -1)
+
+Resistance to penetration of the ground=130.000 kN
\ No newline at end of file diff --git a/3886/CH17/EX17.2/17_2.sce b/3886/CH17/EX17.2/17_2.sce new file mode 100644 index 000000000..62ab5d8b1 --- /dev/null +++ b/3886/CH17/EX17.2/17_2.sce @@ -0,0 +1,13 @@ +//batsman
+//refer fig 17.2 (a) and (b)
+//Let Fx be the horizontal component and Fy be the vertical component
+//Applying impulse momentum equation in horizontal direction
+Fx=(48*cosd(30)+20)/(9.81*0.02) //N
+//Applying impulse momentum equation in vertical direction
+Fy=(48*sind(30))/(9.81*0.02) //N
+//Resultant force
+F=sqrt(((Fx)^2)+((Fy)^2)) //N
+theta=atand(Fy/Fx) //degree
+printf("\nF=%.3f N\ntheta=%.3f degree",F,theta)
+
+
diff --git a/3886/CH17/EX17.2/17_2.txt b/3886/CH17/EX17.2/17_2.txt new file mode 100644 index 000000000..84916dc9d --- /dev/null +++ b/3886/CH17/EX17.2/17_2.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.2.sce', -1)
+
+F=336.807 N
+theta=21.296 degree
+-->
diff --git a/3886/CH17/EX17.20/17_20.sce b/3886/CH17/EX17.20/17_20.sce new file mode 100644 index 000000000..f18560949 --- /dev/null +++ b/3886/CH17/EX17.20/17_20.sce @@ -0,0 +1,7 @@ +//A pile hammer
+h=0.6 //m
+v=sqrt(2*9.81*0.6) //m/sec
+V=(15*3.431)/(22.5) //m/sec
+//Applying work energy equation
+s=(22.5*2.287^2)/(2*9.81*117.5) //m
+printf("\ns=%.3f m",s)
\ No newline at end of file diff --git a/3886/CH17/EX17.20/17_20.txt b/3886/CH17/EX17.20/17_20.txt new file mode 100644 index 000000000..cb9e41d99 --- /dev/null +++ b/3886/CH17/EX17.20/17_20.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.20.sce', -1)
+
+s=0.051 m
\ No newline at end of file diff --git a/3886/CH17/EX17.21/17_21.sce b/3886/CH17/EX17.21/17_21.sce new file mode 100644 index 000000000..97b59b2ba --- /dev/null +++ b/3886/CH17/EX17.21/17_21.sce @@ -0,0 +1,7 @@ +//Hammer
+//refer fig.17.17
+//Applying impulse momentum equation
+V=4.808 //m/sec
+//Applying work energy equation to the system
+R=306.3 //N
+printf("Resistance of the block=%.3f N",R)
diff --git a/3886/CH17/EX17.21/17_21.txt b/3886/CH17/EX17.21/17_21.txt new file mode 100644 index 000000000..64ba847e3 --- /dev/null +++ b/3886/CH17/EX17.21/17_21.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.21.sce', -1)
+Resistance of the block=306.300 N
\ No newline at end of file diff --git a/3886/CH17/EX17.3/17_3.sce b/3886/CH17/EX17.3/17_3.sce new file mode 100644 index 000000000..47c06a9c9 --- /dev/null +++ b/3886/CH17/EX17.3/17_3.sce @@ -0,0 +1,14 @@ +//Block in contact with level plane
+//refer fig. 17.3
+//Normal reaction
+N=1500 //N
+mu=0.1
+F=mu*N //N
+//Applying impulse momentum equation in the horizontal direction
+t=(1500*(16-0))/(9.81*(300-150)) //sec
+//If force is then removed, the only horizontal force is F=150 N
+//Applying impulse momentum equation
+t1=-(1500*(0-16))/(9.81*(300-150)) //sec
+printf("\nThe block takes %.3f sec before comming to rest",t)
+
+
diff --git a/3886/CH17/EX17.3/17_3.txt b/3886/CH17/EX17.3/17_3.txt new file mode 100644 index 000000000..627e5831a --- /dev/null +++ b/3886/CH17/EX17.3/17_3.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.3.sce', -1)
+
+The block takes 16.310 sec before comming to rest
\ No newline at end of file diff --git a/3886/CH17/EX17.4/17_4.sce b/3886/CH17/EX17.4/17_4.sce new file mode 100644 index 000000000..cad50745b --- /dev/null +++ b/3886/CH17/EX17.4/17_4.sce @@ -0,0 +1,14 @@ +//Automobile moving
+//refer fig. 17.4
+//initial velocity
+u=19.44 //m/sec
+//final velocity
+v=0
+//applying impulse momentum equation
+//t=1.982/mu
+//on concrete road
+t1=1.982/0.75 //sec
+//on ice
+t2=1.982/0.08 //sec
+printf("\nOn concrete road t=%.3f sec\nOn ice t=%.3f sec",t1,t2)
+
diff --git a/3886/CH17/EX17.4/17_4.txt b/3886/CH17/EX17.4/17_4.txt new file mode 100644 index 000000000..77aa18007 --- /dev/null +++ b/3886/CH17/EX17.4/17_4.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.4.sce', -1)
+
+On concrete road t=2.643 sec
+On ice t=24.775 sec
\ No newline at end of file diff --git a/3886/CH17/EX17.5/17_5.sce b/3886/CH17/EX17.5/17_5.sce new file mode 100644 index 000000000..cf5fc2e6b --- /dev/null +++ b/3886/CH17/EX17.5/17_5.sce @@ -0,0 +1,15 @@ +//Block on inclined plane
+//refer fig. 17.5
+theta=atand(5/12) //degree
+N=130*cosd(theta) //N
+mu=0.3
+F=mu*N //N
+//Force down the plane
+R=130*sind(theta)-36 //N
+u=2.4 //m/sec
+//v is final velocity
+t=5 //sec
+//applying impulse momentum equation
+v=((14*5*9.81)/(130))+2.4 //m/sec
+printf("\nv=%.3f m/sec",v)
+
diff --git a/3886/CH17/EX17.5/17_5.txt b/3886/CH17/EX17.5/17_5.txt new file mode 100644 index 000000000..b000fd28c --- /dev/null +++ b/3886/CH17/EX17.5/17_5.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.5.sce', -1)
+
+v=7.682 m/sec
\ No newline at end of file diff --git a/3886/CH17/EX17.6/17_6.sce b/3886/CH17/EX17.6/17_6.sce new file mode 100644 index 000000000..6a43fc161 --- /dev/null +++ b/3886/CH17/EX17.6/17_6.sce @@ -0,0 +1,37 @@ +//moving weight
+//refer fig.17.6 (a),(b)and(c)
+//first method
+//For 2000 N block
+W1=2000 //N
+mu=0.2
+N1=W1*cosd(30) //N
+F1=mu*N1 //N
+//For 1800N block
+W2=1800 //N
+N2=W2*cosd(60) //N
+F2=mu*N2 //N
+//Let T be the tension in the chord
+u=0
+v=9.81 //m/sec
+//applying impulse momentum equation for the 2000 N block in upward direction parallel to the plane
+//(T-1346.41)*t=2000
+//Applying impulse momentum equation for 1800 N block
+T=1363.48 //N
+//Thus
+t=117.11 //sec
+printf("\nBy first method-\nT=%.2d N\nt=%.2d sec",T,t)
+//second method
+//Writing impulse momentum equation in the direction of motion
+t1=117.11 //sec
+//To find tension in the chord, consider impulse momentum equation of any block
+T1=1363.48 //N
+printf("\nBy second method-\nt=%.3f sec\nT=%.3f N",t1,T1)
+
+
+
+
+
+
+
+
+
diff --git a/3886/CH17/EX17.6/17_6.txt b/3886/CH17/EX17.6/17_6.txt new file mode 100644 index 000000000..763a024b8 --- /dev/null +++ b/3886/CH17/EX17.6/17_6.txt @@ -0,0 +1,9 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.6.sce', -1)
+
+By first method-
+T=1363 N
+t=117 sec
+By second method-
+t=117.110 sec
+T=1363.480 N
\ No newline at end of file diff --git a/3886/CH17/EX17.7/17_7.sce b/3886/CH17/EX17.7/17_7.sce new file mode 100644 index 000000000..53d4672c7 --- /dev/null +++ b/3886/CH17/EX17.7/17_7.sce @@ -0,0 +1,17 @@ +//Tensions in the strings
+//refer fig. 17.7
+//Case (a)- Initial velocity u=0 t=5 sec
+//Writing impulse momentum equation for 500 N block and 1500 N block and solving obtained equations
+v=7.007 //m/sec
+T=642.86 //N
+//Case (b)-Initial velocity u=3 m/sec
+//Writing impulse momentum equation for 500 N block and 1500 N block and solving obtained equations
+v1=9.15 //m/sec
+T1=655.96 //N
+printf("\nFor case (a)\nv=%.3f m/sec\nT=%.3f N",v,T)
+printf("\nFor case (b)\nv=%.3f m/sec\nT=%.3f N",v1,T1)
+
+
+
+
+
diff --git a/3886/CH17/EX17.7/17_7.txt b/3886/CH17/EX17.7/17_7.txt new file mode 100644 index 000000000..df0bc479b --- /dev/null +++ b/3886/CH17/EX17.7/17_7.txt @@ -0,0 +1,9 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.7.sce', -1)
+
+For case (a)
+v=7.007 m/sec
+T=642.860 N
+For case (b)
+v=9.150 m/sec
+T=655.960 N
\ No newline at end of file diff --git a/3886/CH17/EX17.8/17_8.sce b/3886/CH17/EX17.8/17_8.sce new file mode 100644 index 000000000..a78527e08 --- /dev/null +++ b/3886/CH17/EX17.8/17_8.sce @@ -0,0 +1,10 @@ +//Frictionless pulleys
+//refer fig. 17.8 (a) and (b)
+//consider combined FBD of the system
+N1=500 //N
+F1=0.2*500 //N
+N2=1000*cosd(30) //N
+F2=0.2*N2 //N
+//writing impulse momentum equation
+v=20.19 //m/sec
+printf("\nv=%.3f m/sec",v)
diff --git a/3886/CH17/EX17.8/17_8.txt b/3886/CH17/EX17.8/17_8.txt new file mode 100644 index 000000000..cd36b8836 --- /dev/null +++ b/3886/CH17/EX17.8/17_8.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.8.sce', -1)
+
+v=20.190 m/sec
\ No newline at end of file diff --git a/3886/CH17/EX17.9/17_9.sce b/3886/CH17/EX17.9/17_9.sce new file mode 100644 index 000000000..151af432a --- /dev/null +++ b/3886/CH17/EX17.9/17_9.sce @@ -0,0 +1,10 @@ +//Value of P
+//refer fig. 17.9 (a),(b)and(c)
+//Let t1 be the time required to bring the system to rest
+N=1000 //N
+F=0.2*1000 //N
+//Applying impulse momentum equation upto stationary condition and leftward motion and solving those equations by trial and error method we get
+P=645.74 //N
+printf("Value of P is %.3f N",P)
+
+
diff --git a/3886/CH17/EX17.9/17_9.txt b/3886/CH17/EX17.9/17_9.txt new file mode 100644 index 000000000..7d78b844a --- /dev/null +++ b/3886/CH17/EX17.9/17_9.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\17. Impulse momentum\17.9.sce', -1)
+Value of P is 645.740 N
\ No newline at end of file diff --git a/3886/CH18/EX18.1/18_1.sce b/3886/CH18/EX18.1/18_1.sce new file mode 100644 index 000000000..bec05db58 --- /dev/null +++ b/3886/CH18/EX18.1/18_1.sce @@ -0,0 +1,12 @@ +//Direct central impact
+//Refer fig. 18.3
+u1=6 //m/sec
+u2=-10 //m/sec
+//Principle of conservation of momentum
+//2*v1+v2=2
+//From the defination of coefficient of restitution
+//v2-v1=12.8
+//solving
+v1=-3.6 //m/sec
+v2=12.8-(-v1) //m/sec
+printf("\nv1=%.2f m/sec\nv2=%.2f m/sec",v1,v2)
diff --git a/3886/CH18/EX18.1/18_1.txt b/3886/CH18/EX18.1/18_1.txt new file mode 100644 index 000000000..b151eecff --- /dev/null +++ b/3886/CH18/EX18.1/18_1.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\18. Impact of elastic bodies\18.1.sce', -1)
+
+v1=-3.60 m/sec
+v2=9.20 m/sec
\ No newline at end of file diff --git a/3886/CH18/EX18.11/18_11.sce b/3886/CH18/EX18.11/18_11.sce new file mode 100644 index 000000000..8304cbd87 --- /dev/null +++ b/3886/CH18/EX18.11/18_11.sce @@ -0,0 +1,18 @@ +//Ball dropped from height
+//refer fig. 18.11
+g=9.81 //m/sec^2
+h0=1.2 //m
+uy=sqrt(2*g*h0) //downward
+h1=1 //m
+v1y=sqrt(2*9.81*h1) //upwards
+e=sqrt(1/1.2)
+//Time of flight
+t1=(2*sqrt(2*9.81*1))/(9.81) //sec
+ux=0.4/0.903 //m/sec
+//Vertical component of velocity after second bounce
+v2y=0.903*sqrt(2*9.81*1) //m/sec
+h2=((v2y)^2)/(2*9.81) //m
+t2=(2*4)/(9.81) //sec
+//Horizontal range
+D2=0.443*0.815 //m
+printf("\ne=%.3f \nh2=%.3f m \nD2=%.3f m ",e,h2,D2)
diff --git a/3886/CH18/EX18.11/18_11.txt b/3886/CH18/EX18.11/18_11.txt new file mode 100644 index 000000000..529e046b6 --- /dev/null +++ b/3886/CH18/EX18.11/18_11.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\18. Impact of elastic bodies\18.11.sce', -1)
+
+e=0.913
+h2=0.815 m
+D2=0.361 m
\ No newline at end of file diff --git a/3886/CH18/EX18.12/18_12.sce b/3886/CH18/EX18.12/18_12.sce new file mode 100644 index 000000000..ef9595f61 --- /dev/null +++ b/3886/CH18/EX18.12/18_12.sce @@ -0,0 +1,14 @@ +//Sphere
+//refere fig. 18.12
+u1=3 //m/sec
+u2=0.6 //m/sec
+//From principle of conservation of momentum
+//v1+5*v2=6
+//From the defination of coefficient of restitution
+//v2-v1=1.8
+//solving
+v1=6-1.3*5 //m/sec
+//The velocity of first ball is reversed after impact
+//Loss of K.E.=Initial K.E.-final K.E.
+loss=1.07 //joules
+printf("\nLoss=%.3f joules",loss)
diff --git a/3886/CH18/EX18.12/18_12.txt b/3886/CH18/EX18.12/18_12.txt new file mode 100644 index 000000000..96348bb34 --- /dev/null +++ b/3886/CH18/EX18.12/18_12.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\18. Impact of elastic bodies\18.12.sce', -1)
+
+Loss=1.070 joules
\ No newline at end of file diff --git a/3886/CH18/EX18.13/18_13.sce b/3886/CH18/EX18.13/18_13.sce new file mode 100644 index 000000000..8b364d5fe --- /dev/null +++ b/3886/CH18/EX18.13/18_13.sce @@ -0,0 +1,11 @@ +//Loss in K.E.
+uAX=7.79 //m/sec
+uBX=-6 //m/sec
+vAX=-5.31 //m/sec
+vBX=7.104 //m/sec
+//mass of the ball m
+m=10/9.81
+//Loss of K.E.
+loss=(10*((7.79^2)+(6*6)-(5.31*5.31)-(7.104*7.104)))/(2*9.81) //J
+printf("\nLoss of K.E.=%.3f J",loss)
+//The answer provided in the textbook is wrong
diff --git a/3886/CH18/EX18.13/18_13.txt b/3886/CH18/EX18.13/18_13.txt new file mode 100644 index 000000000..01fca8cb6 --- /dev/null +++ b/3886/CH18/EX18.13/18_13.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\18. Impact of elastic bodies\18.13.sce', -1)
+
+Loss of K.E.=9.185 J
\ No newline at end of file diff --git a/3886/CH18/EX18.2/18_2.sce b/3886/CH18/EX18.2/18_2.sce new file mode 100644 index 000000000..ccd33e2df --- /dev/null +++ b/3886/CH18/EX18.2/18_2.sce @@ -0,0 +1,9 @@ +//Body moving to the right
+v2=3 //m/sec
+u2=-10 //m/sec
+v2=4 //m/sec
+//Applying principles of conservation of momentum
+v1=((80*3)-100-40)/(80) //m/sec
+//Defination of coeff. of restitution gives
+e=(4-1.25)/(3+10)
+printf("\nv1=%.3f m/sec\ne=%.3f ",v1,e)
diff --git a/3886/CH18/EX18.2/18_2.txt b/3886/CH18/EX18.2/18_2.txt new file mode 100644 index 000000000..7c9afcc00 --- /dev/null +++ b/3886/CH18/EX18.2/18_2.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\18. Impact of elastic bodies\18.2.sce', -1)
+
+v1=1.250 m/sec
+e=0.212
\ No newline at end of file diff --git a/3886/CH18/EX18.3/18_3.sce b/3886/CH18/EX18.3/18_3.sce new file mode 100644 index 000000000..74427d055 --- /dev/null +++ b/3886/CH18/EX18.3/18_3.sce @@ -0,0 +1,14 @@ +//A golf ball
+h0=10 //m
+//u1=sqrt(2*g*h0)
+u2=0
+v2=0
+//defination of coefficient of restitution gives
+//v1=%e*sqrt(2*g*h0) in upward direction
+//From kinematic equation
+h1=10*0.894^2 //m
+//After second bounce
+h2=6.388 //m
+//After third bounce
+h3=5.105 //m
+printf("\nh1=%.3f m\nAfter second bounce h2=%.3f m\nAfter third bounce h3=%.3f m",h1,h2,h3)
diff --git a/3886/CH18/EX18.3/18_3.txt b/3886/CH18/EX18.3/18_3.txt new file mode 100644 index 000000000..ff717eafa --- /dev/null +++ b/3886/CH18/EX18.3/18_3.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\18. Impact of elastic bodies\18.3.sce', -1)
+
+h1=7.992 m
+After second bounce h2=6.388 m
+After third bounce h3=5.105 m
\ No newline at end of file diff --git a/3886/CH18/EX18.4/18_4.sce b/3886/CH18/EX18.4/18_4.sce new file mode 100644 index 000000000..97d3dc681 --- /dev/null +++ b/3886/CH18/EX18.4/18_4.sce @@ -0,0 +1,13 @@ +//Ball is dropped from height
+u1=sqrt(2*9.81*1) //m/sec
+v1=-sqrt(2*9.81*0.810) //m/sec
+//There is no movement of the floor before and after striking
+//u2=0
+//v2=0
+//From the defination of coefficient of restitution
+e=(3.987/4.429)
+//Let the velocity of the ball after second bounce be v2
+v2=e*3.987 //m/sec upward
+//Expected height h2
+h2=(3.576^2)/(2*9.81) //m
+printf("\nCoefficient of restitution=%.3f \nExpected height of second bounce h2=%.4f m",e,h2)
diff --git a/3886/CH18/EX18.4/18_4.txt b/3886/CH18/EX18.4/18_4.txt new file mode 100644 index 000000000..ed7b52339 --- /dev/null +++ b/3886/CH18/EX18.4/18_4.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\18. Impact of elastic bodies\18.4.sce', -1)
+
+Coefficient of restitution=0.900
+Expected height of second bounce h2=0.6518 m
\ No newline at end of file diff --git a/3886/CH18/EX18.5/18_5.sce b/3886/CH18/EX18.5/18_5.sce new file mode 100644 index 000000000..c7f2d06c1 --- /dev/null +++ b/3886/CH18/EX18.5/18_5.sce @@ -0,0 +1,22 @@ +//Ball in frictionless tube
+//refer fif. 18.4
+u1=sqrt(2*9.81*2) //m/sec
+u2=0
+//By principle of conservation of momentum
+//v1+2*v2=6.264
+//From defination of coefficient of restitution
+//case(1)-e=1
+//v2-v1=6.264
+//solving
+v2=4.176 //m/sec
+v1=6.264-(2*4.176) //m/sec
+//Let h be the height to which hanging ball will rise
+//Change in K.E=Work Done
+h=(v2^2)/(2*9.81) //m
+//case(2)- e=0.7
+//v2-v1=4.385
+//solving
+bv2=(6.264+4.385)/(3) //m/sec
+//Height to which ball will rise
+h2=(bv2^2)/(2*9.81) //m
+printf("\nCase(1)-\nh=%.4f m\nCase(2)-\nh2=%.4f m ",h,h2)
diff --git a/3886/CH18/EX18.5/18_5.txt b/3886/CH18/EX18.5/18_5.txt new file mode 100644 index 000000000..0c940a833 --- /dev/null +++ b/3886/CH18/EX18.5/18_5.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\18. Impact of elastic bodies\18.5.sce', -1)
+
+Case(1)-
+h=0.8888 m
+Case(2)-
+h2=0.6422 m
\ No newline at end of file diff --git a/3886/CH18/EX18.6/18_6.sce b/3886/CH18/EX18.6/18_6.sce new file mode 100644 index 000000000..ddd12ae7e --- /dev/null +++ b/3886/CH18/EX18.6/18_6.sce @@ -0,0 +1,21 @@ +//Two identical balls
+//refer fig. 18.5 and 18.6 (a) and (b)
+//Before impact
+uAY=4.5 //m/sec
+uAX=7.794 //m/sec
+uBY=10.392 //m/sec
+uBX=-6 //m/sec
+vAY=uAY //m/sec
+vBY=uBY //m/sec
+//Applying principle of conservation of momentum
+//vAX+vBX=1.794
+//From defination of coefficient of restitution
+//vBX-vAX=12.415
+//Solving
+vBX=(12.415+1.794)/(2) //m/sec
+vAX=1.794-7.104 //m/sec
+vA=sqrt((5.31^2)+(4.5^2)) //m/sec
+thetaA=atand(4.5/5.31) //degree
+vB=sqrt((7.104^2)+(10.392^2)) //m/sec
+thetaB=atand(10.392/7.104) //degree
+printf("\nvA=%.3f m/sec\nthetaA=%.3f degree\nvB=%.3f m/sec\nthetaB=%.3f degree",vA,thetaA,vB,thetaB)
diff --git a/3886/CH18/EX18.6/18_6.txt b/3886/CH18/EX18.6/18_6.txt new file mode 100644 index 000000000..f65b21b53 --- /dev/null +++ b/3886/CH18/EX18.6/18_6.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\18. Impact of elastic bodies\18.6.sce', -1)
+
+vA=6.960 m/sec
+thetaA=40.280 degree
+vB=12.588 m/sec
+thetaB=55.643 degree
\ No newline at end of file diff --git a/3886/CH18/EX18.7/18_7.sce b/3886/CH18/EX18.7/18_7.sce new file mode 100644 index 000000000..ff68f5661 --- /dev/null +++ b/3886/CH18/EX18.7/18_7.sce @@ -0,0 +1,15 @@ +//Ball is dropped
+//refer fig. 18.7
+//Normal to line of impact
+u1x=1.986 //m/sec
+//In the line of impact
+u1y=-7.411 //m/sec
+//Let the velocity after impact be v1
+v1x=u1x
+//Initial and final velocities of floor=0
+//From the defination of coefficient of restitution
+v1y=sqrt((1.986^2)+(5.929^2)) //m/sec
+theta=atand(v1x/v1y) //degree to the line of impact
+//Inclination to the plane
+I=90-18.52 //degree The answer provided in the textbook is wrong
+printf("\nv1=%.3f m/sec\nInclination to the plane=%.3f degree",v1,I)
diff --git a/3886/CH18/EX18.7/18_7.txt b/3886/CH18/EX18.7/18_7.txt new file mode 100644 index 000000000..5e8967ca4 --- /dev/null +++ b/3886/CH18/EX18.7/18_7.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\18. Impact of elastic bodies\18.7.sce', -1)
+
+v1=6.253 m/sec
+Inclination to the plane=71.480 degree
\ No newline at end of file diff --git a/3886/CH18/EX18.8/18_8.sce b/3886/CH18/EX18.8/18_8.sce new file mode 100644 index 000000000..71f747103 --- /dev/null +++ b/3886/CH18/EX18.8/18_8.sce @@ -0,0 +1,17 @@ +//Ball falls vertically
+//refer fig. 18.8
+//Velocity of the ball which striking plane=3*g
+//Component of velocity down the plane=3*g*sind(20)
+//Component of velocity in the line of impact before striking
+//vy=-3*g*cosd(20)
+//velocity after the impact after striking plane
+//vy=2.4*g*cosd(20)
+//Acceleration in the line of impact=-g*cosd(20)
+//Using kinematic equation
+t=4.8 //sec
+//vx=3*g*sind(20)
+//Acceleration in this direction=g*sind(20)
+//Distance travelled in 4.8 sec
+s=(3*9.81*t*sind(20))-((9.81*t*t*sind(20))/(2)) //m The answer provided in the textbook is wrong
+printf("\nt=%.4f sec\ns=%.4f m",t,s)
+
diff --git a/3886/CH18/EX18.8/18_8.txt b/3886/CH18/EX18.8/18_8.txt new file mode 100644 index 000000000..b4f27fea6 --- /dev/null +++ b/3886/CH18/EX18.8/18_8.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\18. Impact of elastic bodies\18.8.sce', -1)
+
+t=4.8000 sec
+s=9.6630 m
\ No newline at end of file diff --git a/3886/CH19/EX19.1/19_1.sce b/3886/CH19/EX19.1/19_1.sce new file mode 100644 index 000000000..19f20d038 --- /dev/null +++ b/3886/CH19/EX19.1/19_1.sce @@ -0,0 +1,13 @@ +//Automobile moving on road
+//refer fig. 19.7
+v=13.889 //m/sec
+//case (1)-When vehicle is at A
+CFF1=(25*13.889^2)/(9.81*80) //kN
+//Vertical reaction
+R1=25-6.145 //kN
+//case (2)-When automobile is at B
+CFF2=(25*13.889^2)/(9.81*120) //kN
+R2=25+4.097 //kN
+//case (3)-On level track at C
+R3=25 //kN
+printf("\nWhen vehicle is at A, vertical reaction=%.3f kN\nWhen automobile is at B, vertical reaction=%.3f kN\nOn level track at C, vertical reaction=%.3f kN ",R1,R2,R3)
diff --git a/3886/CH19/EX19.1/19_1.txt b/3886/CH19/EX19.1/19_1.txt new file mode 100644 index 000000000..641ebb6d3 --- /dev/null +++ b/3886/CH19/EX19.1/19_1.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\19. Circular motion of rigid bodies\19.1.sce', -1)
+
+When vehicle is at A, vertical reaction=18.855 kN
+When automobile is at B, vertical reaction=29.097 kN
+On level track at C, vertical reaction=25.000 kN
\ No newline at end of file diff --git a/3886/CH19/EX19.2/19_2.sce b/3886/CH19/EX19.2/19_2.sce new file mode 100644 index 000000000..96fb9fc91 --- /dev/null +++ b/3886/CH19/EX19.2/19_2.sce @@ -0,0 +1,14 @@ +//Car on road
+//refer fig. 19.8
+//Consider dynamic equilibrium of car
+v=sqrt(0.4*9.81*50)*((60*60)/(1000)) //kmph
+//Limiting speed from the consideration of preventing overturning
+//Taking moment about point of contact of outer wheel with road and noting that R1=0 when the vehicle is about to overturn
+//Limiting speed v=50.42 kmph
+//If the vehicle moves with a velocity of 40 kmph
+v=11.111 //m/sec
+//Taking moment about outer wheel
+R1=5.612 //kN
+R2=15-R1 //kN
+printf("\nLimiting speed v=50.42\nR1=%.3f kN\nR2=%.3f kN",R1,R2)
+
diff --git a/3886/CH19/EX19.2/19_2.txt b/3886/CH19/EX19.2/19_2.txt new file mode 100644 index 000000000..76ae615fe --- /dev/null +++ b/3886/CH19/EX19.2/19_2.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\19. Circular motion of rigid bodies\19.2.sce', -1)
+
+Limiting speed v=50.42
+R1=5.612 kN
+R2=9.388 kN
\ No newline at end of file diff --git a/3886/CH19/EX19.3/19_3.sce b/3886/CH19/EX19.3/19_3.sce new file mode 100644 index 000000000..cfab3e396 --- /dev/null +++ b/3886/CH19/EX19.3/19_3.sce @@ -0,0 +1,5 @@ +//Angle of banking
+v=33.33 //m/sec
+//If alpha is the angle of banking then
+alpha=atand((v^2)/(9.81*200)) //degree
+printf("\nalpha=%.3f degree",alpha)
diff --git a/3886/CH19/EX19.3/19_3.txt b/3886/CH19/EX19.3/19_3.txt new file mode 100644 index 000000000..efff762de --- /dev/null +++ b/3886/CH19/EX19.3/19_3.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\19. Circular motion of rigid bodies\19.3.sce', -1)
+
+alpha=29.519 degree
\ No newline at end of file diff --git a/3886/CH19/EX19.4/19_4.sce b/3886/CH19/EX19.4/19_4.sce new file mode 100644 index 000000000..702b91b0f --- /dev/null +++ b/3886/CH19/EX19.4/19_4.sce @@ -0,0 +1,10 @@ +//Vehicle moving round a curve
+r=40 //m
+mu=0.4
+//(1) On level road, limiting speed from the consideration of avoiding skidding
+v=sqrt(0.4*9.81*40) //m/sec
+//(2) On a road banked to an inclination of 1 in 10
+v1=sqrt((9.81*40*(0.4+0.1))/(1-0.4*0.1)) //m/sec
+//If lateral forces are not to be experienced
+v3=sqrt(0.1*9.81*40) //m/sec
+printf("\nOn level road v=%.3f m/sec\nOn a road banked v=%.3f m/sec\nIf lateral forces are not to be experienced v=%.3f m/sec",v,v1,v3)
diff --git a/3886/CH19/EX19.4/19_4.txt b/3886/CH19/EX19.4/19_4.txt new file mode 100644 index 000000000..67eb8e844 --- /dev/null +++ b/3886/CH19/EX19.4/19_4.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\19. Circular motion of rigid bodies\19.4.sce', -1)
+
+On level road v=12.528 m/sec
+On a road banked v=14.296 m/sec
+If lateral forces are not to be experienced v=6.264 m/sec
\ No newline at end of file diff --git a/3886/CH19/EX19.5/19_5.sce b/3886/CH19/EX19.5/19_5.sce new file mode 100644 index 000000000..bb9238806 --- /dev/null +++ b/3886/CH19/EX19.5/19_5.sce @@ -0,0 +1,9 @@ +//Car going around a curve
+//refer fig. 19.9
+v=26.667 //m/sec
+F=20*(((26.667^2)/(9.81*60))-sind(30)) //kN
+//Taking moment about point of contact of outer wheel with road surface, we get
+R1=20*(((0.8*sind(30))/(1.6))+((cosd(30))/(2))+((26.667^2)/(9.81*60))*(((sind(30))/(2))-((0.8*cosd(30))/(1.6)))) //kN
+//Taking summation of forces normal to road surface
+R2=(20*((cosd(30))+(((sind(30))*26.667^2)/(9.81*60))))-9.238 //kN
+printf("\nR1=%.3f kN\nR2=%.3f kN",R1,R2)
diff --git a/3886/CH19/EX19.5/19_5.txt b/3886/CH19/EX19.5/19_5.txt new file mode 100644 index 000000000..ef06351de --- /dev/null +++ b/3886/CH19/EX19.5/19_5.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\19. Circular motion of rigid bodies\19.5.sce', -1)
+
+R1=9.238 kN
+R2=20.164 kN
\ No newline at end of file diff --git a/3886/CH19/EX19.6/19_6.sce b/3886/CH19/EX19.6/19_6.sce new file mode 100644 index 000000000..69353bff5 --- /dev/null +++ b/3886/CH19/EX19.6/19_6.sce @@ -0,0 +1,15 @@ +//Super elevation
+G=1.68 //m
+r=800 //m
+//(1)
+v=16.667 //m/sec
+alpha=atand((v^2)/(9.81*800)) //degree
+//Super elevation
+e=1000*G*tand(alpha) //mm
+//(2)
+v2=22.222 //m/sec
+F2=1000*(((0.99937*22.222^2)/(9.81*800))-0.03537) //kN
+printf("\ne=%.3f mm\nF=%.3f kN",e,F2)
+
+
+
diff --git a/3886/CH19/EX19.6/19_6.txt b/3886/CH19/EX19.6/19_6.txt new file mode 100644 index 000000000..93ea7c503 --- /dev/null +++ b/3886/CH19/EX19.6/19_6.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\19. Circular motion of rigid bodies\19.6.sce', -1)
+
+e=59.466 mm
+F=27.513 kN
\ No newline at end of file diff --git a/3886/CH19/EX19.7/19_7.sce b/3886/CH19/EX19.7/19_7.sce new file mode 100644 index 000000000..aaed16aeb --- /dev/null +++ b/3886/CH19/EX19.7/19_7.sce @@ -0,0 +1,11 @@ +//Aeroplane
+//refer fig. 19.10
+r=1300 //m
+W=8 //kN
+v=(400*1000)/(60*60) //m/sec
+//Angle of bank
+alpha=atand((111.111^2)/(9.81*1300)) //degree
+//Lift under flight condition is
+N=80*((cosd(alpha))+(((sind(alpha))*111.111^2)/(9.81*1300))) //kN
+printf("\nalpha=%.2f degree\nN=%.2f kN",alpha,N)
+
diff --git a/3886/CH19/EX19.7/19_7.txt b/3886/CH19/EX19.7/19_7.txt new file mode 100644 index 000000000..48f1f3158 --- /dev/null +++ b/3886/CH19/EX19.7/19_7.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\19. Circular motion of rigid bodies\19.7.sce', -1)
+
+alpha=44.07 degree
+N=111.34 kN
\ No newline at end of file diff --git a/3886/CH2/EX2.1/2_1.txt b/3886/CH2/EX2.1/2_1.txt new file mode 100644 index 000000000..a27271764 --- /dev/null +++ b/3886/CH2/EX2.1/2_1.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_1.sce', -1)
+Magnitude of forces are :-
+ F1=100 N
+ F2=200 N
+Angle between the forces is :-
+ theta=63.9 degree
\ No newline at end of file diff --git a/3886/CH2/EX2.10/2_10.txt b/3886/CH2/EX2.10/2_10.txt new file mode 100644 index 000000000..1187f80da --- /dev/null +++ b/3886/CH2/EX2.10/2_10.txt @@ -0,0 +1,19 @@ +
+
+Startup execution:
+ loading initial environment
+--> //Determining horizontal force F
+--> //From fig 2.14(b)
+--> //Resolving the forces
+--> //Fy=0 gives
+--> R=1500/cosd(30) //N
+ R =
+
+ 1732.0508
+--> //Fx=0 gives
+--> F=R*sind(30) //N
+ F =
+
+ 866.0254
+--> printf("Horizontal force of F=%.0f N is required to be applied",F)
+Horizontal force of F=866 N is required to be applied
\ No newline at end of file diff --git a/3886/CH2/EX2.10/Ex2_10.sce b/3886/CH2/EX2.10/Ex2_10.sce new file mode 100644 index 000000000..13b004ff6 --- /dev/null +++ b/3886/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,9 @@ +//Determining horizontal force F
+//From fig 2.14(b)
+//Resolving the forces
+//Fy=0 gives
+R=1500/cosd(30) //N
+//Fx=0 gives
+F=R*sind(30) //N
+printf("Horizontal force of F=%.0f N is required to be applied",F)
+
diff --git a/3886/CH2/EX2.11/2_11.txt b/3886/CH2/EX2.11/2_11.txt new file mode 100644 index 000000000..2b8da0e66 --- /dev/null +++ b/3886/CH2/EX2.11/2_11.txt @@ -0,0 +1,14 @@ + --> //Finding forces developed in wires
+--> //applying Lami's theorem
+--> T1=150*sind(90+60)/sind(45+30) //N
+ T1 =
+
+ 77.645714
+--> T2=150*sind(180-45)/sind(45+30) //N
+ T2 =
+
+ 109.80762
+--> printf("The forces in the wires are:-\nT1=%.1f N \nT2=%.1f N",T1,T2)
+The forces in the wires are:-
+T1=77.6 N
+T2=109.8 N
\ No newline at end of file diff --git a/3886/CH2/EX2.11/Ex2_11.sce b/3886/CH2/EX2.11/Ex2_11.sce new file mode 100644 index 000000000..033ee5920 --- /dev/null +++ b/3886/CH2/EX2.11/Ex2_11.sce @@ -0,0 +1,6 @@ +//Finding forces developed in wires
+//applying Lami's theorem
+T1=150*sind(90+60)/sind(45+30) //N
+T2=150*sind(180-45)/sind(45+30) //N
+printf("The forces in the wires are:-\nT1=%.1f N \nT2=%.1f N",T1,T2)
+
diff --git a/3886/CH2/EX2.12/2_12.txt b/3886/CH2/EX2.12/2_12.txt new file mode 100644 index 000000000..5dc08eb40 --- /dev/null +++ b/3886/CH2/EX2.12/2_12.txt @@ -0,0 +1,15 @@ + --> //Determine reactions at contact
+--> //Refer fig.2.16(b)
+--> //applying Lami's Theorem
+--> R1=400*sind(180-45)/sind(60+45) //N
+ R1 =
+
+ 292.82032
+--> R2=400*sind(180-60)/sind(60+45) //N
+ R2 =
+
+ 358.63019
+--> printf("The reactions developed are:-\nR1=%.1f N \nR2=%.1f N",R1,R2)
+The reactions developed are:-
+R1=292.8 N
+R2=358.6 N
\ No newline at end of file diff --git a/3886/CH2/EX2.12/Ex2_12.sce b/3886/CH2/EX2.12/Ex2_12.sce new file mode 100644 index 000000000..c3f8a027f --- /dev/null +++ b/3886/CH2/EX2.12/Ex2_12.sce @@ -0,0 +1,6 @@ +//Determine reactions at contact
+//Refer fig.2.16(b)
+//applying Lami's Theorem
+R1=400*sind(180-45)/sind(60+45) //N
+R2=400*sind(180-60)/sind(60+45) //N
+printf("The reactions developed are:-\nR1=%.1f N \nR2=%.1f N",R1,R2)
diff --git a/3886/CH2/EX2.13/2_13.txt b/3886/CH2/EX2.13/2_13.txt new file mode 100644 index 000000000..28ecd955d --- /dev/null +++ b/3886/CH2/EX2.13/2_13.txt @@ -0,0 +1,13 @@ + --> //To determine force in the bar and floor reaction
+--> //Refer Fig. 2.17(b)
+--> //Equilibrium equation gives
+--> S=(7*cosd(45)-5)/cosd(30) //kN
+ S =
+
+ -0.0580266
+--> R=10+7*sind(45)-S*sind(30) //kN
+ R =
+
+ 14.978761
+--> printf("Tensile force in the bar has magnitude %.3f kN and Reaction from floor is R=%.3f kN",-S,R)
+Tensile force in the bar has magnitude 0.058 kN and Reaction from floor is R=14.979 kN
\ No newline at end of file diff --git a/3886/CH2/EX2.13/Ex2_13.sce b/3886/CH2/EX2.13/Ex2_13.sce new file mode 100644 index 000000000..76eb251fe --- /dev/null +++ b/3886/CH2/EX2.13/Ex2_13.sce @@ -0,0 +1,6 @@ +//To determine force in the bar and floor reaction
+//Refer Fig. 2.17(b)
+//Equilibrium equation gives
+S=(7*cosd(45)-5)/cosd(30) //kN
+R=10+7*sind(45)-S*sind(30) //kN
+printf("Tensile force in the bar has magnitude %.3f kN and Reaction from floor is R=%.3f kN",-S,R)
diff --git a/3886/CH2/EX2.14/2_14.txt b/3886/CH2/EX2.14/2_14.txt new file mode 100644 index 000000000..7a8b93f7d --- /dev/null +++ b/3886/CH2/EX2.14/2_14.txt @@ -0,0 +1,47 @@ + --> // Finding magnitude of F
+--> //When F is applied at point B,refer fig 2.18(a)and(b)
+--> //From triangle AOC
+--> OC=300-150
+ OC =
+
+ 150.
+--> AO=300
+ AO =
+
+ 300.
+--> alpha=acosd(OC/AO) //degree
+ alpha =
+
+ 60.
+--> //from triangle AOB using geometry we get angle OBA=30 degree
+--> //Resolving the forces we get
+--> R=2000/cosd(30) //N
+ R =
+
+ 2309.4011
+--> F=R*sind(30) //N
+ F =
+
+ 1154.7005
+--> printf("Least force through point B is F=%.1f N",F)
+Least force through point B is F=1154.7 N--> //Least force required through the centre of roller
+--> //Assume that F makes an angle theta with the horizontal
+--> //Refer fig. 2.19 (a) and (b)
+--> //Resolving the forces we get
+--> //F*cosd(theta)=R*sind(60)...(1)
+--> //F*sind(theta)+R*cosd(60)=W...(2)
+--> //Solving (1) and (2) we get
+--> //sind(theta)+cotd(60)*cosd(theta)=W/F
+--> //For obtaining maximum value of W/F we differentiate W/F w.r.t. theta and we get
+--> theta=acotd(cotd(60)) //degree
+ theta =
+
+ 60.
+--> //Least value of F is observed when it is at right angle to reaction R
+--> Fmin=2000*sind(60) //N
+ Fmin =
+
+ 1732.0508
+--> printf("\nLeast force through the centre of roller is Fmin=%.0f N",Fmin)
+
+Least force through the centre of roller is Fmin=1732 N
\ No newline at end of file diff --git a/3886/CH2/EX2.14/Ex2_14.sce b/3886/CH2/EX2.14/Ex2_14.sce new file mode 100644 index 000000000..4d9c976e3 --- /dev/null +++ b/3886/CH2/EX2.14/Ex2_14.sce @@ -0,0 +1,25 @@ +// Finding magnitude of F
+//When F is applied at point B,refer fig 2.18(a)and(b)
+//From triangle AOC
+OC=300-150
+AO=300
+alpha=acosd(OC/AO) //degree
+//from triangle AOB using geometry we get angle OBA=30 degree
+//Resolving the forces we get
+R=2000/cosd(30) //N
+F=R*sind(30) //N
+printf("Least force through point B is F=%.1f N",F)
+//Least force required through the centre of roller
+//Assume that F makes an angle theta with the horizontal
+//Refer fig. 2.19 (a) and (b)
+//Resolving the forces we get
+//F*cosd(theta)=R*sind(60)...(1)
+//F*sind(theta)+R*cosd(60)=W...(2)
+//Solving (1) and (2) we get
+//sind(theta)+cotd(60)*cosd(theta)=W/F
+//For obtaining maximum value of W/F we differentiate W/F w.r.t. theta and we get
+theta=acotd(cotd(60)) //degree
+//Least value of F is observed when it is at right angle to reaction R
+Fmin=2000*sind(60) //N
+printf("\nLeast force through the centre of roller is Fmin=%.0f N",Fmin)
+
diff --git a/3886/CH2/EX2.15/2_15.txt b/3886/CH2/EX2.15/2_15.txt new file mode 100644 index 000000000..c4e3258fb --- /dev/null +++ b/3886/CH2/EX2.15/2_15.txt @@ -0,0 +1,14 @@ + --> //Determining the forces in bars AB and AC
+--> //Refer fig 2.20(a) and (b)
+--> //Select AB and AC as x and y axes
+--> //Resolving the forces we get
+--> F1=0 //N
+ F1 =
+
+ 0.
+--> F2=40*cosd(30) //N
+ F2 =
+
+ 34.641016
+--> printf("Force in bar AB is F1=%.0f N and force in bar AC is F2=%.1f N",F1,F2)
+Force in bar AB is F1=0 N and force in bar AC is F2=34.6 N
\ No newline at end of file diff --git a/3886/CH2/EX2.15/Ex2_15.sce b/3886/CH2/EX2.15/Ex2_15.sce new file mode 100644 index 000000000..45826bb0f --- /dev/null +++ b/3886/CH2/EX2.15/Ex2_15.sce @@ -0,0 +1,7 @@ +//Determining the forces in bars AB and AC
+//Refer fig 2.20(a) and (b)
+//Select AB and AC as x and y axes
+//Resolving the forces we get
+F1=0 //N
+F2=40*cosd(30) //N
+printf("Force in bar AB is F1=%.0f N and force in bar AC is F2=%.1f N",F1,F2)
diff --git a/3886/CH2/EX2.16/2_16.txt b/3886/CH2/EX2.16/2_16.txt new file mode 100644 index 000000000..46f93d825 --- /dev/null +++ b/3886/CH2/EX2.16/2_16.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_16.sce', -1)
+
+the various forces are:-
+T1=224.1 N
+T2=183.0 N
+T3=336.6 N
+T4=326.8 N
\ No newline at end of file diff --git a/3886/CH2/EX2.16/Ex2_16.sce b/3886/CH2/EX2.16/Ex2_16.sce new file mode 100644 index 000000000..36811d549 --- /dev/null +++ b/3886/CH2/EX2.16/Ex2_16.sce @@ -0,0 +1,11 @@ +//Forces in various segments of cable
+//Refer fig. 2.21 (a) and (b)
+//Apply Lami's theorem at point D
+T1=250*sind(180-60)/sind(60+45) //N
+T2=250*sind(90+45)/sind(60+45) //N
+//Now consider system of forces acting at B
+//Resolving the forces we get
+T3=(T2*cosd(60)+200)/cosd(30) //N
+T4=T3*sind(30)+T2*sind(60) //N
+printf("\nthe various forces are:-\nT1=%.1f N\nT2=%.1f N\nT3=%.1f N\nT4=%.1f N",T1,T2,T3,T4)
+
diff --git a/3886/CH2/EX2.17/2_17.txt b/3886/CH2/EX2.17/2_17.txt new file mode 100644 index 000000000..c48f19d89 --- /dev/null +++ b/3886/CH2/EX2.17/2_17.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_17.sce', -1)
+The load required to be connected at point D is W=2863.6 N
\ No newline at end of file diff --git a/3886/CH2/EX2.17/Ex2_17.sce b/3886/CH2/EX2.17/Ex2_17.sce new file mode 100644 index 000000000..8def98ef6 --- /dev/null +++ b/3886/CH2/EX2.17/Ex2_17.sce @@ -0,0 +1,12 @@ +//Load required at point D
+//Refer fig. 2.22 (a),(b) and (c)
+//Using simple geometry we have
+alpha=acosd(1.3125/1.5) //degree
+Beta=acosd(2-1.3125) //degree
+//Applying Lami's Theorem at point C
+T1=1500*sind(90)/sind(180-alpha) //N
+T2=1500*sind(90+alpha)/sind(180-alpha) //N
+//Applying Lami's Theorem at point B
+T3=T2*sind(90)/sind(90+Beta) //N
+W=T2*sind(180-Beta)/sind(90+Beta) //N
+printf("The load required to be connected at point D is W=%.1f N",W)
diff --git a/3886/CH2/EX2.18/2_18.txt b/3886/CH2/EX2.18/2_18.txt new file mode 100644 index 000000000..45993c3de --- /dev/null +++ b/3886/CH2/EX2.18/2_18.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_18.sce', -1)
+
+The required values are:-
+T1=44.80 kN
+T2=29.24 kN
+T3=25.04 kN
\ No newline at end of file diff --git a/3886/CH2/EX2.18/Ex2_18.sce b/3886/CH2/EX2.18/Ex2_18.sce new file mode 100644 index 000000000..04b2f780d --- /dev/null +++ b/3886/CH2/EX2.18/Ex2_18.sce @@ -0,0 +1,17 @@ +//finding tension and inclination
+//refer fig. 2.23 (a),(b) and (c)
+//Applying Lami's theorem at B
+T1=20*sind(50)/sind(180+30-50) //kN
+T2=20*sind(180-30)/sind(180+30-50) //kN
+//now consider equilibrium of forces at point C we get
+//T3*sind(theta)=22.4...(1)
+//T3*cosd(theta)=11.20...(2)
+//from (1) and (2) we get
+theta=atand(2) //degree
+//then (1) gives
+T3=T2*sind(50)/sind(theta) //kN
+printf("\nThe required values are:-\nT1=%.2f kN\nT2=%.2f kN\nT3=%.2f kN",T1,T2,T3)
+
+
+
+
diff --git a/3886/CH2/EX2.19/2_19.txt b/3886/CH2/EX2.19/2_19.txt new file mode 100644 index 000000000..45ecb131a --- /dev/null +++ b/3886/CH2/EX2.19/2_19.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_19.sce', -1)
+
+The required values are:-
+T1=38.97 kN
+T2=23.85 kN
+T3=22.50 kN
+theta=54.77 degree
\ No newline at end of file diff --git a/3886/CH2/EX2.19/Ex2_19.sce b/3886/CH2/EX2.19/Ex2_19.sce new file mode 100644 index 000000000..609b37097 --- /dev/null +++ b/3886/CH2/EX2.19/Ex2_19.sce @@ -0,0 +1,21 @@ +//determine tension and inclination
+//Refer fig. 2.24 (a),(b) and (c)
+//consider equilibrium at point B,we get
+//T2*sind(theta)=T1*sind(30)...(1)
+//T2*cosd(theta)=T1*sind(30)-20...(2)
+//consider equilibrium at point C,we get
+//T2*sind(theta)=T3*sind(60)...(3)
+//T2*cosd(theta)=-T3*cosd(60)+25...(4)
+//solving (1) and (3) we get
+//T1=T3*sqrt(3)...(5)
+//solving (2) and (4) and substituting (5) we get
+T3=45/2 //kN
+T1=T3*sqrt(3) //kN
+//then (1)/(2) gives
+theta=atand(1.416) //degree
+T2=19.48/sind(theta) //kN
+printf("\nThe required values are:-\nT1=%.2f kN\nT2=%.2f kN\nT3=%.2f kN\ntheta=%.2f degree",T1,T2,T3,theta)
+
+
+
+
diff --git a/3886/CH2/EX2.2/2_2.txt b/3886/CH2/EX2.2/2_2.txt new file mode 100644 index 000000000..52ef601ef --- /dev/null +++ b/3886/CH2/EX2.2/2_2.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_2.sce', -1)
+Horizontal and vertical components respectively are:-
+ Fx=10.00 kN (towards left)
+ Fy=17.32 kN (Downward)
\ No newline at end of file diff --git a/3886/CH2/EX2.20/2_20.txt b/3886/CH2/EX2.20/2_20.txt new file mode 100644 index 000000000..ea1a888a6 --- /dev/null +++ b/3886/CH2/EX2.20/2_20.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_20.sce', -1)
+
+The reactions are:-
+RA=433.0 N
+RB=250.0 N
+RC=721.7 N
+RD=577.4 N
\ No newline at end of file diff --git a/3886/CH2/EX2.20/Ex2_20.sce b/3886/CH2/EX2.20/Ex2_20.sce new file mode 100644 index 000000000..92234f9ba --- /dev/null +++ b/3886/CH2/EX2.20/Ex2_20.sce @@ -0,0 +1,15 @@ +//Reactions developed at contacts
+//Refer fig. 2.25(a),(b and (c)
+//consider equilibrium of cylinder 1
+//using conditions of equilibrium we get
+RA=500*cosd(30) //N
+RB=500*sind(30) //N
+//Consider equilibrium of cylinder 2
+//using conditions of equilibrium we get
+RC=(500+250*sind(30))/cosd(30) //N
+RD=RC*sind(30)+250*cosd(30) //N
+printf("\nThe reactions are:-\nRA=%.1f N\nRB=%.1f N\nRC=%.1f N\nRD=%.1f N",RA,RB,RC,RD)
+
+
+
+
diff --git a/3886/CH2/EX2.21/2_21.txt b/3886/CH2/EX2.21/2_21.txt new file mode 100644 index 000000000..d83233754 --- /dev/null +++ b/3886/CH2/EX2.21/2_21.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_21.sce', -1)
+
+The reactions are:-
+RA=1066.7 N
+RB=1333.3 N
+RC=3066.7 N
+RD=2828.4 N
\ No newline at end of file diff --git a/3886/CH2/EX2.21/Ex2_21.sce b/3886/CH2/EX2.21/Ex2_21.sce new file mode 100644 index 000000000..06c4ef72b --- /dev/null +++ b/3886/CH2/EX2.21/Ex2_21.sce @@ -0,0 +1,14 @@ +//reactions developed at contact surfaces
+//Refer fig. 2.26 (a),(b) and (c)
+//using geometry
+theta=acosd(0.8) //degree
+//consider equilibrium of cylinder 1
+//Using equilibrium conditions
+RB=800/sind(theta) //N
+RA=RB*cosd(theta) //N
+//consider equilibrium of cylinder 2
+//Using equilibrium conditions
+RD=((RB*sind(theta))+1200)/cosd(45) //N
+RC=RD*sind(45)+RB*cosd(theta) //N
+printf("\nThe reactions are:-\nRA=%.1f N\nRB=%.1f N\nRC=%.1f N\nRD=%.1f N",RA,RB,RC,RD)
+
diff --git a/3886/CH2/EX2.22/2_22.txt b/3886/CH2/EX2.22/2_22.txt new file mode 100644 index 000000000..23d53e936 --- /dev/null +++ b/3886/CH2/EX2.22/2_22.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_22.sce', -1)
+
+The reactions are:-
+RA=200.0 N
+RB=600.0 N
+RC=200.0 N
+RD=632.5 N
\ No newline at end of file diff --git a/3886/CH2/EX2.22/Ex2_22.sce b/3886/CH2/EX2.22/Ex2_22.sce new file mode 100644 index 000000000..76e7694fb --- /dev/null +++ b/3886/CH2/EX2.22/Ex2_22.sce @@ -0,0 +1,11 @@ +//determine the reactions developed at contact points
+//refer fig. 2.27 (a),(b) and (c)
+//considering the equilibrium conditions of cylinders we have
+RB=600 //N
+alpha=atand(450/150) //degree
+RD=RB/sind(alpha) //N
+RC=RD*cosd(alpha) //N
+RA=RC //N
+printf("\nThe reactions are:-\nRA=%.1f N\nRB=%.1f N\nRC=%.1f N\nRD=%.1f N",RA,RB,RC,RD)
+//The answers vary due to round off error
+
diff --git a/3886/CH2/EX2.23/2_23.txt b/3886/CH2/EX2.23/2_23.txt new file mode 100644 index 000000000..a07c5897b --- /dev/null +++ b/3886/CH2/EX2.23/2_23.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_23.sce', -1)
+P=1071.8 N is required to hold the system in given position.
\ No newline at end of file diff --git a/3886/CH2/EX2.23/Ex2_23.sce b/3886/CH2/EX2.23/Ex2_23.sce new file mode 100644 index 000000000..e3bb15dd9 --- /dev/null +++ b/3886/CH2/EX2.23/Ex2_23.sce @@ -0,0 +1,8 @@ +//Force P required to hold the system
+//Refer to fig. 2.28 (a),(b)&(c)
+//Applying Lami's Theorem at A we get
+C=4000*sind(180-60)/sind(60+90-15) //N
+//Applying equilibrium conditions B
+P=(-2000*cosd(45)+C*cosd(60))/cosd(15) //N
+printf("P=%.1f N is required to hold the system in given position.",P)
+
\ No newline at end of file diff --git a/3886/CH2/EX2.3/2_3.txt b/3886/CH2/EX2.3/2_3.txt new file mode 100644 index 000000000..e8cf16c3a --- /dev/null +++ b/3886/CH2/EX2.3/2_3.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_3.sce', -1)
+The normal and parallel components respectively are :-
+ Wn=9.40 kN
+ Wp=3.42 kN
\ No newline at end of file diff --git a/3886/CH2/EX2.3/Ex2_3.sce b/3886/CH2/EX2.3/Ex2_3.sce new file mode 100644 index 000000000..f25a1debc --- /dev/null +++ b/3886/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,6 @@ +//Components of block normal to and parallel to inclined plane
+//Let Wn be the normal component and Wp be the parallel component
+//Refer fig. 2.5(b),triangle ABC
+Wn=10*cosd(20) //kN
+Wp=10*sind(20) //kN
+printf("The normal and parallel components respectively are :-\n Wn=%.2f kN\n Wp=%.2f kN",Wn,Wp)
diff --git a/3886/CH2/EX2.4/2_4.txt b/3886/CH2/EX2.4/2_4.txt new file mode 100644 index 000000000..36d030bf0 --- /dev/null +++ b/3886/CH2/EX2.4/2_4.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_4.sce', -1)
+
+The resultant is R=161.5 N
+The inclination of resultant w.r.t. positive x-axis is alpha=18.81 degree
\ No newline at end of file diff --git a/3886/CH2/EX2.4/Ex2_4.sce b/3886/CH2/EX2.4/Ex2_4.sce new file mode 100644 index 000000000..76affa39e --- /dev/null +++ b/3886/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,8 @@ +//Resultant of three forces that are acting on a hook
+//Resolving all forces along x and y axis gives
+Fx=70*cosd(50)+80*cosd(25)+50*cosd(45) //N
+Fy=70*sind(50)+80*sind(25)-50*sind(45) //N
+R=sqrt(Fx^2+Fy^2) //N
+alpha=atand(Fy/Fx) //degree
+printf("\nThe resultant is R=%.1f N \nThe inclination of resultant w.r.t. positive x-axis is alpha=%.2f degree",R,alpha)
+//The answers vary due to round off error
diff --git a/3886/CH2/EX2.5/2_5.txt b/3886/CH2/EX2.5/2_5.txt new file mode 100644 index 000000000..c8117022d --- /dev/null +++ b/3886/CH2/EX2.5/2_5.txt @@ -0,0 +1,4 @@ + The resultant of given forces :-
+R=160.2 N.
+Inclination of resultant w.r.t X-axis :-
+alpha=24.1 degree.
\ No newline at end of file diff --git a/3886/CH2/EX2.5/Ex2_5.sce b/3886/CH2/EX2.5/Ex2_5.sce new file mode 100644 index 000000000..c7c6c6239 --- /dev/null +++ b/3886/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,17 @@ +//Determining the resultant
+clc
+//Given-
+//inclination of 200N force with x axis calculated by using slope of 200N force as shown in fig.2.8
+theta1=atand(1/2) //degree
+//inclination of 120N force with x axis calculated by using slope of 120N force as shown in fig.2.8
+theta2=atand(4/3) //degree
+//summation of forces in X direction
+Fx=200*cosd(26.565)-120*cosd(53.13)-50*cosd(60)+100*sind(40) //N
+//summation of forces in Y direction
+Fy=200*sind(26.565)+120*sind(53.13)-50*sind(60)-100*cosd(40) //N
+//Resultant
+R=sqrt((Fx)^2+(Fy)^2) //N
+//inclination of resultant w.r.t X axis
+alpha=atand(65.5/146.2) //degree
+printf("The resultant of given forces :-\nR=%.1f N.\n",R)
+printf("Inclination of resultant w.r.t X-axis :-\nalpha=%.1f degree.",alpha)
diff --git a/3886/CH2/EX2.6/2_6.txt b/3886/CH2/EX2.6/2_6.txt new file mode 100644 index 000000000..748e11388 --- /dev/null +++ b/3886/CH2/EX2.6/2_6.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_6.sce', -1)
+The resultant has magnitude R=234 N directed up the plane
\ No newline at end of file diff --git a/3886/CH2/EX2.6/Ex2_6.sce b/3886/CH2/EX2.6/Ex2_6.sce new file mode 100644 index 000000000..f47c7d20f --- /dev/null +++ b/3886/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,12 @@ +//Determine Resultant Force
+//From given data
+T=1200 //N
+F=100 //N
+N=500 //N
+W=1000 //N
+theta=60 //degree
+//Taking co-ordinate system parallel and perpendicular to plane as x and y axis and resolving the forces
+Fx=T-F-W*sind(theta) //N
+Fy=N-W*cosd(theta) //N
+R=sqrt(Fx^2+Fy^2)
+printf("The resultant has magnitude R=%.0f N directed up the plane",R)
diff --git a/3886/CH2/EX2.7/2_7.txt b/3886/CH2/EX2.7/2_7.txt new file mode 100644 index 000000000..52fc22652 --- /dev/null +++ b/3886/CH2/EX2.7/2_7.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_7.sce', -1)
+The third force is F=467.2 N and makes an angle of theta=61.08 degree
\ No newline at end of file diff --git a/3886/CH2/EX2.7/Ex2_7.sce b/3886/CH2/EX2.7/Ex2_7.sce new file mode 100644 index 000000000..71b465ef7 --- /dev/null +++ b/3886/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,10 @@ +//Finding the third force F
+//Assume that the third force F makes an angle theta with x-axis
+//Resolving the forces we get
+//F*cosd(theta)=-225.9...(1) //N
+//F*sind(theta)=-408.9...(2) //N
+//Then (2)/(1) gives
+theta=atand(-408.9/-225.9) //degree
+F=sqrt(225.9^2+408.9^2) //N
+printf("The third force is F=%.1f N and makes an angle of theta=%.2f degree",F,theta)
+
diff --git a/3886/CH2/EX2.8/2_8.txt b/3886/CH2/EX2.8/2_8.txt new file mode 100644 index 000000000..ee13f3330 --- /dev/null +++ b/3886/CH2/EX2.8/2_8.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_8.sce', -1)
+Required value of theta=6.31 degree
\ No newline at end of file diff --git a/3886/CH2/EX2.8/Ex2_8.sce b/3886/CH2/EX2.8/Ex2_8.sce new file mode 100644 index 000000000..398e033aa --- /dev/null +++ b/3886/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,6 @@ +//Determining the value of theta
+//x and y axes are selected as shown in fig. 2.11
+//As the resultant is directed along x-axis,component of resultant in y-direction is zero
+//Fy=0 gives
+theta=(asind(0.833/(2*cosd(20))))-20 //degree
+printf("Required value of theta=%.2f degree",theta)
diff --git a/3886/CH2/EX2.9/2_9.txt b/3886/CH2/EX2.9/2_9.txt new file mode 100644 index 000000000..950590c99 --- /dev/null +++ b/3886/CH2/EX2.9/2_9.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_9.sce', -1)
+
+The required values are:-
+T=103.5 N
+R=26.8 N
\ No newline at end of file diff --git a/3886/CH2/EX2.9/Ex2_9.sce b/3886/CH2/EX2.9/Ex2_9.sce new file mode 100644 index 000000000..6a5b0b890 --- /dev/null +++ b/3886/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,5 @@ +//Finding T and R
+//applying Lami's Theorem we get
+T=(100*sind(90))/sind(90+15) //N
+R=(100*sind(180-15))/sind(90+15) //N
+printf("\nThe required values are:-\nT=%.1f N \nR=%.1f N",T,R)
diff --git a/3886/CH20/EX20.1/20_1.sce b/3886/CH20/EX20.1/20_1.sce new file mode 100644 index 000000000..f2ee73e0f --- /dev/null +++ b/3886/CH20/EX20.1/20_1.sce @@ -0,0 +1,14 @@ +//Fly wheel
+//omega=3*t^3-2*t+2
+//theta=t^3-t^2+2*t+C
+//When t=1 theta=4
+C=2
+//theta=t^3-t^2+2*t+2
+//When t=3
+theta=3*3*3-3*3+2*3+2 //radian
+omega=3*3*3-2*3+2 //rad/sec
+//angular acceleration alpha
+//alpha=6*t-2
+//when t=3
+alpha=6*3-2 //rad/sec^2
+printf("\ntheta=%.3f radian\nomega=%.3f rad/sec\nalpha=%.3f rad/sec^2",theta,omega,alpha)
diff --git a/3886/CH20/EX20.1/20_1.txt b/3886/CH20/EX20.1/20_1.txt new file mode 100644 index 000000000..879cb21be --- /dev/null +++ b/3886/CH20/EX20.1/20_1.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.1.sce', -1)
+
+theta=26.000 radian
+omega=23.000 rad/sec
+alpha=16.000 rad/sec^2
\ No newline at end of file diff --git a/3886/CH20/EX20.10/20_10.sce b/3886/CH20/EX20.10/20_10.sce new file mode 100644 index 000000000..54161666c --- /dev/null +++ b/3886/CH20/EX20.10/20_10.sce @@ -0,0 +1,19 @@ +//welded cylinder
+//refer fig. 20.8 (a) and (b)
+//Mass moment of inertia of the bar about A
+IBA=((200)/(2*9.81))+((200*0.5^2)/(9.81))
+//Moment of inertia of the cylinder about A
+ICA=((500*0.2*0.2)/(2*9.81))+((500*1.2*1.2)/(9.81))
+//mass moment of inertia of the system about A
+I=6.7958+74.41
+//Rotational moment about A
+Mt=200*0.5+500*1.2 //N-m
+//Equating it to I*alpha
+alpha=((700)/(81.2097)) //rad/sec
+//Instantaneous acceleration of rod AB is vertical with magnitude
+Iaccnrod=0.5*8.6197 //m/sec
+//Instantaneous acceleration of cylinder is vertical with magnitude
+Iaccncylinder=1.2*8.6197 //m/sec
+//Applying D'Alembert's dynamic equilibrium equation to the system of forces
+RA=200+500-((200*4.3100)/(9.81))-((500*10.344)/(9.81)) //N
+printf("\nalpha=%.3f rad/sec\nRA=%.3f N",alpha,RA)
diff --git a/3886/CH20/EX20.10/20_10.txt b/3886/CH20/EX20.10/20_10.txt new file mode 100644 index 000000000..99b2ee1db --- /dev/null +++ b/3886/CH20/EX20.10/20_10.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.10.sce', -1)
+
+alpha=8.620 rad/sec
+RA=84.913 N
\ No newline at end of file diff --git a/3886/CH20/EX20.11/20_11.sce b/3886/CH20/EX20.11/20_11.sce new file mode 100644 index 000000000..455b3d26d --- /dev/null +++ b/3886/CH20/EX20.11/20_11.sce @@ -0,0 +1,16 @@ +//Rods welded
+//refer fig. 20.9 (a) and (b)
+//Mass moment of inertia of AB about axis of rotation
+AB=((200*1.2*1.2)/(12*9.81))+((200*0.6*0.6)/(9.81))
+//Mass moment of inertia of rod CD about A
+CD=((100*0.6*0.6)/(12*9.81))+((100*1.2*1.2)/(9.81))
+//Total mass moment of the system about A
+I=9.786+147.0
+//Let alpha be the instantaneous angular acceleration
+//Kinetic equation for motion gives
+alpha=(300*0.75)/(156.786) //rad/sec
+//Writing the dynamic equilibrium condition
+VA=200+100 //N
+HA=300-((200*0.6*alpha)/(9.81))-((100*1.2*alpha)/(9.81)) //N
+printf("\nVA=%.3f N\nHA=%.3f N",VA,HA)
+
diff --git a/3886/CH20/EX20.11/20_11.txt b/3886/CH20/EX20.11/20_11.txt new file mode 100644 index 000000000..c53337f4a --- /dev/null +++ b/3886/CH20/EX20.11/20_11.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.11.sce', -1)
+
+VA=300.000 N
+HA=264.891 N
\ No newline at end of file diff --git a/3886/CH20/EX20.2/20_2.sce b/3886/CH20/EX20.2/20_2.sce new file mode 100644 index 000000000..49b9e8bf2 --- /dev/null +++ b/3886/CH20/EX20.2/20_2.sce @@ -0,0 +1,14 @@ +//Flywheel
+//alpha=12-t
+//omega=12*t-(t^2)/2+C
+//When t=4 sec omega=60 rad/sec
+C1=20
+//When t=6 sec
+omega=12*6-((6*6)/(2))+20 //rad/sec
+//theta=6*t^2-(t^3)/6+20*t+C2
+//When t=0 theta0=C2
+//When t=6 sec theta6=180+C2
+//Angular displacement during 6 seconds=180 rad
+//Number of revolution
+N=180/(2*%pi)
+printf("\nomega=%.3f rad/sec\nNumber of revolution=%.3f ",omega,N)
diff --git a/3886/CH20/EX20.2/20_2.txt b/3886/CH20/EX20.2/20_2.txt new file mode 100644 index 000000000..613895fa4 --- /dev/null +++ b/3886/CH20/EX20.2/20_2.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.2.sce', -1)
+
+omega=74.000 rad/sec
+Number of revolution=28.648
\ No newline at end of file diff --git a/3886/CH20/EX20.3/20_3.sce b/3886/CH20/EX20.3/20_3.sce new file mode 100644 index 000000000..dd394d594 --- /dev/null +++ b/3886/CH20/EX20.3/20_3.sce @@ -0,0 +1,14 @@ +//Wheel rotating about fixed axis
+//Initial velocity
+omega0=2.0944 //rad/sec
+t=70 //sec
+//Angular displacement
+theta=100*%pi //radian
+//Using kinematic equation
+alpha=0.06839 //rad/sec^2
+//Angular velocity at the end of 70 seconds interval
+omega=2.0944+0.06839*70 //rad/sec
+//Let the time required for the velocity to reach 100 rpm be t
+t=(((200*%pi)/(60))-(2.0944))*((1)/(0.06839)) //sec
+printf("\nomega=%.3f rad/sec\nt=%.3f sec",omega,t)
+
diff --git a/3886/CH20/EX20.3/20_3.txt b/3886/CH20/EX20.3/20_3.txt new file mode 100644 index 000000000..4bd4b6eb1 --- /dev/null +++ b/3886/CH20/EX20.3/20_3.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.3.sce', -1)
+
+omega=6.882 rad/sec
+t=122.497 sec
\ No newline at end of file diff --git a/3886/CH20/EX20.4/20_4.sce b/3886/CH20/EX20.4/20_4.sce new file mode 100644 index 000000000..0c24701ba --- /dev/null +++ b/3886/CH20/EX20.4/20_4.sce @@ -0,0 +1,12 @@ +//Fly-wheel
+//theta=200*%pi //radian
+omega0=(120*2*%pi)/(60) //rad/sec
+omega=(160*2*%pi)/(60) //rad/sec
+//Using kinematic relation
+alpha=0.0977 //rad/sec^2
+//Also
+t=(16.755-4*%pi)/0.0977 //sec
+//theta' be the total angular displacement in reaching the velocity of 160 rpm
+theta=(1436.1)/(2*%pi) //revolution
+printf("\nt=%.3f sec\ntheta=%.3f revolution",t,theta)
+
diff --git a/3886/CH20/EX20.4/20_4.txt b/3886/CH20/EX20.4/20_4.txt new file mode 100644 index 000000000..067f65a7e --- /dev/null +++ b/3886/CH20/EX20.4/20_4.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.4.sce', -1)
+
+t=42.872 sec
+theta=228.562 revolution
\ No newline at end of file diff --git a/3886/CH20/EX20.5/20_5.sce b/3886/CH20/EX20.5/20_5.sce new file mode 100644 index 000000000..65728eafe --- /dev/null +++ b/3886/CH20/EX20.5/20_5.sce @@ -0,0 +1,9 @@ +//Power driven wheel
+omega0=30*%pi
+omega=0
+theta=720*%pi //rad
+//thus using kinematic equations
+alpha=-1.9635 //rad/sec^2
+//Also
+t=(30*%pi)/(1.9635) //sec
+printf("\nRetardation is %.3f rad/sec^2\nt=%.3f sec",-alpha,t)
diff --git a/3886/CH20/EX20.5/20_5.txt b/3886/CH20/EX20.5/20_5.txt new file mode 100644 index 000000000..5c5eb3091 --- /dev/null +++ b/3886/CH20/EX20.5/20_5.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.5.sce', -1)
+
+Retardation is 1.964 rad/sec^2
+t=48.000 sec
\ No newline at end of file diff --git a/3886/CH20/EX20.6/20_6.sce b/3886/CH20/EX20.6/20_6.sce new file mode 100644 index 000000000..135af1d5a --- /dev/null +++ b/3886/CH20/EX20.6/20_6.sce @@ -0,0 +1,11 @@ +//The step pulley
+//refer fig. 20.3
+theta=20 //radian
+alpha=2 //rad/sec^2
+omega0=0
+//Using kinematic relation
+t=sqrt(20) //sec
+//Velocity of A
+vA=8.944 //m/sec
+vB=0.6*8.944 //m/sec
+printf("\nt=%.3f sec\nvA=%.3f m/sec\nvB=%.3f m/sec",t,vA,vB)
diff --git a/3886/CH20/EX20.6/20_6.txt b/3886/CH20/EX20.6/20_6.txt new file mode 100644 index 000000000..d451d6ba3 --- /dev/null +++ b/3886/CH20/EX20.6/20_6.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.6.sce', -1)
+
+t=4.472 sec
+vA=8.944 m/sec
+vB=5.366 m/sec
\ No newline at end of file diff --git a/3886/CH20/EX20.7/20_7.sce b/3886/CH20/EX20.7/20_7.sce new file mode 100644 index 000000000..9797210e8 --- /dev/null +++ b/3886/CH20/EX20.7/20_7.sce @@ -0,0 +1,20 @@ +//A flywheel
+omega0=41.888 //rad/sec
+omega=29.322 //rad/sec
+t=120 //sec
+//Kinematic equation gives
+alpha=(29.3224-41.888)/(120) //rad/sec^2
+//Weight of flywheel
+Wf=50000 //N
+//Radius of gyration
+k=1 //m
+I=(50000/9.81) //kg-m^2
+//(1) Retarding torque acting on the flywheel Tr
+Tr=5096.84*0.1047 //N-m
+//(2) Change in K.E.
+C.K.E=(5096.84*((41.888^2)-(27.322^2)))/(2) //N-m The answer provided in the textbook is wrong
+//(3) Change in its angular momentum
+C.A.M=5096.84*(41.888-29.322) //N-sec
+printf("\nRetarding torque acting on the flywheel Tr=%.3f N-m\nChange in K.E.=%.3f N-m\nChange in its angular momentum=%.3f N-sec",Tr,C.K.E,C.A.M)
+
+
diff --git a/3886/CH20/EX20.7/20_7.txt b/3886/CH20/EX20.7/20_7.txt new file mode 100644 index 000000000..f475b5e6a --- /dev/null +++ b/3886/CH20/EX20.7/20_7.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.7.sce', -1)
+
+Retarding torque acting on the flywheel Tr=533.639 N-m
+Change in K.E.=2569094.975 N-m
+Change in its angular momentum=64046.891 N-sec
\ No newline at end of file diff --git a/3886/CH20/EX20.8/20_8.sce b/3886/CH20/EX20.8/20_8.sce new file mode 100644 index 000000000..43c40ae5d --- /dev/null +++ b/3886/CH20/EX20.8/20_8.sce @@ -0,0 +1,11 @@ +//Pulley
+//refer fig. 20.6
+//Let a be the resulting acceleration and T be the tension in the rope
+//Angular acceleration of pulley
+//alpha=1.667*a rad/sec^2
+//Dynamic equilibrium condition for the block gives
+//T=(600-(600*a)/(9.81))
+//From kinetic equation for pulley
+T=(200*7.358)/(9.81) //N
+a=(600*9.81)/(800) //m/sec^2
+printf("\nT=%.3f N\na=%.3f m/sec^2",T,a)
diff --git a/3886/CH20/EX20.8/20_8.txt b/3886/CH20/EX20.8/20_8.txt new file mode 100644 index 000000000..8f083a423 --- /dev/null +++ b/3886/CH20/EX20.8/20_8.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.8.sce', -1)
+
+T=150.010 N
+a=7.357 m/sec^2
\ No newline at end of file diff --git a/3886/CH20/EX20.9/20_9.sce b/3886/CH20/EX20.9/20_9.sce new file mode 100644 index 000000000..a075f95f6 --- /dev/null +++ b/3886/CH20/EX20.9/20_9.sce @@ -0,0 +1,10 @@ +//Composite pulley
+//refer fig. 20.7 (a) and (b)
+//Let aA be acceleration of 4000 N block and aB that of 2000 N block,and alpha be the angular velocity of pulley, then
+//aA=0.5*alpha
+//aB=0.75*alpha
+//Writing dynamic equilibrium equation for the two blocks and from kinetic equation of pulley
+alpha=500/245.97 //rad/sec^2
+TA=4000*(1-(0.5*2.033)/(9.81)) //N
+TB=2000*(1+(0.75*2.033)/(9.81)) //N
+printf("\nalpha=%.3f rad/sec^2\nTA=%.3f N\nTB=%.3f N",alpha,TA,TB)
diff --git a/3886/CH20/EX20.9/20_9.txt b/3886/CH20/EX20.9/20_9.txt new file mode 100644 index 000000000..807d4fa4c --- /dev/null +++ b/3886/CH20/EX20.9/20_9.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.9.sce', -1)
+
+alpha=2.033 rad/sec^2
+TA=3585.525 N
+TB=2310.856 N
\ No newline at end of file diff --git a/3886/CH21/EX21.1/21_1.sce b/3886/CH21/EX21.1/21_1.sce new file mode 100644 index 000000000..2c244ffc2 --- /dev/null +++ b/3886/CH21/EX21.1/21_1.sce @@ -0,0 +1,7 @@ +//S.H.M
+//a=-25*s
+omega=5
+//Period
+T=(2*%pi)/(5) //sec
+f=(1/T) //osc. per second
+printf("\nT=%.3f sec\nf=%.3f osc. per second",T,f)
diff --git a/3886/CH21/EX21.1/21_1.txt b/3886/CH21/EX21.1/21_1.txt new file mode 100644 index 000000000..9310f2774 --- /dev/null +++ b/3886/CH21/EX21.1/21_1.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\21. Mechanical vibration\21.1.sce', -1)
+
+T=1.257 sec
+f=0.796 osc. per second
\ No newline at end of file diff --git a/3886/CH21/EX21.10/21_10.sce b/3886/CH21/EX21.10/21_10.sce new file mode 100644 index 000000000..788f9a734 --- /dev/null +++ b/3886/CH21/EX21.10/21_10.sce @@ -0,0 +1,9 @@ +//Circular ring
+//refer fig. 21.9 (a) and (b)
+//Radius of gyration about centre of the ring
+//Kz^2=(((R1^2)+(R2^2))/(2))+R2^2
+//thus
+Kz=sqrt(((1+0.75*0.75)/(2))+(0.75^2))
+T=2*%pi*sqrt(1.34375/(9.81*0.75)) //sec
+printf("\nT=%.3f sec",T)
+
diff --git a/3886/CH21/EX21.10/21_10.txt b/3886/CH21/EX21.10/21_10.txt new file mode 100644 index 000000000..65f4edaaf --- /dev/null +++ b/3886/CH21/EX21.10/21_10.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\21. Mechanical vibration\21.10.sce', -1)
+
+T=2.685 sec
\ No newline at end of file diff --git a/3886/CH21/EX21.2/21_2.sce b/3886/CH21/EX21.2/21_2.sce new file mode 100644 index 000000000..b1c6b101b --- /dev/null +++ b/3886/CH21/EX21.2/21_2.sce @@ -0,0 +1,9 @@ +//S.H.M
+r=0.75 //m
+T=1.2 //sec
+omega=((2*%pi)/(1.2)) //rad/sec
+vxmax=0.75*5.236 //m/sec
+axmax=0.75*5.236^2 //m/sec^2
+printf("\nvx(max)=%.3f m/sec\nax(max)=%.3f m/sec^2",vxmax,axmax)
+
+
diff --git a/3886/CH21/EX21.2/21_2.txt b/3886/CH21/EX21.2/21_2.txt new file mode 100644 index 000000000..4b66afe8a --- /dev/null +++ b/3886/CH21/EX21.2/21_2.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\21. Mechanical vibration\21.2.sce', -1)
+
+vx(max)=3.927 m/sec
+ax(max)=20.562 m/sec^2
\ No newline at end of file diff --git a/3886/CH21/EX21.3/21_3.sce b/3886/CH21/EX21.3/21_3.sce new file mode 100644 index 000000000..828ad6d5d --- /dev/null +++ b/3886/CH21/EX21.3/21_3.sce @@ -0,0 +1,9 @@ +//Displacement
+//After 0.5 sec
+theta=5.236*0.5*(180/%pi)
+//displacement
+x=0.75*sind(150) //m
+//Velocity
+vx=0.75*5.236*cosd(150)
+ax=0.375*5.236^2 //m/sec^2
+printf("\nx=%.3f m\nvx=%.3f m/sec\nax=%.2f ",x,vx,ax)
diff --git a/3886/CH21/EX21.3/21_3.txt b/3886/CH21/EX21.3/21_3.txt new file mode 100644 index 000000000..e1c2d6949 --- /dev/null +++ b/3886/CH21/EX21.3/21_3.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\21. Mechanical vibration\21.3.sce', -1)
+
+x=0.375 m
+vx=-3.401 m/sec
+ax=10.28
\ No newline at end of file diff --git a/3886/CH21/EX21.4/21_4.sce b/3886/CH21/EX21.4/21_4.sce new file mode 100644 index 000000000..2617b6d9d --- /dev/null +++ b/3886/CH21/EX21.4/21_4.sce @@ -0,0 +1,13 @@ +//SHM
+//r*sind(theta1)=0.2
+//r*omega*cosd(theta1)=0.5
+//r*sind(theta2)=0.3
+//r*omega*cosd(theta2)=0.35
+//thus
+theta1=asind(0.44)
+r=(0.2)/(sind(26.1)) //m
+omega=1.225 //rad/sec^2
+vmax=0.454*1.225 //m/sec
+amax=-0.454*1.225^2 //m/sec^2
+f=(1.225)/(2*%pi) //osc. per sec
+printf("\namax=%.3f m/sec^2\nf=%.3f osc. per. sec\nvmax=%.3f m/sec",amax,f,vmax)
diff --git a/3886/CH21/EX21.4/21_4.txt b/3886/CH21/EX21.4/21_4.txt new file mode 100644 index 000000000..95cc9d2db --- /dev/null +++ b/3886/CH21/EX21.4/21_4.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\21. Mechanical vibration\21.4.sce', -1)
+
+amax=-0.681 m/sec^2
+f=0.195 osc. per. sec
+vmax=0.556 m/sec
\ No newline at end of file diff --git a/3886/CH21/EX21.6/21_6.sce b/3886/CH21/EX21.6/21_6.sce new file mode 100644 index 000000000..f2914d433 --- /dev/null +++ b/3886/CH21/EX21.6/21_6.sce @@ -0,0 +1,3 @@ +//Period of simple pendulum
+T=2*%pi*sqrt(1.5/9.81) //sec
+printf("\nT=%.2f sec",T)
diff --git a/3886/CH21/EX21.6/21_6.txt b/3886/CH21/EX21.6/21_6.txt new file mode 100644 index 000000000..325e9f963 --- /dev/null +++ b/3886/CH21/EX21.6/21_6.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\21. Mechanical vibration\21.6.sce', -1)
+
+T=2.46 sec
\ No newline at end of file diff --git a/3886/CH21/EX21.7/21_7.sce b/3886/CH21/EX21.7/21_7.sce new file mode 100644 index 000000000..1808899a1 --- /dev/null +++ b/3886/CH21/EX21.7/21_7.sce @@ -0,0 +1,3 @@ +//Length of pendulum
+L=(1*9.81)/(4*%pi*%pi) //m
+printf("\nL=%.3f m",L)
diff --git a/3886/CH21/EX21.7/21_7.txt b/3886/CH21/EX21.7/21_7.txt new file mode 100644 index 000000000..102a9d197 --- /dev/null +++ b/3886/CH21/EX21.7/21_7.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\21. Mechanical vibration\21.7.sce', -1)
+
+L=0.248 m
\ No newline at end of file diff --git a/3886/CH21/EX21.9/21_9.sce b/3886/CH21/EX21.9/21_9.sce new file mode 100644 index 000000000..a7fc58991 --- /dev/null +++ b/3886/CH21/EX21.9/21_9.sce @@ -0,0 +1,16 @@ +//Compound pendulum
+//refer fig. 21.8
+//Length of uniform rod
+l=0.6 //m
+//Radius of uniform disc
+r=0.3 //m
+//Mass moment of inertia about centre of suspention
+Iz=((25*0.6^2)/(9.81*12))+((25*0.6^2)/(9.81*2*2))+((40*0.15^2)/(2*9.81))+((40*(0.6+0.15)^2)/(9.81)) //units
+M=((25)/(9.81))+((40)/(9.81))
+//Kzz^2=0.3992
+//Distance of centre of gravity of compound pendulum from centre of suspension
+r=(25*0.3+40*0.75)/(25+40) //m
+T=2*%pi*sqrt((0.3992)/(0.5769)) //sec
+//equivalent length
+Le=(0.3992/0.5769) //m
+printf("\T=%.3f sec\nLe=%.3f m",T,Le)
diff --git a/3886/CH21/EX21.9/21_9.txt b/3886/CH21/EX21.9/21_9.txt new file mode 100644 index 000000000..6a0ea1909 --- /dev/null +++ b/3886/CH21/EX21.9/21_9.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\21. Mechanical vibration\21.9.sce', -1)
+T=5.227 sec
+Le=0.692 m
\ No newline at end of file diff --git a/3886/CH22/EX22.10/22_10.sce b/3886/CH22/EX22.10/22_10.sce new file mode 100644 index 000000000..59bab8f89 --- /dev/null +++ b/3886/CH22/EX22.10/22_10.sce @@ -0,0 +1,24 @@ +//rotating crank
+//refer fig. 22.16 (a),(b),(c),(d),(e),(f),(g),(h) and (i)
+//from sine rule
+theta=asind((80*sind(60))/(200)) //degree
+//angular velocity of crank
+omega=(2*%pi*1800)/(60) //rad/sec
+vB=0.08*omega //m/sec
+//it is at right angles to BC
+aB=(15.0796^2/0.8) //m/sec^2
+vA=15.8436 //m/sec
+aB=2842.4292 //making 60 degree with horizontal
+alpha=13120.457 //rad/sec^2
+aAB=0.2*alpha
+aA=512.2027 //m/sec^2
+aX=1118.2109 //m/sec^2
+aY=-1174.862 //m/sec^2 downward
+//Consider dynamic equilibrium of piston A
+HA=(4000)-((50*512.2027)/(9.81)) //kN The answer provided in the textbook is wrong
+//Taking moment about B
+vA=813.95 //N
+vB=2001.57 //N
+HB=2598.51 //N
+printf("\nHA=%.2f kN\nHB=%.2f N\nvA=%.2f N\nvB=%.2f N",HA,HB,vA,vB)
+
diff --git a/3886/CH22/EX22.10/22_10.txt b/3886/CH22/EX22.10/22_10.txt new file mode 100644 index 000000000..127e5fa16 --- /dev/null +++ b/3886/CH22/EX22.10/22_10.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\22.General plane motion of rigid bodies\22.10.sce', -1)
+
+HA=1389.38 kN
+HB=2598.51 N
+vA=813.95 N
+vB=2001.57 N
\ No newline at end of file diff --git a/3886/CH22/EX22.2/22_2.sce b/3886/CH22/EX22.2/22_2.sce new file mode 100644 index 000000000..b27f951e3 --- /dev/null +++ b/3886/CH22/EX22.2/22_2.sce @@ -0,0 +1,20 @@ +//1 m radius wheel
+//refer fig. 22.4(a),(b),(c),(d),(e) and (f)
+vA=1*5 //m/sec
+aA=1*4 //m/sec^2
+vBA=1*5 //m/sec
+vB=vA+vBA //m/sec
+aBA=1*4 //m/sec^2
+an=5^2 //m/sec^2
+aB=sqrt((8^2)+(25^2)) //m/sec^2
+theta=atand(25/8) //degree
+//Consider rotation of point D
+vDx=5+3*sind(60) //m/sec
+vDy=3*cosd(60) //m/sec
+vD=7.745 //m/sec
+//inclination to horizontal
+theta2=atand(1.5/7.598) //degree
+vDA=0.6*5 //m/sec^2
+aD=sqrt((14.190^2)+(1.422^2)) //m/sec^2
+theta3=atand(14.190/1.422) //degree
+printf("\nAt B\naB=%.3f m/sec^2\ntheta=%.2f degree\nvB=%.3f m/sec\nAt D\nvD=%.3f m/sec^2\ntheta2=%.2f degree\naD=%.3f m/sec^2\ntheta3=%.2f degree",aB,theta,vB,vD,theta2,aD,theta3)
diff --git a/3886/CH22/EX22.2/22_2.txt b/3886/CH22/EX22.2/22_2.txt new file mode 100644 index 000000000..cd5aaf0cc --- /dev/null +++ b/3886/CH22/EX22.2/22_2.txt @@ -0,0 +1,12 @@ +
+--> exec('E:\My program EM\22.General plane motion of rigid bodies\22.2.sce', -1)
+
+At B
+aB=26.249 m/sec^2
+theta=72.26 degree
+vB=10.000 m/sec
+At D
+vD=7.745 m/sec^2
+theta2=11.17 degree
+aD=14.261 m/sec^2
+theta3=84.28 degree
\ No newline at end of file diff --git a/3886/CH22/EX22.3/22_3.sce b/3886/CH22/EX22.3/22_3.sce new file mode 100644 index 000000000..7842cd8d7 --- /dev/null +++ b/3886/CH22/EX22.3/22_3.sce @@ -0,0 +1,9 @@ +//slender beam
+//refer fig. 22.5 (a),(b) and (c)
+//from vector diagram
+vB=2*cotd(60) //m/sec
+vBA=(2/sind(60)) //m/sec
+printf("\nvB=%.3f m/sec^2\nvB/A=%.3f m/sec",vB,vBA)
+//acceleration of point B
+aB=(1.778*sind(60))+(3*0.958*sind(30))
+printf("\naB=%.3f m/sec^2",aB)
diff --git a/3886/CH22/EX22.3/22_3.txt b/3886/CH22/EX22.3/22_3.txt new file mode 100644 index 000000000..4755147cc --- /dev/null +++ b/3886/CH22/EX22.3/22_3.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\22.General plane motion of rigid bodies\22.3.sce', -1)
+
+vB=1.155 m/sec^2
+vB/A=2.309 m/sec
+aB=2.977 m/sec^2
\ No newline at end of file diff --git a/3886/CH22/EX22.4/22_4.sce b/3886/CH22/EX22.4/22_4.sce new file mode 100644 index 000000000..ec7aa2d40 --- /dev/null +++ b/3886/CH22/EX22.4/22_4.sce @@ -0,0 +1,17 @@ +//Length of crank
+//refer fig. 22.6 (a)
+//angular velocity
+omega=(1500*2*%pi)/(60) //rad/sec
+r=0.100
+//Tangential velocity of end B
+vB=r*omega //m/sec
+//Consider motion of connecting rod BC
+theta=asind((100*sind(30))/(250)) //degree
+//Refer fig. 22.6
+//Let omega' be the angular velocity of BC
+omega1=13.6035/0.244 //rad/sec
+//Considering horizontal component of velocities
+vC=15.7080*cosd(60)+0.25*55.547*sind(11.5378) //m/sec
+printf("\nomega1=%.3f rad/sec\nvC=%.2f m/sec",omega1,vC)
+
+
diff --git a/3886/CH22/EX22.4/22_4.txt b/3886/CH22/EX22.4/22_4.txt new file mode 100644 index 000000000..275ff3aa0 --- /dev/null +++ b/3886/CH22/EX22.4/22_4.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\22.General plane motion of rigid bodies\22.4.sce', -1)
+
+omega1=55.752 rad/sec
+vC=10.63 m/sec
\ No newline at end of file diff --git a/3886/CH22/EX22.5/22_5.sce b/3886/CH22/EX22.5/22_5.sce new file mode 100644 index 000000000..ef6f7a228 --- /dev/null +++ b/3886/CH22/EX22.5/22_5.sce @@ -0,0 +1,13 @@ +//Velocities of point B and D
+//refer fig. 22.8
+vA=5*1 //m/sec
+//Instantaneous centre in vertically downward direction
+Ic=5/5 //m
+vB=2*5 //m/sec
+CP=1+0.6*sind(60) //m
+PD=0.6*cosd(60) //m
+CD=sqrt((1.520^2)+(0.3^2)) //m
+vD=1.549*5 //m/sec
+//Inclination to horizontal
+theta=atand((0.3)/(1.520)) //degree
+printf("\nvD=%.2f m/sec\ntheta=%.2f degree",vD,theta)
diff --git a/3886/CH22/EX22.5/22_5.txt b/3886/CH22/EX22.5/22_5.txt new file mode 100644 index 000000000..1ad356b5b --- /dev/null +++ b/3886/CH22/EX22.5/22_5.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\22.General plane motion of rigid bodies\22.5.sce', -1)
+
+vD=7.74 m/sec
+theta=11.16 degree
\ No newline at end of file diff --git a/3886/CH22/EX22.6/22_6.sce b/3886/CH22/EX22.6/22_6.sce new file mode 100644 index 000000000..997567380 --- /dev/null +++ b/3886/CH22/EX22.6/22_6.sce @@ -0,0 +1,5 @@ +//Velocity of B
+//refer fig.2.9
+omega=(2)/(3*sind(60)) //rad/sec
+vB=3*0.770*cosd(60) //m/sec
+printf("\nvB=%.3f m/sec",vB)
diff --git a/3886/CH22/EX22.6/22_6.txt b/3886/CH22/EX22.6/22_6.txt new file mode 100644 index 000000000..3e17969aa --- /dev/null +++ b/3886/CH22/EX22.6/22_6.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\22.General plane motion of rigid bodies\22.6.sce', -1)
+
+vB=1.155 m/sec
\ No newline at end of file diff --git a/3886/CH22/EX22.7/22_7.sce b/3886/CH22/EX22.7/22_7.sce new file mode 100644 index 000000000..462153e31 --- /dev/null +++ b/3886/CH22/EX22.7/22_7.sce @@ -0,0 +1,19 @@ +//Solid cylinder acted upon by force P
+//refer fig. 22.11 (a) and (b)
+//(1)Maximum Value of P for Rolling without slipping
+//aA=0.8*alpha
+I=(1200*0.8^2)/(2*9.81)
+W=1200 //N
+N=W //N
+//From law of friction
+F=0.2*1200 //N
+//Consider moment equilibrium equation about C
+//on solving
+alpha=(240)/(97.859-73.394) //rad/sec
+P=73.394*9.81 //N
+//(2) When P=1000 N
+F2=0.15*1200 //N
+//Taking moment about A
+alpha2=(1000*0.8-180*0.8)/(39.144) //rad/sec^2
+aA=((1000+180)*9.81)/(1200) //rad/sec^2
+printf("\nMaximum Value of P for Rolling without slipping \nP=%.2f N\nWhen P=1000 N\nalpha=%.2f rad/sec^2\naA=%.3f rad/sec^2",P,alpha2,aA)
diff --git a/3886/CH22/EX22.7/22_7.txt b/3886/CH22/EX22.7/22_7.txt new file mode 100644 index 000000000..c2d03e256 --- /dev/null +++ b/3886/CH22/EX22.7/22_7.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\22.General plane motion of rigid bodies\22.7.sce', -1)
+
+Maximum Value of P for Rolling without slipping
+P=720.00 N
+When P=1000 N
+alpha=16.76 rad/sec^2
+aA=9.647 rad/sec^2
\ No newline at end of file diff --git a/3886/CH22/EX22.9/22_9.sce b/3886/CH22/EX22.9/22_9.sce new file mode 100644 index 000000000..996d5d633 --- /dev/null +++ b/3886/CH22/EX22.9/22_9.sce @@ -0,0 +1,12 @@ +//Uniform bar
+//refer fig. 22.15(a),(b),(c),(d),(e) and (f)
+beta=atand(0.6928/1.7856) //degree
+//aG=alpha*sqrt((1.7856^2)+(0.6928^2))
+I=(300*1.6^2)/(12*9.81)
+//Equating
+alpha=((300*0.6928)/(82.3160)) //rad/sec^2
+//Taking horizontal components of the forces
+NB=(300*1.7856*2.5249)/(9.81*cosd(45)) //N
+//Taking vertical components of the forces
+NA=(300)-(194.98*sind(45))+((300*0.6928*2.5249)/(9.81)) //N (Printing mistake in text book)
+printf("\nNA=%.2f N\nNB=%.2f N\nalpha=%.2f rad/sec^2",NA,NB,alpha)
diff --git a/3886/CH22/EX22.9/22_9.txt b/3886/CH22/EX22.9/22_9.txt new file mode 100644 index 000000000..cc49cf166 --- /dev/null +++ b/3886/CH22/EX22.9/22_9.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\22.General plane motion of rigid bodies\22.9.sce', -1)
+
+NA=215.62 N
+NB=194.98 N
+alpha=2.52 rad/sec^2
\ No newline at end of file diff --git a/3886/CH3/EX3.1/3_1.txt b/3886/CH3/EX3.1/3_1.txt new file mode 100644 index 000000000..0931c10a8 --- /dev/null +++ b/3886/CH3/EX3.1/3_1.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_1.sce', -1)
+MA=28301.270189 N-mm,Anticlockwise
\ No newline at end of file diff --git a/3886/CH3/EX3.1/Ex3_1.sce b/3886/CH3/EX3.1/Ex3_1.sce new file mode 100644 index 000000000..4ad0268ac --- /dev/null +++ b/3886/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,6 @@ +//Determine the moment
+//Refer fig. 3.5
+//Take clockwise moment as positive
+//Apply Varignon's Theorem
+MA=100*300*cosd(60)-100*500*sind(60) //N-mm
+printf("MA=%f N-mm,Anticlockwise",-MA)
diff --git a/3886/CH3/EX3.10/3_10.txt b/3886/CH3/EX3.10/3_10.txt new file mode 100644 index 000000000..140103f95 --- /dev/null +++ b/3886/CH3/EX3.10/3_10.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_10.sce', -1)
+tension T in cable BC is T=100.38 kN.
+RA=114.543 kN
+alpha=32.17 degree
\ No newline at end of file diff --git a/3886/CH3/EX3.10/Ex3_10.sce b/3886/CH3/EX3.10/Ex3_10.sce new file mode 100644 index 000000000..a25fea531 --- /dev/null +++ b/3886/CH3/EX3.10/Ex3_10.sce @@ -0,0 +1,10 @@ +//tension in the cable and reaction at a point
+//refer fig. 3.21 (a),(b)&(c)
+//Taking moment about A we get
+T=((25*12*cosd(30))+(10*6*cosd(30)))/(12*sind(15)) //kN
+//applying equilibrium conditions
+HA=T*cosd(15) //kN
+VA=10+25+T*sind(15) //kN
+RA=sqrt(HA^2+VA^2) //kN
+alpha=atand(VA/HA) //degree
+printf("tension T in cable BC is T=%.2f kN.\nRA=%.3f kN\nalpha=%.2f degree",T,RA,alpha)
diff --git a/3886/CH3/EX3.11/3_11.txt b/3886/CH3/EX3.11/3_11.txt new file mode 100644 index 000000000..53f1ee37e --- /dev/null +++ b/3886/CH3/EX3.11/3_11.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_11.sce', -1)
+The required force is F=298.3 N
\ No newline at end of file diff --git a/3886/CH3/EX3.11/Ex3_11.sce b/3886/CH3/EX3.11/Ex3_11.sce new file mode 100644 index 000000000..ba7b746cc --- /dev/null +++ b/3886/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,7 @@ +//Horizontal force required at bottom to avoid slipping
+//refer fig 3.22 (a)&(b)
+//Taking moment about A
+RB=(700*2*cotd(60)+100*1.5*cotd(60))/3 //N
+F=RB //N
+printf("The required force is F=%.1f N",F)
+
diff --git a/3886/CH3/EX3.12/3_12.txt b/3886/CH3/EX3.12/3_12.txt new file mode 100644 index 000000000..4b8daeb6d --- /dev/null +++ b/3886/CH3/EX3.12/3_12.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_12.sce', -1)
+Required force is F=447.45 N
\ No newline at end of file diff --git a/3886/CH3/EX3.12/Ex3_12.sce b/3886/CH3/EX3.12/Ex3_12.sce new file mode 100644 index 000000000..53212d39e --- /dev/null +++ b/3886/CH3/EX3.12/Ex3_12.sce @@ -0,0 +1,10 @@ +//Horizontal force required at certain height to avoid slipping
+//Refer fig. 3.23
+//applying equilibrium conditions we get
+//F=RB...(1)
+//-RB*3+700*2*cotd(60)+100*1.5*cotd(60)+F=0...(2)
+//Solving (1)&(2) we get
+F=(700*2+100*1.5)*cotd(60)/2 //N
+printf("Required force is F=%.2f N",F)
+
+
diff --git a/3886/CH3/EX3.13/3_13.txt b/3886/CH3/EX3.13/3_13.txt new file mode 100644 index 000000000..2ab7a6aed --- /dev/null +++ b/3886/CH3/EX3.13/3_13.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_13.sce', -1)
+The reactions are:-
+RD=1466.7 N
+HC=1154.7 N
\ No newline at end of file diff --git a/3886/CH3/EX3.13/Ex3_13.sce b/3886/CH3/EX3.13/Ex3_13.sce new file mode 100644 index 000000000..e539d1129 --- /dev/null +++ b/3886/CH3/EX3.13/Ex3_13.sce @@ -0,0 +1,17 @@ +//Reactions at supports C and D
+//Refer fig.3.24(a),(b)&(c)
+//applying equilibrium conditions for roller
+R2=2000/cosd(30) //N
+//consider equilibrium of bar CD
+//Taking moment about C
+RD=((800*2.5*cosd(30))+(R2*2))/(5*cosd(30)) //N
+VC=800+R2*cosd(30)-RD //N
+HC=R2*sind(30) //N
+printf("The reactions are:-\nRD=%.1f N\nHC=%.1f N",RD,HC)
+
+
+
+
+
+
+
diff --git a/3886/CH3/EX3.14/3_14.txt b/3886/CH3/EX3.14/3_14.txt new file mode 100644 index 000000000..e21cc4dd6 --- /dev/null +++ b/3886/CH3/EX3.14/3_14.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_14.sce', -1)
+Required values are:-
+T=51.962 kN
+R1=23.660 kN
+R2=6.34 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.14/Ex3_14.sce b/3886/CH3/EX3.14/Ex3_14.sce new file mode 100644 index 000000000..87699a2ea --- /dev/null +++ b/3886/CH3/EX3.14/Ex3_14.sce @@ -0,0 +1,10 @@ +//Tension in the cable and reaction at axles
+//Refer fig.3.25 (a)&(b)
+//assume T as tension in the rope parallel to track
+//applying equilibrium conditions
+T=60*sind(60) //kN
+//applying moment equilibrium condition about upper axle reaction point we get
+R1=(-T*600+60*800*sind(60)+60*600*cosd(60))/1200 //kN
+R2=60*cosd(60)-R1 //kN
+printf("Required values are:-\nT=%.3f kN\nR1=%.3f kN\nR2=%.2f kN",T,R1,R2)
+
diff --git a/3886/CH3/EX3.15/3_15.txt b/3886/CH3/EX3.15/3_15.txt new file mode 100644 index 000000000..ab230155a --- /dev/null +++ b/3886/CH3/EX3.15/3_15.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_15.sce', -1)
+Minimum weight of hollow cylinder is W=0.75 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.15/Ex3_15.sce b/3886/CH3/EX3.15/Ex3_15.sce new file mode 100644 index 000000000..245f221d0 --- /dev/null +++ b/3886/CH3/EX3.15/Ex3_15.sce @@ -0,0 +1,14 @@ +//Minimum weight of hollow cylinder
+//Refer fig.3.26 (a),(b)&(c)
+O1O2=400+600 //mm
+O2D=1600-400-600 //mm
+alpha=acosd(O2D/O1O2) //degree
+//considering equilibrium of spheres
+//Taking moment about O2
+R1=(600)/(1000*sind(alpha)) //kN
+R2=R1 //kN
+R3=1+3 //kN
+//Assume minimum weight W.During tipping there will be no reaction at point B
+//Taking moment about A
+W=(0.75*1000*sind(53.13))/(800) //kN
+printf("Minimum weight of hollow cylinder is W=%.2f kN",W)
diff --git a/3886/CH3/EX3.16/3_16.txt b/3886/CH3/EX3.16/3_16.txt new file mode 100644 index 000000000..24d0518e7 --- /dev/null +++ b/3886/CH3/EX3.16/3_16.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_16.sce', -1)
+The tension is:-
+T=374.0 N
\ No newline at end of file diff --git a/3886/CH3/EX3.16/Ex3_16.sce b/3886/CH3/EX3.16/Ex3_16.sce new file mode 100644 index 000000000..12502f76f --- /dev/null +++ b/3886/CH3/EX3.16/Ex3_16.sce @@ -0,0 +1,10 @@ +//Tension in horizontal rope
+//Refer fig. 3.27 (a),(b)&(c)
+//Considering equilibrium of the entire system
+RB=500/2 //N
+RA=RB //N (symmetry)
+R1=500/(2*cosd(60)) //N
+R2=R1 //N
+//Taking moment about C
+T=((500*0.866)+(250*1.2*0.5))/(1.8*sind(60)) //N
+printf("The tension is:-\nT=%.1f N",T)
diff --git a/3886/CH3/EX3.17/3_17.txt b/3886/CH3/EX3.17/3_17.txt new file mode 100644 index 000000000..815be1e70 --- /dev/null +++ b/3886/CH3/EX3.17/3_17.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_17.sce', -1)
+Required values:-
+VA=52.32 kN
+HA=10.00 kN
+MA=99.64 kN-m
\ No newline at end of file diff --git a/3886/CH3/EX3.17/Ex3_17.sce b/3886/CH3/EX3.17/Ex3_17.sce new file mode 100644 index 000000000..f232296ec --- /dev/null +++ b/3886/CH3/EX3.17/Ex3_17.sce @@ -0,0 +1,9 @@ +//Reactions developed in cantilever beam
+//Refer fig. 3.44 (a)&(b)
+//assumptions are made as shown in fig. 3.44 (a)&(b)
+//applying equilibrium conditions
+VA=15+(10*2)+(20*sind(60)) //kN
+HA=20*cosd(60) //kN
+//Taking moment about A
+MA=10*2*1+20*2*sind(60)+15*3 //kN-m
+printf("Required values:-\nVA=%.2f kN\nHA=%.2f kN\nMA=%.2f kN-m",VA,HA,MA)
diff --git a/3886/CH3/EX3.18/3_18.txt b/3886/CH3/EX3.18/3_18.txt new file mode 100644 index 000000000..adbcaa93b --- /dev/null +++ b/3886/CH3/EX3.18/3_18.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_18.sce', -1)
+Required values:-
+VA=105.00 kN
+HA=0.00 kN
+MA=180.00 kN-m
\ No newline at end of file diff --git a/3886/CH3/EX3.18/Ex3_18.sce b/3886/CH3/EX3.18/Ex3_18.sce new file mode 100644 index 000000000..cd08724c7 --- /dev/null +++ b/3886/CH3/EX3.18/Ex3_18.sce @@ -0,0 +1,10 @@ +//Reactions developed in cantilever beam
+//Refer fig. 3.45 (a)&(b)
+//Make assumptions as shown in fig. 3.45(a) and(b)
+//applying equilibrium conditions
+VA=60+45*2/2 //kN
+HA=0 //kN
+//Taking moment about A
+MA=((45*2*2)/(3*2))+(60*2.5) //kN-m
+printf("Required values:-\nVA=%.2f kN\nHA=%.2f kN\nMA=%.2f kN-m",VA,HA,MA)
+
diff --git a/3886/CH3/EX3.19/3_19.txt b/3886/CH3/EX3.19/3_19.txt new file mode 100644 index 000000000..b5cd7e401 --- /dev/null +++ b/3886/CH3/EX3.19/3_19.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_19.sce', -1)
+The reactions are:-
+RA=44.44 kN
+RB=115.56 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.19/Ex3_19.sce b/3886/CH3/EX3.19/Ex3_19.sce new file mode 100644 index 000000000..938f7e1b7 --- /dev/null +++ b/3886/CH3/EX3.19/Ex3_19.sce @@ -0,0 +1,7 @@ +//Reactions developed in simply supported beam
+//Refer fig. 3.46 (a)&(b)
+//make assumptions as shown in fig. 3.46 (a)&(b)
+//Taking moment about B
+RA=((20*4*2)+((4*40*4)/(3*2)))/(6) //kN
+RB=80+80-RA //kN
+printf("The reactions are:-\nRA=%.2f kN\nRB=%.2f kN",RA,RB)
diff --git a/3886/CH3/EX3.2/3_2.txt b/3886/CH3/EX3.2/3_2.txt new file mode 100644 index 000000000..a132c6a87 --- /dev/null +++ b/3886/CH3/EX3.2/3_2.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_2.sce', -1)
+y-intercept will be y=2 m
\ No newline at end of file diff --git a/3886/CH3/EX3.2/Ex3_2.sce b/3886/CH3/EX3.2/Ex3_2.sce new file mode 100644 index 000000000..de00c41c7 --- /dev/null +++ b/3886/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,7 @@ +//Finding y-intercept
+//Apply law of Transmissibility and resolve 5000 N force at B
+Fx=5000*4/5 //N
+Fy=5000*3/5 //N
+//Apply Varignon's Theorem
+y=8000/4000 //m
+printf("y-intercept will be y=%.0d m",y)
diff --git a/3886/CH3/EX3.20/3_20.txt b/3886/CH3/EX3.20/3_20.txt new file mode 100644 index 000000000..ee913bc7b --- /dev/null +++ b/3886/CH3/EX3.20/3_20.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_20.sce', -1)
+The reactions developed are:-
+HA=27.13 kN
+VA=12.77 kN
+RB=18.87 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.20/Ex3_20.sce b/3886/CH3/EX3.20/Ex3_20.sce new file mode 100644 index 000000000..048e9e8ce --- /dev/null +++ b/3886/CH3/EX3.20/Ex3_20.sce @@ -0,0 +1,9 @@ +//Reactions developed at A and B
+//Refer fig. 3.47 (a) and (b)
+//Make proper assumptions from this fig.
+//applying equilibrium conditions
+HA=15*cosd(30)+20*cosd(45) //kN
+//Taking moment about A
+RB=(10*4+15*6*sind(30)+20*10*sind(45))/12 //kN
+VA=-RB+10+15*sind(30)+20*sind(45) //kN
+printf("The reactions developed are:-\nHA=%.2f kN\nVA=%.2f kN\nRB=%.2f kN",HA,VA,RB)
diff --git a/3886/CH3/EX3.21/3_21.txt b/3886/CH3/EX3.21/3_21.txt new file mode 100644 index 000000000..361bccb4c --- /dev/null +++ b/3886/CH3/EX3.21/3_21.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_21.sce', -1)
+The reactions are:-
+RA=87.02 kN
+RB=100.45 kN
+As shown in fig. 3.48
\ No newline at end of file diff --git a/3886/CH3/EX3.21/Ex3_21.sce b/3886/CH3/EX3.21/Ex3_21.sce new file mode 100644 index 000000000..75eab1950 --- /dev/null +++ b/3886/CH3/EX3.21/Ex3_21.sce @@ -0,0 +1,9 @@ +//determine magnitude and direction of support reactions
+//Refer fig. 3.48 (a)&(b)
+//Taking moment about A
+RB=((60*1*sind(60))+(80*3*sind(75))+(50*5.5*sind(60)))/(6*sind(60)) //kN (At 60 degree to the horizontal)
+HA=(-60*cosd(60))+(80*cosd(75))-(50*cosd(60))+(100.45*cosd(60)) //kN
+VA=(-100.45*sind(60))+(60*sind(60))+(80*sind(75))+(50*sind(60)) //kN
+RA=sqrt((HA^2)+(VA^2)) //kN
+alphaA=atand(VA/HA) //Degree
+printf("The reactions are:-\nRA=%.2f kN \nRB=%.2f kN \nAs shown in fig. 3.48",RA,RB)
diff --git a/3886/CH3/EX3.22/3_22.txt b/3886/CH3/EX3.22/3_22.txt new file mode 100644 index 000000000..c36db50db --- /dev/null +++ b/3886/CH3/EX3.22/3_22.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_22.sce', -1)
+The reactions are:-
+RA=91.65 kN
+HB=42.43 kN
+VB=90.78 kN
+As shown in fig.3.49
\ No newline at end of file diff --git a/3886/CH3/EX3.22/Ex3_22.sce b/3886/CH3/EX3.22/Ex3_22.sce new file mode 100644 index 000000000..ebd0d264f --- /dev/null +++ b/3886/CH3/EX3.22/Ex3_22.sce @@ -0,0 +1,8 @@ +//determine support reactions
+//Refer fig. 3.49
+//Taking moment about B
+RA=(20*7+30*4*5+60*2*sind(45))/9 //kN
+HB=60*cosd(45) //kN
+VB=20+30*4+60*sind(45)-RA //kN
+printf("The reactions are:-\nRA=%.2f kN \nHB=%.2f kN \nVB=%.2f kN \nAs shown in fig.3.49",RA,HB,VB)
+
diff --git a/3886/CH3/EX3.23/3_23.txt b/3886/CH3/EX3.23/3_23.txt new file mode 100644 index 000000000..ddeb18ea2 --- /dev/null +++ b/3886/CH3/EX3.23/3_23.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_23.sce', -1)
+The support reactions are:-
+RA=42.60 kN
+RB=89.40 kN
+As shown in fig. 3.50
\ No newline at end of file diff --git a/3886/CH3/EX3.23/Ex3_23.sce b/3886/CH3/EX3.23/Ex3_23.sce new file mode 100644 index 000000000..9664daed5 --- /dev/null +++ b/3886/CH3/EX3.23/Ex3_23.sce @@ -0,0 +1,7 @@ +//determine support reactions
+//Refer fig. 3.50
+//Taking moment about A
+RB=(30*1+24*3*(2+1.5)+(1.5*40/2)*(5+1.5/3))/5 //kN
+RA=30+72+30-RB //kN
+printf("The support reactions are:-\nRA=%.2f kN \nRB=%.2f kN \nAs shown in fig. 3.50",RA,RB)
+
diff --git a/3886/CH3/EX3.24/3_24.txt b/3886/CH3/EX3.24/3_24.txt new file mode 100644 index 000000000..092a184cf --- /dev/null +++ b/3886/CH3/EX3.24/3_24.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_24.sce', -1)
+Required values are:-
+RB=71.01 kN
+HA=21.21 kN
+VA=9.80 kN downwards
\ No newline at end of file diff --git a/3886/CH3/EX3.24/Ex3_24.sce b/3886/CH3/EX3.24/Ex3_24.sce new file mode 100644 index 000000000..b2c6e39ec --- /dev/null +++ b/3886/CH3/EX3.24/Ex3_24.sce @@ -0,0 +1,8 @@ +//determine support reactions
+//Refer fig. 3.51
+//Taking moment about A
+RB=(40+30*5*sind(45)+20*2*7)/6 //kN
+HA=30*cosd(45) //kN
+VA=30*sind(45)-RB+40 //kN (downwards)
+printf("Required values are:-\nRB=%.2f kN\nHA=%.2f kN\nVA=%.2f kN downwards",RB,HA,-VA)
+
diff --git a/3886/CH3/EX3.25/3_25.txt b/3886/CH3/EX3.25/3_25.txt new file mode 100644 index 000000000..0ca325774 --- /dev/null +++ b/3886/CH3/EX3.25/3_25.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_25.sce', -1)
+Required values are:-
+RA=26.00 kN
+RB=34.00 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.25/Ex3_25.sce b/3886/CH3/EX3.25/Ex3_25.sce new file mode 100644 index 000000000..1a96052a4 --- /dev/null +++ b/3886/CH3/EX3.25/Ex3_25.sce @@ -0,0 +1,9 @@ +//determine support reactions
+//Refer fig. 3.52
+//Taking moment about B
+RA=((10/2)*(1/3+5)+(2*10/2)*(5-2/3)+10*3*1.5+3*10*1/2)/5 //kN
+RB=-26+5+10+30+15 //kN
+printf("Required values are:-\nRA=%.2f kN\nRB=%.2f kN",RA,RB)
+
+
+
diff --git a/3886/CH3/EX3.26/3_26.txt b/3886/CH3/EX3.26/3_26.txt new file mode 100644 index 000000000..8b2b6b301 --- /dev/null +++ b/3886/CH3/EX3.26/3_26.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_26.sce', -1)
+The required distance is x=5 m
\ No newline at end of file diff --git a/3886/CH3/EX3.26/Ex3_26.sce b/3886/CH3/EX3.26/Ex3_26.sce new file mode 100644 index 000000000..475823bf9 --- /dev/null +++ b/3886/CH3/EX3.26/Ex3_26.sce @@ -0,0 +1,14 @@ +//determine distance of support C from end A
+//Refer fig. 3.53
+//assume that the support at C is at a distance of x metres from end A
+//applying equilibrium conditions
+RC=(30+120+50)/2 //kN
+RD=RC //kN (Given)
+//Taking moment about A
+x=(1000+1200-100*12)/200 //m
+printf("The required distance is x=%.0f m",x)
+
+
+
+
+
diff --git a/3886/CH3/EX3.28/3_28.txt b/3886/CH3/EX3.28/3_28.txt new file mode 100644 index 000000000..e9ebddfa9 --- /dev/null +++ b/3886/CH3/EX3.28/3_28.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_28.sce', -1)
+Required reactions are:-
+RA=12.86 kN
+RB=8.57 kN,
+RD=16.07 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.28/Ex3_28.sce b/3886/CH3/EX3.28/Ex3_28.sce new file mode 100644 index 000000000..e96a19817 --- /dev/null +++ b/3886/CH3/EX3.28/Ex3_28.sce @@ -0,0 +1,15 @@ +//determine reactions at A,B and D
+//Refer fig. 3.55
+//Taking moment about C
+RD=(3*5*2.5+(5*9*2*5)/(2*3))/7 //kN
+RC=15+22.5-RD //kN
+//Taking moment about A
+RB=(2*RC/5) //kN
+RA=RC-RB //kN
+printf("Required reactions are:-\nRA=%.2f kN\nRB=%.2f kN,\nRD=%.2f kN",RA,RB,RD)
+
+
+
+
+
+
diff --git a/3886/CH3/EX3.29/3_29.txt b/3886/CH3/EX3.29/3_29.txt new file mode 100644 index 000000000..24f8e9d25 --- /dev/null +++ b/3886/CH3/EX3.29/3_29.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_29.sce', -1)
+Required Values are:-
+HA=28.28 kN
+VA=9.43 kN downward
+RD=40.91 kN
+HC=10.00 kN
+VC=34.12 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.29/Ex3_29.sce b/3886/CH3/EX3.29/Ex3_29.sce new file mode 100644 index 000000000..c81d38777 --- /dev/null +++ b/3886/CH3/EX3.29/Ex3_29.sce @@ -0,0 +1,18 @@ +//determine reactions at A,C and D
+//Refer fig. 3.56
+//Taking moment about A
+RE=(20*3+40*4*sind(45))/3 //kN
+HA=40*cosd(45) //kN
+VA=20+40*sind(45)-RE //kN (Downwards)
+//Taking moment about C
+RD=((20*sind(60))-(10)+(57.71*2))/3 //kN
+HC=20*cosd(60) //kN
+VC=20*sind(60)+RE-RD //kN
+printf("Required Values are:-\nHA=%.2f kN\nVA=%.2f kN downward\nRD=%.2f kN\nHC=%.2f kN\nVC=%.2f kN",HA,-VA,RD,HC,VC)
+
+
+
+
+
+
+
diff --git a/3886/CH3/EX3.3/3_3.txt b/3886/CH3/EX3.3/3_3.txt new file mode 100644 index 000000000..e7c451717 --- /dev/null +++ b/3886/CH3/EX3.3/3_3.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_3.sce', -1)
+R=68.06 kN as shown in fig. 3.12(a)
diff --git a/3886/CH3/EX3.3/Ex3_3.sce b/3886/CH3/EX3.3/Ex3_3.sce new file mode 100644 index 000000000..9115560e8 --- /dev/null +++ b/3886/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,11 @@ +//Determine Resultant
+//horizontal direction is assumed as x-axis and vertical as y-axis
+Rx=-20*cosd(60) //kN (towards left)
+Ry=-20-30-20*sind(60) //kN (downwards)
+R=sqrt(Rx^2+Ry^2) //kN
+alpha=atand(Ry/Rx) //degree (as shown in fig. 3.12(b))
+//Taking moment about A
+MA=20*1.5+30*3+20*6*sind(60) //kN-m
+//x-intercept of the resultant is
+x=MA/Ry //m (shown in fig.)
+printf("R=%.2f kN as shown in fig. 3.12(a)",R)
diff --git a/3886/CH3/EX3.4/3_4.txt b/3886/CH3/EX3.4/3_4.txt new file mode 100644 index 000000000..d2a876941 --- /dev/null +++ b/3886/CH3/EX3.4/3_4.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_4.sce', -1)
+R=108 kN is the resultant as shown in fig. 3.13 (a)
\ No newline at end of file diff --git a/3886/CH3/EX3.4/Ex3_4.sce b/3886/CH3/EX3.4/Ex3_4.sce new file mode 100644 index 000000000..18472dba7 --- /dev/null +++ b/3886/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,11 @@ +//Find resultant
+Rx=60-100*cosd(60)-120*cosd(30) //kN (towards left)
+Ry=-80+100*sind(60)-120*sind(30) //kN (downwards)
+R=sqrt(Rx^2+Ry^2) //kN
+alpha=atand(Ry/Rx) //degree (shown in fig. 3.13(b))
+MA=(80*100*cosd(60)+60*100*sind(60)+120*100*sind(30)) //kN-mm
+//intercept on x-axis is
+x=MA/Ry //mm (as shown in fig. 3.13(a))
+printf("R=%.0f kN is the resultant as shown in fig. 3.13 (a)",R)
+
+
diff --git a/3886/CH3/EX3.5/3_5.txt b/3886/CH3/EX3.5/3_5.txt new file mode 100644 index 000000000..7f9684413 --- /dev/null +++ b/3886/CH3/EX3.5/3_5.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_5.sce', -1)
+The resultant of the system is R=4.656 kN as shown in fig. 3.14(b)
\ No newline at end of file diff --git a/3886/CH3/EX3.5/Ex3_5.sce b/3886/CH3/EX3.5/Ex3_5.sce new file mode 100644 index 000000000..d74c2e7a4 --- /dev/null +++ b/3886/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,17 @@ +//Find the resultant
+//refer fig. 3.14 (a) and (b)
+theta1=atand(10/10)
+theta2=atand(30/40)
+theta3=atand(10/20)
+Rx=2*cosd(theta1)+5*cosd(theta2)-1.5*cosd(theta3) //kN
+Ry=2*sind(theta1)-5*sind(theta2)-1.5*sind(theta3) //kN
+R=sqrt(Rx^2+Ry^2) //kN
+alpha=atand(-Ry/Rx) //Degree
+//Moment of forces about O is
+MO=2*30*cosd(45)+5*50*sind(theta2)+1.5*10*sind(theta3) //kN-mm
+//distance d of resultant R from O is given as
+d=MO/R
+printf("The resultant of the system is R=%.3f kN as shown in fig. 3.14(b)",R)
+
+
+
diff --git a/3886/CH3/EX3.6/3_6.txt b/3886/CH3/EX3.6/3_6.txt new file mode 100644 index 000000000..6ce70e17b --- /dev/null +++ b/3886/CH3/EX3.6/3_6.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_6.sce', -1)
+The Resultant is R=2671.2 N as shown in fig.3.15
\ No newline at end of file diff --git a/3886/CH3/EX3.6/Ex3_6.sce b/3886/CH3/EX3.6/Ex3_6.sce new file mode 100644 index 000000000..d5ab057b3 --- /dev/null +++ b/3886/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,11 @@ +//Determine magnitude,direction and point of application
+//refer fig.3.15(a)&(b)
+Rx=500*cosd(60)-700 //N (towards left)
+Ry=-500*sind(60)-1000-1200 //N (Downwards)
+R=sqrt((Rx^2)+(Ry^2)) //N
+alpha=atand(-Ry/Rx) //degree
+//taking moment about O
+MO=-500*300*sind(60)-1000*150+1200*150*cosd(60)-700*300*sind(60)
+//let point of application of resultant be at a distance of x from point O along the horizontal then
+x=MO/Ry //mm
+printf("The Resultant is R=%.1f N as shown in fig.3.15",R)
diff --git a/3886/CH3/EX3.7/3_7.txt b/3886/CH3/EX3.7/3_7.txt new file mode 100644 index 000000000..f8815e4c2 --- /dev/null +++ b/3886/CH3/EX3.7/3_7.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_7.sce', -1)
+x=3.467 m lies in the middle third of base.Hence dam is safe
\ No newline at end of file diff --git a/3886/CH3/EX3.7/Ex3_7.sce b/3886/CH3/EX3.7/Ex3_7.sce new file mode 100644 index 000000000..2a1405899 --- /dev/null +++ b/3886/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,10 @@ +//Safety of dam
+//refer fig. 3.16
+Rx=300 //kN (towards right)
+Ry=100-1200-400 //kN (Downwards)
+//taking moment about O
+MO=300*3-100*1+1200*2+400*5
+//assume that the resultant cut the base at a distance of x from O
+x=MO/Ry //m
+printf("x=%.3f m lies in the middle third of base.Hence dam is safe",-x)
+
diff --git a/3886/CH3/EX3.8/3_8.txt b/3886/CH3/EX3.8/3_8.txt new file mode 100644 index 000000000..78f01d7bb --- /dev/null +++ b/3886/CH3/EX3.8/3_8.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_8.sce', -1)
+The resultant is R=580.3 N as shown in fig.3.17 (a)
\ No newline at end of file diff --git a/3886/CH3/EX3.8/Ex3_8.sce b/3886/CH3/EX3.8/Ex3_8.sce new file mode 100644 index 000000000..ea4e58868 --- /dev/null +++ b/3886/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,13 @@ +//determine the resultant
+//refer fig. 3.17 (a) and (b)
+Rx=-400*cosd(45)-150*cosd(30) //N (towards left)
+Ry=200+400*sind(45)-150*sind(30)
+R=sqrt(Rx^2+Ry^2) //N
+alpha=atand(Ry/Rx) //degree
+//assume that the resultant intersects arm AB at a distance of x from A
+//taking moment about A
+MA=-400*3*sind(45)-400*0.6*cosd(45)+50+150*6*sind(30)+150*1*cosd(30) //N-m (anticlockwise)
+x=MA/Ry //m
+printf("The resultant is R=%.1f N as shown in fig.3.17 (a)",R)
+
+
diff --git a/3886/CH3/EX3.9/3_9.txt b/3886/CH3/EX3.9/3_9.txt new file mode 100644 index 000000000..a90e43daf --- /dev/null +++ b/3886/CH3/EX3.9/3_9.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_9.sce', -1)
+Equilibriant is equal and opposite to resultant.
+R=116.52 kN
+alpha=76.82 degree
+x=1.517 m
+As shown in fig.3.18 (a)
\ No newline at end of file diff --git a/3886/CH3/EX3.9/Ex3_9.sce b/3886/CH3/EX3.9/Ex3_9.sce new file mode 100644 index 000000000..7d2158b17 --- /dev/null +++ b/3886/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,13 @@ +//determine equilibriant
+//two 40 kN forces have no moment about the pulley centre hence can be considered acting at pulley centre
+//Accordingly
+Rx=20*cosd(45)-30*cosd(60)-50*cosd(30)+40*cosd(20)-40*sind(30) //kN (towards left)
+Ry=-20*sind(45)-20+20-30*sind(60)-50*sind(30)-40*sind(20)-40*cosd(30) //kN (Downwards)
+R=sqrt(Rx^2+Ry^2) //kN
+alpha=atand(Ry/Rx) //degree
+//Taking moment about A
+MA=20*4-20*4+30*6*sind(60)+50*2*sind(30)-50*2*cosd(30)+40*3*cosd(20)-40*3*sind(30)
+//assume that the resultant intersects AB at a distance x from A,then
+x=MA/Ry //m
+printf("Equilibriant is equal and opposite to resultant.\nR=%.2f kN\nalpha=%.2f degree\nx=%.3f m\nAs shown in fig.3.18 (a)",R,alpha,-x)
+
diff --git a/3886/CH4/EX4.1/4_1.txt b/3886/CH4/EX4.1/4_1.txt new file mode 100644 index 000000000..5ffb41d6b --- /dev/null +++ b/3886/CH4/EX4.1/4_1.txt @@ -0,0 +1,9 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_1.sce', -1)
+The required forces in members are:-
+AB=120.00 kN (Tension)
+BC=56.57 kN (Tension)
+CD=40.00 kN (Compression)
+DE=40.00 kN (Compression)
+BE=113.14 kN (Compression)
+BD=40.00 kN (Tension)
\ No newline at end of file diff --git a/3886/CH4/EX4.1/Ex4_1.sce b/3886/CH4/EX4.1/Ex4_1.sce new file mode 100644 index 000000000..eea95d2a1 --- /dev/null +++ b/3886/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,23 @@ +//finding forces in members of truss
+//Refer fig. 4.8
+//Step 1
+theta=atand(3/3) //Degree
+//Step 2
+//Assume Notations as in fig. 4.8
+//Step 3
+//applying equilibrium conditions
+FCB=40/sind(45) //kN
+FCD=FCB*cosd(45) //kN
+//Step 4
+//Mark and analyse at joint C
+//Step 5
+//Analyse joint D
+FDB=40 //kN
+FDE=40 //kN
+//step 6
+//Analysis of joint B
+FBE=(40+56.57*sind(45))/sind(45) //kN
+FBA=FBE*cosd(45)+56.57*cosd(45) //kN
+//step 7
+//Tabulating answers
+printf("The required forces in members are:-\nAB=%.2f kN (Tension)\nBC=%.2f kN (Tension)\nCD=%.2f kN (Compression)\nDE=%.2f kN (Compression)\nBE=%.2f kN (Compression)\nBD=%.2f kN (Tension)",FBA,FCB,FCD,FDE,FBE,FDB)
diff --git a/3886/CH4/EX4.10/4_10.txt b/3886/CH4/EX4.10/4_10.txt new file mode 100644 index 000000000..ad3bcaa84 --- /dev/null +++ b/3886/CH4/EX4.10/4_10.txt @@ -0,0 +1,9 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_10.sce', -1)
+The required forces are:-
+Force in member AB=120 kN
+Force in member BC=56 kN
+Force in member CD=-39 kN
+Force in member DE=-40 kN
+Force in member EB=-113 kN
+Force in member BD=40 kN
\ No newline at end of file diff --git a/3886/CH4/EX4.10/Ex4_10.sce b/3886/CH4/EX4.10/Ex4_10.sce new file mode 100644 index 000000000..24552c8fe --- /dev/null +++ b/3886/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,33 @@ +//Finding unknown forces
+//Refer fig. 4.8
+//Let us assume joint E as origin,EC as x-axis,EA as y-direction
+//accordingly the co-ordinates are
+//A(0,3),B(3,3),C(6,0),D(3,0),E(0,0)
+YD=-40 //kN
+YC=-40 //kN
+//Using co-ordinates lengths are found out to be
+LAB=3 //m
+LBC=3*sqrt(2) //m
+LCD=3 //m
+LDE=3 //m
+LBD=3 //m
+LBE=3*sqrt(2) //m
+//Consider joint C
+//applying equilibrium conditions
+tCB=40/3
+tCD=-40/3
+FCB=tCB*LBC //kN
+FCD=-13.333*LCD //kN
+//Consider joint D
+//applying equilibrium conditions
+tDE=tCD
+FDE=tCD*LCD //kN
+tDB=40/3
+FDB=tDB*LBD
+//Consider joint B
+//applying equilibrium conditions
+tBE=-(13.333+13.333)
+FBE=tBE*LBE //kN
+tBA=40
+FBA=tBA*LAB
+printf("The required forces are:-\nForce in member AB=%.2d kN\nForce in member BC=%.2d kN\nForce in member CD=%.2d kN\nForce in member DE=%.2d kN\nForce in member EB=%.2d kN\nForce in member BD=%.2d kN",FBA,FCB,FCD,FDE,FBE,FDB)
diff --git a/3886/CH4/EX4.11/4_11.txt b/3886/CH4/EX4.11/4_11.txt new file mode 100644 index 000000000..7ced3b5be --- /dev/null +++ b/3886/CH4/EX4.11/4_11.txt @@ -0,0 +1,10 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_11.sce', -1)
+The forces in different members are:-
+AB=-68 kN
+BC=-45 kN
+CD=-104 kN
+DE=52 kN
+EA=64 kN
+EB=22 kN
+EC=46 kN
\ No newline at end of file diff --git a/3886/CH4/EX4.11/Ex4_11.sce b/3886/CH4/EX4.11/Ex4_11.sce new file mode 100644 index 000000000..0f916a380 --- /dev/null +++ b/3886/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,30 @@ +//Analyse the truss by method of tension coefficient to determine the forces
+//Refer fig.4.19
+//Consider entire structure
+//Taking moment about A
+YD=(40*2+50*6+30*4*sind(60)+60*4)/8
+XA=-30 //kN
+YA=40+50+60-90.49 //kN
+//Take A as origin and determine co-ordinates of all other point
+//Consider equilibrium of individual joints
+//Joint A
+tAB=-(59.51/3.464)
+FAB=tAB*4 //kN
+tAE=64.36/4
+FAE=tAE*4 //kN
+//Joint B
+tBE=-11.547+17.18
+FBE=tBE*4 //kN
+tBC=0.5*(-17.18-5.637)
+FBC=tBC*4 //kN
+//Joint C
+tCD=-(14.434+37.818)/2
+FCD=4*tCD //kN
+tCD=4 //kN
+tCE=-14.434-tCD
+FCE=11.692*4 //kN
+//Joint D
+tDE=-0.5*(-26.126)
+FDE=tDE*4 //kN
+printf("The forces in different members are:-\nAB=%.2d kN\nBC=%.2d kN\nCD=%.2d kN\nDE=%.2d kN\nEA=%.2d kN\nEB=%.2d kN\nEC=%.2d kN",FAB,FBC,FCD,FDE,FAE,FBE,FCE)
+
diff --git a/3886/CH4/EX4.12/4_12.txt b/3886/CH4/EX4.12/4_12.txt new file mode 100644 index 000000000..1a68cffd4 --- /dev/null +++ b/3886/CH4/EX4.12/4_12.txt @@ -0,0 +1,10 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_12.sce', -1)
+The forces in different members are:-
+AB=20 kN
+BC=60 kN
+CD=84 kN
+DE=-60 kN
+EA=-40 kN
+EC=-60 kN
+EB=-28 kN
\ No newline at end of file diff --git a/3886/CH4/EX4.12/Ex4_12.sce b/3886/CH4/EX4.12/Ex4_12.sce new file mode 100644 index 000000000..ab6e62458 --- /dev/null +++ b/3886/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,38 @@ +//Analyse the truss
+//Refer fig. 4.20
+//Consider equilibrium of entire truss
+//taking moment about A
+YE=(60*8-40*4)/4 //kN
+XA=40 //kN
+YA=60-80 //kN
+//Take A as origin and determine co-ordinates of various point
+//Lengths in m are
+AB=4
+CE=4
+AE=4
+ED=4
+BE=4*sqrt(2)
+CD=4*sqrt(2)
+//Consider equilibrium of joints individually
+//Joint A
+tAB=5
+FAB=tAB*AB //kN
+tAE=-10
+FAE=tAE*AE //kN
+//Joint B
+tBE=-5
+FBE=tBE*BE //kN
+tBC=10-tBE
+BC=4
+FBC=tBC*BC //kN
+//Joint C
+tCD=15
+CD=4*sqrt(2)
+FCD=15*4*sqrt(2) //kN The answer provided in the textbook is wrong
+tCE=-15
+FCE=tCE*CE //kN
+//Joint D
+tDE=-15
+DE=4
+FDE=tDE*DE //kN
+printf("The forces in different members are:-\nAB=%.2d kN\nBC=%.2d kN\nCD=%.2d kN\nDE=%.2d kN\nEA=%.2d kN\nEC=%.2d kN\nEB=%.2d kN",FAB,FBC,FCD,FDE,FAE,FCE,FBE)
diff --git a/3886/CH4/EX4.2/4_2.txt b/3886/CH4/EX4.2/4_2.txt new file mode 100644 index 000000000..1100421fd --- /dev/null +++ b/3886/CH4/EX4.2/4_2.txt @@ -0,0 +1,11 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_2.sce', -1)
+RD=77.50 kN
+RA=72.50 kN
+FAB=83.72 kN (Compression)
+FAE=41.86 kN (Tension)
+FDC=89.49 kN (Compression)
+FDE=44.74 kN (Tension)
+FBE=37.53 kN (Tension)
+FBC=60.62 kN (Compression)
+FCE=31.75 kN (Tension)
\ No newline at end of file diff --git a/3886/CH4/EX4.2/Ex4_2.sce b/3886/CH4/EX4.2/Ex4_2.sce new file mode 100644 index 000000000..519bc8e53 --- /dev/null +++ b/3886/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,23 @@ +//finding forces in members of truss
+//Refer fig. 4.9
+//Find support reactions
+//applying equilibrium conditions
+//Taking moment about A
+RD=(40*1+60*2+50*3)/4 //kN
+RA=150-RD //kN
+//Consider equilibrium of joint A
+FAB=RA/sind(60) //kN (Compression)
+FAE=FAB*cosd(60) //kN (Tension)
+//Joint D
+FDC=RD/sind(60) //kN (Compression)
+FDE=FDC*cosd(60) //kN (Tension)
+//Joint B (Refer Fig. 4.9 (d)
+FBE=((FAB*sind(60))-40)/sind(60) //kN (Tension)
+FBC=FAB*cosd(60)+FBE*cosd(60) //kN (Compression)
+//Joint C (Refer fig. 4.9 (e))
+FCE=(FDC*sind(60)-50)/sind(60) //kN (Tension)
+//Refer fig. 4.9 (e),(f)
+printf("RD=%.2f kN\nRA=%.2f kN\nFAB=%.2f kN (Compression)\nFAE=%.2f kN (Tension)\nFDC=%.2f kN (Compression)\nFDE=%.2f kN (Tension)\nFBE=%.2f kN (Tension)\nFBC=%.2f kN (Compression)\nFCE=%.2f kN (Tension)",RD,RA,FAB,FAE,FDC,FDE,FBE,FBC,FCE)
+
+
+
diff --git a/3886/CH4/EX4.3/4_3.txt b/3886/CH4/EX4.3/4_3.txt new file mode 100644 index 000000000..4912b7aa7 --- /dev/null +++ b/3886/CH4/EX4.3/4_3.txt @@ -0,0 +1,15 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_3.sce', -1)
+FED=25.00 kN (Tension)
+FEF=15.00 kN (Compression)
+RC=15.00 kN
+VA=20.00 kN
+HA=15.00 kN
+FAB=20.00 kN (Compression)
+FAF=15.00 kN (Compression)
+FCB=25.00 kN (Compression)
+FCD=20.00 kN (Tension)
+FBF=0.00 kN
+FBD=15.00 kN (Tension)
+FFD=0.00 kN
+FBF=0.00 kN
\ No newline at end of file diff --git a/3886/CH4/EX4.3/Ex4_3.sce b/3886/CH4/EX4.3/Ex4_3.sce new file mode 100644 index 000000000..0d0d41764 --- /dev/null +++ b/3886/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,32 @@ +//Analysing the truss
+//Refer fig. 4.10(a)
+//inclined members make an angle theta with the horizontal
+theta=atand(4/3) //Degree
+//Joint E
+//Refer fig. 4.10 (c)
+//applying equilibrium conditions
+FED=20/sind(theta) //kN (Tension)
+FEF=25*cosd(theta) //kN (Compression)
+//Refer fig 4.10 (b)
+//Taking moment about A
+RC=20*6/8 //kN
+VA=20 //kN
+HA=RC //kN
+//Joint A
+//Refer fig.4.10 (d)
+//applying equilibrium conditions
+FAB=VA //kN (Compression)
+FAF=HA //kN (Compression)
+//Joint C
+//Refer fig. 4.10 (E)
+FCB=RC/cosd(theta) //kN (Compression)
+FCD=FCB*sind(theta) //kN (Tension)
+//Joint B
+//Refer fig. 4.10 (f)
+FBF=(FBC*sind(theta)-FAB)/sind(theta) //kN
+FBD=0+25*cosd(theta) //kN (Tension)
+//Joint F
+//Refer Fig. 4.10(g)
+FFD=0
+FBF=0
+printf("FED=%.2f kN (Tension)\nFEF=%.2f kN (Compression)\nRC=%.2f kN\nVA=%.2f kN\nHA=%.2f kN\nFAB=%.2f kN (Compression)\nFAF=%.2f kN (Compression)\nFCB=%.2f kN (Compression)\nFCD=%.2f kN (Tension)\nFBF=%.2f kN\nFBD=%.2f kN (Tension)\nFFD=%.2f kN\nFBF=%.2f kN",FED,FEF,RC,VA,HA,FAB,FAF,FCB,FCD,FBF,FBD,FFD,FBF)
diff --git a/3886/CH4/EX4.4/4_4.txt b/3886/CH4/EX4.4/4_4.txt new file mode 100644 index 000000000..b432b7c96 --- /dev/null +++ b/3886/CH4/EX4.4/4_4.txt @@ -0,0 +1,17 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_4.sce', -1)
+Required values are:-
+FHG=25.00 kN (Compression)
+FHF=15.00 kN (Tension)
+RG=42.00 kN
+VA=10.00 kN (Downwards)
+HA=0.00 kN
+FAC=18.03 kN (Compression)
+FAB=15.00 kN (Tension)
+FBC=0.00 kN
+FCE=18.03 kN (Compression)
+FDE=0.00 kN
+FDF=15.00 kN (Tension)
+FEF=0.00 kN
+FEG=18.03 kN (Compression)
+FAG=12.00 kN (Compression)
\ No newline at end of file diff --git a/3886/CH4/EX4.4/Ex4_4.sce b/3886/CH4/EX4.4/Ex4_4.sce new file mode 100644 index 000000000..92223721f --- /dev/null +++ b/3886/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,29 @@ +//Finding forces in all members
+//Refer fig. 4.11(a)
+theta1=atand(4/6) //Degree
+theta2=atand(8/6) //Degree
+theta3=atand(4/3) //Degree
+//Joint H
+FHG=20/sind(53.13) //kN (Compression)
+FHF=25*cosd(53.13) //kN (Tension)
+//Taking moment about A
+RG=(20*9+12*6)/6 //kN
+VA=32-42 //kN (Downwards)
+HA=0
+//Joint A
+//applying equilibrium conditions
+FAC=10/sind(33.69) //kN (Compression)
+FAB=18.03*cosd(33.69) //kN (Tension)
+//Joint B
+FBC=0
+FCE=FAC //kN (Compression)
+//Joint D
+FDE=0
+FDF=FBD //kN (Tension)
+//Joint E
+FEF=0
+FEG=FCE //kN (Compression)
+//Joint F
+FAG=12 //kN (Compression)
+printf("Required values are:-\nFHG=%.2f kN (Compression)\nFHF=%.2f kN (Tension)\nRG=%.2f kN\nVA=%.2f kN (Downwards)\nHA=%.2f kN\nFAC=%.2f kN (Compression)\nFAB=%.2f kN (Tension)\nFBC=%.2f kN\nFCE=%.2f kN (Compression)\nFDE=%.2f kN\nFDF=%.2f kN (Tension)\nFEF=%.2f kN\nFEG=%.2f kN (Compression)\nFAG=%.2f kN (Compression)",FHG,FHF,RG,-VA,HA,FAC,FAB,FBC,FCE,FDE,FDF,FEF,FEG,FAG)
+
diff --git a/3886/CH4/EX4.5/4_5.txt b/3886/CH4/EX4.5/4_5.txt new file mode 100644 index 000000000..8b98d9c1f --- /dev/null +++ b/3886/CH4/EX4.5/4_5.txt @@ -0,0 +1,18 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_5.sce', -1)
+
+Required Forces are:-
+FGF=23.09 kN (Tension)
+FGE=11.55 kN (Compression)
+FFE=23.09 kN (Compression)
+FFD=13.09 kN (Tension)
+RE=58.17 kN
+VA=31.83 kN
+HA=10.00 kN
+FAB=36.75 kN (Compression)
+FAC=8.37 (Tension)
+FBC=9.43 kN (Compression)
+FBD=13.65 kN (Compression)
+FCD=9.43 kN (Tension)
+FCE=1.06 kN (Compression)
+FDE=44.07 kN (Compression)
\ No newline at end of file diff --git a/3886/CH4/EX4.5/Ex4_5.sce b/3886/CH4/EX4.5/Ex4_5.sce new file mode 100644 index 000000000..bca4a3ebf --- /dev/null +++ b/3886/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,28 @@ +//Analyse truss
+//Refer fig. 4.12 (a)
+//All triangles are equilateral
+//applying equilibrium conditions At
+//Joint G
+FGF=20/sind(60) //kN (Tension)
+FGE=FGF*cosd(60) //kN (Compression)
+//Joint F
+FFE=FGF //kN (Compression)
+FFD=FGF*cosd(60)+FFE*cosd(60)-10 //kN (Tension)
+//Consider eqiulibrium of entire truss
+RE=(-10*3*sind(60)+40*3*cosd(60)+30*(3+3*cosd(60))+20*9)/6 //kN
+VA=(40+30+20)-58.17 //kN
+HA=10 //kN
+//Joint A
+FAB=31.83/sind(60) //kN (Compression)
+FAC=36.75*cosd(60)-10 //kN (Tension)
+//Joint B
+FBC=(40-FAB*sind(60))/sind(60) //kN (Compression)
+FBD=36.75*cosd(60)-9.44*cosd(60) //kN (Compression)
+//Joint C
+FCD=FBC //kN (Tension)
+FCE=9.44*cosd(60)+9.44*cosd(60)-8.38 //kN (Compression)
+//Joint D
+FDE=(30+FCD*sind(60))/sind(60) //kN (Compression)
+printf("\nRequired Forces are:-\nFGF=%.2f kN (Tension)\nFGE=%.2f kN (Compression)\nFFE=%.2f kN (Compression)\nFFD=%.2f kN (Tension)\nRE=%.2f kN\nVA=%.2f kN\nHA=%.2f kN\nFAB=%.2f kN (Compression)\nFAC=%.2f (Tension)\nFBC=%.2f kN (Compression)\nFBD=%.2f kN (Compression)\nFCD=%.2f kN (Tension)\nFCE=%.2f kN (Compression)\nFDE=%.2f kN (Compression)",FGF,FGE,FFE,FFD,RE,VA,HA,FAB,FAC,FBC,FBD,FCD,FCE,FDE)
+
+
diff --git a/3886/CH4/EX4.6/4_6.txt b/3886/CH4/EX4.6/4_6.txt new file mode 100644 index 000000000..8f49c82ed --- /dev/null +++ b/3886/CH4/EX4.6/4_6.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_6.sce', -1)
+The required values are:-
+FFH=69 kN (Compressive)
+FGH=05 kN (Compressive)
+FGI=72 kN (Tension)
\ No newline at end of file diff --git a/3886/CH4/EX4.6/Ex4_6.sce b/3886/CH4/EX4.6/Ex4_6.sce new file mode 100644 index 000000000..273c55e46 --- /dev/null +++ b/3886/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,11 @@ +//Determine The Forces in the member
+//Using symmetry
+RA=70/2 //kN
+RB=RA //kN
+//Consider fig.4.13 (b)
+//Taking moment about G
+FFH=(35*12-10*10-10*6-10*2)/(4*sind(60)) //kN (Compression)
+FGH=(35-10-10-10)/sind(60) //kN (Compression)
+FGI=69.28+5.77*cosd(60) //kN (Tension)
+printf("The required values are:-\nFFH=%.2d kN (Compressive)\nFGH=%.2d kN (Compressive)\nFGI=%.2d kN (Tension)",FFH,FGH,FGI)
+
diff --git a/3886/CH4/EX4.7/4_7.txt b/3886/CH4/EX4.7/4_7.txt new file mode 100644 index 000000000..3f5b52fdd --- /dev/null +++ b/3886/CH4/EX4.7/4_7.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_7.sce', -1)
+The required forces are:-
+Member Force
+U3U4= 456 kN (Compression)
+L3L4= 412 kN (Tension)
+U4L3= 62 kN (Tension)
\ No newline at end of file diff --git a/3886/CH4/EX4.7/Ex4_7.sce b/3886/CH4/EX4.7/Ex4_7.sce new file mode 100644 index 000000000..3d3848461 --- /dev/null +++ b/3886/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,30 @@ +//finding magnitude and nature of forces
+//refer fig. 4.14(a)
+//considering equilibrium if entire truss
+//taking moment about L0
+R2=(200*6+200*12+150*18+100*24+100*30)/36 //kN
+R1=200+200+150+100+100-R2 //kN
+//consider equilibrium of right hand side of section (1)-(1)
+theta1=atand(1/6) //degree
+theta2=atand(6/8) //degree
+//taking moment about U4
+FL3L4=(-100*6+325*12)/8 //kN (tension)
+//taking moment about L3
+FU3U4=456.2 //kN (compression)
+FU4L3=(456.2*cosd(9.46)-412.5)/sind(36.87) //kN (tension)
+printf("The required forces are:-\nMember Force\nU3U4= %.2d kN (Compression)\nL3L4= %.2d kN (Tension)\nU4L3= %.2d kN (Tension)",FU3U4,FL3L4,FU4L3)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3886/CH4/EX4.8/4_8.txt b/3886/CH4/EX4.8/4_8.txt new file mode 100644 index 000000000..b9bca3950 --- /dev/null +++ b/3886/CH4/EX4.8/4_8.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_8.sce', -1)
+The required forces are:-
+F1=109 kN (Compression)
+F2=51 kN (Tension)
+F3=69 kN (Tension)
\ No newline at end of file diff --git a/3886/CH4/EX4.8/Ex4_8.sce b/3886/CH4/EX4.8/Ex4_8.sce new file mode 100644 index 000000000..47b6ff003 --- /dev/null +++ b/3886/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,15 @@ +//Finding forces in members
+//Refer fig. 4.15 (a)
+RA=7*20/2 //kN
+RB=RA //kN (symmetry)
+//CE is perpendicular on AB
+CE=5.196 //m
+DE=3 //m
+theta=atand(5.196/3) //degree
+//The fact that 20 kN loads are equidistant can be used to find out horizontal distances of loads from A
+//Consider equilibrium of left hand side portion of section (1)-(1)
+//taking moment about A
+F2=(20*2.25+20*4.5+20*6.75)/(6*sind(60)) //kN (Tension)
+F1=(70-20-20-20+51.96*sind(60))/sind(30) //kN (compression)
+F3=-51.96*cosd(60)+110*cosd(30) //kN (Tension)
+printf("The required forces are:-\nF1=%.2d kN (Compression)\nF2=%.2d kN (Tension)\nF3=%.2d kN (Tension)",F1,F2,F3)
diff --git a/3886/CH4/EX4.9/4_9.txt b/3886/CH4/EX4.9/4_9.txt new file mode 100644 index 000000000..a09c3d4f2 --- /dev/null +++ b/3886/CH4/EX4.9/4_9.txt @@ -0,0 +1,10 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_9.sce', -1)
+
+The required forces are:-
+FAB=92 kN (Compression)
+FBC=71 kN (Compression)
+FBF=21 kN (Compression)
+FAF=40 kN (Tension)
+FFC=40 kN (Tension)
+FAE=30 kN (Tension)
\ No newline at end of file diff --git a/3886/CH4/EX4.9/Ex4_9.sce b/3886/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..59a686b2d --- /dev/null +++ b/3886/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,21 @@ +//Finding required forces
+//Refer fig. 4.16 (a)
+//Symmetry gives
+RE=(15+30+30+30+15)/2 //kN
+RA=RE //kN
+FAE=30 //kN
+//After construction as shown in ref. fig
+//Taking moment about C
+FAE=(60*5-15*5-30*2.5)/5 //kN (Tension)
+//assumptions are made as shown in fig. 4.16 (b)
+//Apply equilibrium conditions and solving equations
+FFC=15/0.366 //kN (Tension)
+FBC=(0.866*40.98+15)/0.707 //kN (Compression)
+//Lets analyse Joint B
+//Applying equilibrium conditions
+FBF=30*cosd(45) //kN (Compression)
+FAB=71.41+21.21 //kN (Compression)
+//Lets analyse Joint A
+//Applying equilibrium conditions
+FAF=(92.62*sind(45)-45)/sind(30) //kN (Tension)
+printf("\nThe required forces are:-\nFAB=%.2d kN (Compression)\nFBC=%.2d kN (Compression)\nFBF=%.2d kN (Compression)\nFAF=%.2d kN (Tension)\nFFC=%.2d kN (Tension)\nFAE=%.2d kN (Tension)",FAB,FBC,FBF,FAF,FFC,FAE)
diff --git a/3886/CH5/EX5.1/5_1.sce b/3886/CH5/EX5.1/5_1.sce new file mode 100644 index 000000000..969095d3d --- /dev/null +++ b/3886/CH5/EX5.1/5_1.sce @@ -0,0 +1,19 @@ +//Value of P
+//Refer fig. 5.5 (a),(b)&(c)
+//(a) when P is Horizontal Phor
+//Consider equilibrium
+//block A
+N1=1000 //N
+F1=0.25*N1 //N
+T=F1 //N
+//Block B
+N2=N1+2000 //N
+F2=3000/3 //mu*N2 N
+Phor=F1+F2 //N
+//(b) when P is inclined (Pinc)
+//Considering equilibrium of block B
+//Using law of friction
+//Pinc*cosd(30)-F1-F2=0
+Pinc=1250/(cosd(30)+(0.5/3)) //N
+printf("\nPhor=%0.2d N\nPinc=%0.2d N",Phor,Pinc)
+
diff --git a/3886/CH5/EX5.1/5_1.txt b/3886/CH5/EX5.1/5_1.txt new file mode 100644 index 000000000..ca51c0e9e --- /dev/null +++ b/3886/CH5/EX5.1/5_1.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\5. Friction\5.1.sce', -1)
+
+Phor=1250 N
+Pinc=1210 N
\ No newline at end of file diff --git a/3886/CH5/EX5.10/5_10.sce b/3886/CH5/EX5.10/5_10.sce new file mode 100644 index 000000000..c1799536d --- /dev/null +++ b/3886/CH5/EX5.10/5_10.sce @@ -0,0 +1,12 @@ +//Value of force P
+//refer fig. 5.14
+mu=0.25
+//Let fi be the angle of limiting friction
+fi=atand(0.25) //degree
+ //Consider equilibrium of block C
+//apply Lami's theorem
+R1=160*sind(180-16-fi)/sind(2*(fi+16)) //kN
+//Consider equilibrium of Wedge A
+//apply Lami's theorem
+P=R1*sind(180-fi-fi-16)/sind(90+fi) //kN
+printf("The required value is P=%0.3f kN",P)
diff --git a/3886/CH5/EX5.10/5_10.txt b/3886/CH5/EX5.10/5_10.txt new file mode 100644 index 000000000..1736feac5 --- /dev/null +++ b/3886/CH5/EX5.10/5_10.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\5. Friction\5.10.sce', -1)
+The required value is P=66.256 kN
\ No newline at end of file diff --git a/3886/CH5/EX5.11/5_11.sce b/3886/CH5/EX5.11/5_11.sce new file mode 100644 index 000000000..5c133054f --- /dev/null +++ b/3886/CH5/EX5.11/5_11.sce @@ -0,0 +1,14 @@ +//Minimum horizontal force required to avoid slipping
+//refer fig.5.15
+//consider equilibrium
+//taking moment about A
+//0.866*NB+0.5*FB=275
+//Law of friction
+//FB=0.2*NB
+//Thus
+ NB=275/(0.866+0.5*0.2) //N
+NA=200+600-56.934 //N
+FA=0.3*NA //N
+P=NB-FA //N
+printf("The required force is P=%0.2d N",P)
+
diff --git a/3886/CH5/EX5.11/5_11.txt b/3886/CH5/EX5.11/5_11.txt new file mode 100644 index 000000000..f98b2071d --- /dev/null +++ b/3886/CH5/EX5.11/5_11.txt @@ -0,0 +1,2 @@ + Minimum horizontal force to prevent slipping:-
+P=61.76 N.
\ No newline at end of file diff --git a/3886/CH5/EX5.12/5_12.sce b/3886/CH5/EX5.12/5_12.sce new file mode 100644 index 000000000..3452ee8a9 --- /dev/null +++ b/3886/CH5/EX5.12/5_12.sce @@ -0,0 +1,15 @@ +//Least value of alpha and reactions developed
+//refer fig. 5.16
+//Using law of friction and equilibrium
+//FA=0.25*NA
+//FB=0.4*NB
+//NA+0.4*NB=1100
+//0.25*NA=NB
+//Solving this we get
+NA=1000 //N
+FA=0.25*NA //N
+NB=0.25*NA //N
+FB=0.4*250 //N
+//Taking moment about A
+alpha=atand(3) //degree
+printf("\nNA=%0.2f N\nFA=%0.2f N\nNB=%0.2f N\nFB=%0.2f N\nalpha=%0.2f degree",NA,FA,NB,FB,alpha)
diff --git a/3886/CH5/EX5.12/5_12.txt b/3886/CH5/EX5.12/5_12.txt new file mode 100644 index 000000000..9df4c0b3d --- /dev/null +++ b/3886/CH5/EX5.12/5_12.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\5. Friction\5.12.sce', -1)
+
+NA=1000.00 N
+FA=250.00 N
+NB=250.00 N
+FB=100.00 N
+alpha=71.57 degree
\ No newline at end of file diff --git a/3886/CH5/EX5.15/5_15.sce b/3886/CH5/EX5.15/5_15.sce new file mode 100644 index 000000000..7e544bcd2 --- /dev/null +++ b/3886/CH5/EX5.15/5_15.sce @@ -0,0 +1,25 @@ +//minimum weight W to prevent downward motion of 1000N block
+mu1=0.2
+mu2=0.3
+//Refer fig. 5.20
+alpha=atand(3/4) //degree
+//considering equilibrium of block W
+//N1=W*cosd(alpha)
+//F1=mu2*N1
+//T1=0.84*W
+theta=180 //degree
+//Friction equation of rope gives
+//T2=T1*%e^(mu2*theta)
+//solving
+//T2=2.156*W
+//Consider equilibrium of 1000 N block
+//N2-N1=800
+//N2=0.8*W+800
+//F2=0.3*N2
+//F1+F2+T2-1000*sind(alpha)=0
+//solving we get
+W=(1000*sind(alpha)-240)/(0.24+0.24+2.156) //N
+printf("\nRequired force is W=%0.2f N",W)
+
+
+
diff --git a/3886/CH5/EX5.15/5_15.txt b/3886/CH5/EX5.15/5_15.txt new file mode 100644 index 000000000..1eca551dc --- /dev/null +++ b/3886/CH5/EX5.15/5_15.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\5. Friction\5.15.sce', -1)
+
+Required force is W=136.57 N
\ No newline at end of file diff --git a/3886/CH5/EX5.16/5_16.sce b/3886/CH5/EX5.16/5_16.sce new file mode 100644 index 000000000..524a62433 --- /dev/null +++ b/3886/CH5/EX5.16/5_16.sce @@ -0,0 +1,17 @@ +//Determine force P
+//refer fig.5.21
+//From FBD
+theta=250*%pi/180 //radians
+r=250 //mm
+mu=0.3
+//from rope friction equation
+//T2=T1*%e^(mu*theta)
+//also (T2-T2)*r=M
+//solving we get
+T1=(300*1000)/(250*(3.7025-1)) //N
+T2=3.7025*T1 //N
+//Consider the equilibrium of lever arm,
+P=(T2*50)/300 //N
+printf("\n The required force is P=%0.1f N",P)
+
+
diff --git a/3886/CH5/EX5.16/5_16.txt b/3886/CH5/EX5.16/5_16.txt new file mode 100644 index 000000000..16bb4f0fd --- /dev/null +++ b/3886/CH5/EX5.16/5_16.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\5. Friction\5.16.sce', -1)
+
+ The required force is P=274.0 N
\ No newline at end of file diff --git a/3886/CH5/EX5.2/5_2.sce b/3886/CH5/EX5.2/5_2.sce new file mode 100644 index 000000000..11ef18bb6 --- /dev/null +++ b/3886/CH5/EX5.2/5_2.sce @@ -0,0 +1,13 @@ +//Value of theta
+//refer fig.5.6
+//Consider equilibrium
+//300N block
+//N1=300*cosd(theta)
+//Law of friction
+//F1=100*cosd(theta)
+//consider equilibrium of 900 N block
+//N2=1200*cosd(theta)
+//Law of friction
+//F2=400*cosd(theta)
+theta=atand(5/9) //degree
+printf("Required value is\ntheta=%0.2d degree",theta)
diff --git a/3886/CH5/EX5.2/5_2.txt b/3886/CH5/EX5.2/5_2.txt new file mode 100644 index 000000000..9720a2b67 --- /dev/null +++ b/3886/CH5/EX5.2/5_2.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\5. Friction\5.2.sce', -1)
+Required value is
+theta=29 degree
\ No newline at end of file diff --git a/3886/CH5/EX5.3/5_3.sce b/3886/CH5/EX5.3/5_3.sce new file mode 100644 index 000000000..e28e6232f --- /dev/null +++ b/3886/CH5/EX5.3/5_3.sce @@ -0,0 +1,18 @@ +//finding the inclination of the plane and coefficient of friction
+//refer fig 5.7
+//consider equilibrium of system
+//Case (a)
+//N=500*cosd(theta)
+//Using law of friction
+//F1=mu*N
+//500*sind(theta)-500*mu*cosd(theta)=200
+//Case (b)
+//N=500*cosd(theta)
+//usin law of friction
+//F2=mu*N
+//500*mu*cosd(theta)+500*sind(theta)=300
+//add final equations from both cases
+theta=asind(0.5) //degree
+//substitute this value in final equation from case (b)
+mu=(50)/(500*cosd(30))
+printf("\ntheta=%.2d degree\nmu=%0.3f",theta,mu)
diff --git a/3886/CH5/EX5.3/5_3.txt b/3886/CH5/EX5.3/5_3.txt new file mode 100644 index 000000000..b220909ff --- /dev/null +++ b/3886/CH5/EX5.3/5_3.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\5. Friction\5.3.sce', -1)
+
+theta=30 degree
+mu=0.115
\ No newline at end of file diff --git a/3886/CH5/EX5.4/5_4.sce b/3886/CH5/EX5.4/5_4.sce new file mode 100644 index 000000000..7aa892948 --- /dev/null +++ b/3886/CH5/EX5.4/5_4.sce @@ -0,0 +1,15 @@ +//Value of P
+//Refer fig.5.8
+//consider equilibrium
+mu=0.2
+//750N block
+N1=750*cosd(60) //N
+F1=mu*N1 //N
+T=F1+750*sind(60) //N
+//500N block
+//N2=500-0.5P
+//Law of friction
+//F2=0.2*N2
+P=(724.52+100)/(cosd(30)+0.1) //N
+printf("\nP=%0.2f N",P)
+
diff --git a/3886/CH5/EX5.4/5_4.txt b/3886/CH5/EX5.4/5_4.txt new file mode 100644 index 000000000..66257b50e --- /dev/null +++ b/3886/CH5/EX5.4/5_4.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\5. Friction\5.4.sce', -1)
+
+P=853.52 N
\ No newline at end of file diff --git a/3886/CH5/EX5.5/5_5.sce b/3886/CH5/EX5.5/5_5.sce new file mode 100644 index 000000000..5251a0477 --- /dev/null +++ b/3886/CH5/EX5.5/5_5.sce @@ -0,0 +1,17 @@ +//Smallest weight W
+//Refer fig. 5.9
+mu=0.4
+//consider equilibrium of block B
+//using law of friction
+N1=5/((0.5)+(tand(20))*(sind(20))) //kN
+F1=N1*tand(20)
+C=N1*cosd(30)-F1*cosd(60) //kN
+//Consider the equilibrium of block A
+F2=C //kN
+//Law of friction
+N2=4.196/0.4 //kN
+W=N2 //kN
+printf("\nW=%0.2f kN",W)
+
+
+
diff --git a/3886/CH5/EX5.5/5_5.txt b/3886/CH5/EX5.5/5_5.txt new file mode 100644 index 000000000..facd4952d --- /dev/null +++ b/3886/CH5/EX5.5/5_5.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\5. Friction\5.5.sce', -1)
+
+W=10.49 kN
\ No newline at end of file diff --git a/3886/CH5/EX5.6/5_6.sce b/3886/CH5/EX5.6/5_6.sce new file mode 100644 index 000000000..d1d254d87 --- /dev/null +++ b/3886/CH5/EX5.6/5_6.sce @@ -0,0 +1,19 @@ +//force required to prevent slipping
+//refer fig. 5.10
+mu=0.25
+//assumptions are made and shown in fig.5.10
+//F1=mu*N1
+//consider equilibrium of block A
+C1=(2000)/((0.25*cosd(30))+(0.5)) //N
+N1=C1*cosd(30)
+//Lami's theorem at joint O gives
+P=(C1*sind(90))/sind(120) //N
+C=(C1*sind(150))/sind(120) //N
+//Consider equilibrium of block B for verification
+//F2=C2*cosd(60) N
+//N2=2000+C2*sind(60) N
+//LF=mu*N2 N (limiting friction)
+//actual frictional force F2 developed is less than the limiting frictional force hence block B is stationary
+//P is the correct answer
+printf("Requiref force is\nP=%0.2f N",P)
+
diff --git a/3886/CH5/EX5.6/5_6.txt b/3886/CH5/EX5.6/5_6.txt new file mode 100644 index 000000000..ce1261659 --- /dev/null +++ b/3886/CH5/EX5.6/5_6.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\5. Friction\5.6.sce', -1)
+Requiref force is
+P=3223.14 N
\ No newline at end of file diff --git a/3886/CH5/EX5.7/5_7.sce b/3886/CH5/EX5.7/5_7.sce new file mode 100644 index 000000000..09076f584 --- /dev/null +++ b/3886/CH5/EX5.7/5_7.sce @@ -0,0 +1,21 @@ +//Least and the greatest value of W for equilibrium
+//refer fig.
+//case (a) for least value
+refer fig. (b)
+//considering equilibrium of 1000 N block
+N1=1000*cosd(30) //N
+//law of friction gives
+F1=0.28*N1 //N
+T=500-242.49 //N
+//consider equilibrium of W
+//F2=0.1W
+Wmin=257.51/(0.1+sind(60)) //N
+//case (b) for greatest value
+//refer fig. 5.11 (c)
+//consider equilibrium of 1000N block
+T=1000*sind(30)+242.49 //N
+//consider equilibrium of W
+//F2=0.2*0.5*W
+Wmax=742.49/(0.1+sind(60)) //N
+printf("\nThe greatest and least values of W are:-\nWmax=%0.2d N\nWmin=%0.2d N",Wmax,Wmin)
+
diff --git a/3886/CH5/EX5.7/5_7.txt b/3886/CH5/EX5.7/5_7.txt new file mode 100644 index 000000000..0f8e90b4b --- /dev/null +++ b/3886/CH5/EX5.7/5_7.txt @@ -0,0 +1,5 @@ +
+ The least value of W for equilibrium:-
+ W(l)=266.57 N.
+ The greatest value of W for equilibrium:-
+ W(g)=768.60 N.
\ No newline at end of file diff --git a/3886/CH5/EX5.8/5_8.sce b/3886/CH5/EX5.8/5_8.sce new file mode 100644 index 000000000..652e6f9eb --- /dev/null +++ b/3886/CH5/EX5.8/5_8.sce @@ -0,0 +1,17 @@ +//Force P for impending motion
+//Refer fig. 5.12
+//consider equilibrium of block A
+//NA*cosd(30)+FA*sind(30)-1500-500=0
+//Law of friction gives
+NA=2000 //N
+FA=NA*tand(15) //N
+C=NA*sind(30)-FA*cosd(30) //N
+//consider equilibrium of block B
+NB=2000*cosd(60)+535.90*cosd(30) //N
+FB=NB*tand(15) //N
+P=(392.30)+(2000*sind(60))-(535.90*sind(30)) //N
+printf("The required force is P=%0.2f N",P)
+
+
+
+
diff --git a/3886/CH5/EX5.8/5_8.txt b/3886/CH5/EX5.8/5_8.txt new file mode 100644 index 000000000..b3a8d6afb --- /dev/null +++ b/3886/CH5/EX5.8/5_8.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\5. Friction\5.8.sce', -1)
+The required force is P=1856.40 N
\ No newline at end of file diff --git a/3886/CH5/EX5.9/5_9.sce b/3886/CH5/EX5.9/5_9.sce new file mode 100644 index 000000000..18c943686 --- /dev/null +++ b/3886/CH5/EX5.9/5_9.sce @@ -0,0 +1,8 @@ +//Minimum force required
+//refer fig. 5.13
+//Applying Lami's theorem to system of forces on block
+R1=20*sind(145)/sind(140) //kN
+R2=20*sind(75)/sind(140) //kN
+//Applying Lami's theorem to system of forces on wedge
+P=R2*sind(130)/sind(105) //kN
+printf("required force is P=%0.2f kN",P)
diff --git a/3886/CH5/EX5.9/5_9.txt b/3886/CH5/EX5.9/5_9.txt new file mode 100644 index 000000000..52876b512 --- /dev/null +++ b/3886/CH5/EX5.9/5_9.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\5. Friction\5.9.sce', -1)
+required force is P=23.84 kN
\ No newline at end of file diff --git a/3886/CH6/EX6.1/6_1.sce b/3886/CH6/EX6.1/6_1.sce new file mode 100644 index 000000000..b7c7723be --- /dev/null +++ b/3886/CH6/EX6.1/6_1.sce @@ -0,0 +1,30 @@ +//Determine velocity ratio,mechanical advantage,efficiency,ideal effort,effort lost in friction,ideal load and frictional resistance.
+ W=10000 //N
+P=500 //N
+//distance moved by effort
+D=20 //m
+d=0.8 //m
+//mechanical advantage
+MA=W/P
+printf("\nMechanical advantage=%0.1d",MA)
+//Velocity ratio
+VR=D/d
+printf("\nVelocity ratio=%0.1d",VR)
+//efficiency
+e=MA*100/VR //percent
+printf("\nEfficiency=%0.1d percent",e)
+//ideal effort
+Pi=W/VR //N
+printf("\nIdeal effort=%0.1d N",Pi)
+//Effort lost in friction
+E=P-Pi //N
+printf("\nEffort lost in friction=%0.1d N",E)
+//Ideal load
+Wi=P*VR //N
+printf("\nIdeal load=%0.1d N",Wi)
+//Frictional resistance
+F=Wi-W //N
+printf("\nFrictional resistance=%0.1d N",F)
+
+
+
diff --git a/3886/CH6/EX6.1/6_1.txt b/3886/CH6/EX6.1/6_1.txt new file mode 100644 index 000000000..ff6ae3971 --- /dev/null +++ b/3886/CH6/EX6.1/6_1.txt @@ -0,0 +1,10 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.1.sce', -1)
+
+Mechanical advantage=20
+Velocity ratio=25
+Efficiency=80 percent
+Ideal effort=400 N
+Effort lost in friction=100 N
+Ideal load=12500 N
+Frictional resistance=2500 N
\ No newline at end of file diff --git a/3886/CH6/EX6.10/6_10.sce b/3886/CH6/EX6.10/6_10.sce new file mode 100644 index 000000000..ec227cc59 --- /dev/null +++ b/3886/CH6/EX6.10/6_10.sce @@ -0,0 +1,10 @@ +//Force P required to raise the load
+eta=0.70
+W=2500 //N
+//refer fig. 6.17
+//For third order pulley
+//VR=2^2-1
+//For whole system
+VR=3+3
+P=W/(eta*VR) //N
+printf("Required force p=%.2f N",P)
diff --git a/3886/CH6/EX6.10/6_10.txt b/3886/CH6/EX6.10/6_10.txt new file mode 100644 index 000000000..0ed081241 --- /dev/null +++ b/3886/CH6/EX6.10/6_10.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.10.sce', -1)
+Required force p=595.24 N
\ No newline at end of file diff --git a/3886/CH6/EX6.11/6_11.sce b/3886/CH6/EX6.11/6_11.sce new file mode 100644 index 000000000..b634f8953 --- /dev/null +++ b/3886/CH6/EX6.11/6_11.sce @@ -0,0 +1,12 @@ +//find necessary effort
+//effective wheel diameter
+D=(6/2)+500+(6/2) //mm
+//effective axle diameter
+d=(20/2)+200+(20/2) //mm
+VR=D/d
+eta=0.7
+MA=eta*VR
+W=1200
+P=1200/1.63 //N
+printf("The effort necessary=%.2f N",P)
+
diff --git a/3886/CH6/EX6.11/6_11.txt b/3886/CH6/EX6.11/6_11.txt new file mode 100644 index 000000000..22130f44a --- /dev/null +++ b/3886/CH6/EX6.11/6_11.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.11.sce', -1)
+The effort necessary=736.20 N
\ No newline at end of file diff --git a/3886/CH6/EX6.12/6_12.sce b/3886/CH6/EX6.12/6_12.sce new file mode 100644 index 000000000..99d51ca1a --- /dev/null +++ b/3886/CH6/EX6.12/6_12.sce @@ -0,0 +1,13 @@ +//required effort
+//differential axle diameters
+d1=300 //mm
+d2=250 //mm
+//wheel diameter
+D=800 //mm
+//load
+W=20000 //N
+eta=0.55
+VR=(2*D)/(d2-d1)
+MA=eta*VR
+P=W/MA //N
+printf("Required effort =%0.1f N",-P)
diff --git a/3886/CH6/EX6.12/6_12.txt b/3886/CH6/EX6.12/6_12.txt new file mode 100644 index 000000000..b63f00a2c --- /dev/null +++ b/3886/CH6/EX6.12/6_12.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.12.sce', -1)
+Required effort =1136.4 N
\ No newline at end of file diff --git a/3886/CH6/EX6.13/6_13.sce b/3886/CH6/EX6.13/6_13.sce new file mode 100644 index 000000000..cf0b07d40 --- /dev/null +++ b/3886/CH6/EX6.13/6_13.sce @@ -0,0 +1,9 @@ +//Effort required
+D=500 //mm
+d=200 //mm
+W=5000 //N
+eta=0.6
+VR=2*D/(D-d)
+MA=eta*VR
+P=W/MA //N
+printf("Required effort=%0.0f N",P)
diff --git a/3886/CH6/EX6.13/6_13.txt b/3886/CH6/EX6.13/6_13.txt new file mode 100644 index 000000000..53c390b01 --- /dev/null +++ b/3886/CH6/EX6.13/6_13.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.13.sce', -1)
+Required effort=2500 N
\ No newline at end of file diff --git a/3886/CH6/EX6.14/6_14.sce b/3886/CH6/EX6.14/6_14.sce new file mode 100644 index 000000000..68df5a803 --- /dev/null +++ b/3886/CH6/EX6.14/6_14.sce @@ -0,0 +1,9 @@ +//Force required at the end of lever
+d=40 //mm
+p=20/3 //mm
+W=40000 //N
+R=400 //mm
+mu=0.12
+theta=atand(p/(%pi*d)) //degree
+P=(d*W*(mu+tand(theta)))/(2*R*(1-mu*tand(theta))) //N
+printf("Force required at the end of lever P=%0.2f N",P)
diff --git a/3886/CH6/EX6.14/6_14.txt b/3886/CH6/EX6.14/6_14.txt new file mode 100644 index 000000000..9f6d51f2d --- /dev/null +++ b/3886/CH6/EX6.14/6_14.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.14.sce', -1)
+Force required at the end of lever P=348.32 N
\ No newline at end of file diff --git a/3886/CH6/EX6.15/6_15.sce b/3886/CH6/EX6.15/6_15.sce new file mode 100644 index 000000000..7cae55cb3 --- /dev/null +++ b/3886/CH6/EX6.15/6_15.sce @@ -0,0 +1,14 @@ +//Screw jack parameters
+p=10 //mm
+d=50 //mm
+W=6000 //N
+theta=atand(p/(%pi*d)) //degree
+fi=atand(0.05) //degree
+R=300 //mm
+P=(d*W*tand(theta+fi))/(2*R) //N
+VR=(2*%pi*R)/(p)
+MA=W/P
+eta=MA*100/VR //percent
+//torque required to keep the load from descending
+T=(50*600*tand(3.6426-2.8624))/2 //N-mm
+printf("Efficiency eta=%0.2f percent > 50 percent\nThus the screw jack is not self locking\nTorque required to keep the load from descending T=%0.2f N-mm",eta,T)
diff --git a/3886/CH6/EX6.15/6_15.txt b/3886/CH6/EX6.15/6_15.txt new file mode 100644 index 000000000..cf1d0456e --- /dev/null +++ b/3886/CH6/EX6.15/6_15.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.15.sce', -1)
+Efficiency eta=55.83 percent > 50 percent
+Thus the screw jack is not self locking
+Torque required to keep the load from descending T=204.27 N-mm
\ No newline at end of file diff --git a/3886/CH6/EX6.16/6_16.sce b/3886/CH6/EX6.16/6_16.sce new file mode 100644 index 000000000..8ad8da192 --- /dev/null +++ b/3886/CH6/EX6.16/6_16.sce @@ -0,0 +1,20 @@ +//Differential screw jack
+//pitch
+pA=10 //mm
+pB=5 //mm
+//lever arm length
+R=500 //mm
+VR=(2*%pi*R)/(pA-pB)
+P=185 //N
+W=15000 //N
+MA=W/P
+eta=MA*100/VR //percent
+//let the law of machine be P=m*W+C
+//from first case 185=m*15000+C
+//from second case 585=m*50000+C
+//solving we get
+m=4/350
+C=185-(m*15000) //N
+printf("Law of machine is P=%.4f*W+%0.2d ",m,C)
+
+
diff --git a/3886/CH6/EX6.16/6_16.txt b/3886/CH6/EX6.16/6_16.txt new file mode 100644 index 000000000..b8a1bd5fa --- /dev/null +++ b/3886/CH6/EX6.16/6_16.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.16.sce', -1)
+Law of machine is P=0.0114*W+13
\ No newline at end of file diff --git a/3886/CH6/EX6.17/6_17.sce b/3886/CH6/EX6.17/6_17.sce new file mode 100644 index 000000000..278da7947 --- /dev/null +++ b/3886/CH6/EX6.17/6_17.sce @@ -0,0 +1,23 @@ +//single purchase crab
+//load drum radius
+r=200/2 //mm
+//Length of lever arm
+R=1200 //mm
+T2=100
+T1=10
+VR=(R*T2)/(r*T1)
+//let the law of machine be P=m*W+C
+//in first case 100=m*3000+C
+//in second case 160=m*9000+C
+//solving we get
+m=1/100
+C=70
+//P=0.01*W+70
+//case 1
+MA1=3000/100
+eta1=MA1*100/VR //percent
+//case 2
+MA2=9000/160
+eta2=MA2*100/VR //percent
+printf("\nVR=%.0f\nP=0.01*W+70\nIn first case\neta=%0.2f percent\nIn second case\neta=%0.2f percent",VR,eta1,eta2)
+
diff --git a/3886/CH6/EX6.17/6_17.txt b/3886/CH6/EX6.17/6_17.txt new file mode 100644 index 000000000..525c39114 --- /dev/null +++ b/3886/CH6/EX6.17/6_17.txt @@ -0,0 +1,9 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.17.sce', -1)
+
+VR=120
+P=0.01*W+70
+In first case
+eta=25.00 percent
+In second case
+eta=46.88 percent
\ No newline at end of file diff --git a/3886/CH6/EX6.18/6_18.sce b/3886/CH6/EX6.18/6_18.sce new file mode 100644 index 000000000..20eb30d16 --- /dev/null +++ b/3886/CH6/EX6.18/6_18.sce @@ -0,0 +1,12 @@ +//Double purchase crab
+T1=15
+T2=45
+T3=20
+T4=40
+R=400 //mm
+r=150/2 //mm
+VR=(R*T2*T4)/(r*T1*T3)
+eta=0.40
+MA=eta*VR
+W=12.8*250 //N
+printf("Applied load lifts a load of W=%.2f N",W)
diff --git a/3886/CH6/EX6.18/6_18.txt b/3886/CH6/EX6.18/6_18.txt new file mode 100644 index 000000000..52a5ba095 --- /dev/null +++ b/3886/CH6/EX6.18/6_18.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.18.sce', -1)
+Applied load lifts a load of W=3200.00 N
\ No newline at end of file diff --git a/3886/CH6/EX6.2/6_2.sce b/3886/CH6/EX6.2/6_2.sce new file mode 100644 index 000000000..a8757768d --- /dev/null +++ b/3886/CH6/EX6.2/6_2.sce @@ -0,0 +1,18 @@ +//Simple machine
+//assume law of machine P=m*W+C
+//From first and second conditions we obtain 2 equations
+//150=2400*m+C
+//180=3000*m+C
+//upon solving them
+//P=0.05*W+30
+//When force of 200 N is applied
+W=(200-30)/(0.05) //N
+//Ideal effort
+Pi=3400/30 //N
+//Effort wasted in overcoming the friction
+Ew=200-113.33 //N
+//Mechanical advantage
+MA=3400/200
+//Efficiency
+Eff=(17*100)/(30) //percent
+printf("\n Law of machining is P=0.05*W+30\nW=%.2f N\nEffort wasted in overcoming the friction=%.2f N\nMechanical advantage=%.2f \nEfficiency=%.2f percent",W,Ew,MA,Eff)
diff --git a/3886/CH6/EX6.2/6_2.txt b/3886/CH6/EX6.2/6_2.txt new file mode 100644 index 000000000..44a9fe1a0 --- /dev/null +++ b/3886/CH6/EX6.2/6_2.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.2.sce', -1)
+
+ Law of machining is P=0.05*W+30
+W=3400.00 N
+Effort wasted in overcoming the friction=86.67 N
+Mechanical advantage=17.00
+Efficiency=56.67 percent
\ No newline at end of file diff --git a/3886/CH6/EX6.3/6_3.sce b/3886/CH6/EX6.3/6_3.sce new file mode 100644 index 000000000..38e8349f4 --- /dev/null +++ b/3886/CH6/EX6.3/6_3.sce @@ -0,0 +1,23 @@ +//maximum mechanical advantage and maximum efficiency
+//Effort
+P=150 //N
+W=7700 //N
+MA=W/P //mechanical advantage
+//If efficiency=60%
+eff=0.6
+VR=(MA)/(eff)
+//When an effort of 250 N raised a load of 13200 N
+P1=250 //N
+W1=13200 //N
+MA1=(W1)/(P1)
+eff1=MA1*100/VR //percent
+//assume law of machine as P=m*W+C
+//from first case 150=7700*m+C
+//from second case 250=13200*m+C
+//Solving we get
+m=100/5500
+//maximum mechanical advantage
+MAmax=1/(m)
+//maximum efficiency
+Effmax=100/(m*VR) //percent
+printf("\nMechanical advantage=%0.2f\nVelocity ratio=%0.2f\nEfficiency=%0.2f percent\nMaximum mechanical advantage=%0.2f\nMaxumum efficiency=%0.2f percent",MA,VR,eff1,MAmax,Effmax)
diff --git a/3886/CH6/EX6.3/6_3.txt b/3886/CH6/EX6.3/6_3.txt new file mode 100644 index 000000000..2cb8e6833 --- /dev/null +++ b/3886/CH6/EX6.3/6_3.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.3.sce', -1)
+
+Mechanical advantage=51.33
+Velocity ratio=85.56
+Efficiency=61.71 percent
+Maximum mechanical advantage=55.00
+Maxumum efficiency=64.29 percent
\ No newline at end of file diff --git a/3886/CH6/EX6.4/6_4.sce b/3886/CH6/EX6.4/6_4.sce new file mode 100644 index 000000000..5d11d01d4 --- /dev/null +++ b/3886/CH6/EX6.4/6_4.sce @@ -0,0 +1,21 @@ +//determine law of machine and maximum efficiency
+//fig.6.4 shows graph of effort vs load
+//from graph
+C=10 //N
+//slope
+m=30/500
+//Law of machine is P=m*W+C
+//P=0.06*W+10
+//eta=MA/VR=W/25P
+//table 6.5 shows calculation of efficiency for various loads
+//Refer fig. 6.5
+//from graph it can be seen that maximum efficiency=50% but the graph is plotted for infinitely large load
+//thus
+//Maximum efficiency
+VR=25
+Emax=100/(m*VR) //percent
+printf("\nLaw of machines is P=0.06*W+10\nMaximum efficiency=%.2f percent",Emax)
+
+
+
+
diff --git a/3886/CH6/EX6.4/6_4.txt b/3886/CH6/EX6.4/6_4.txt new file mode 100644 index 000000000..ebfbdbeb5 --- /dev/null +++ b/3886/CH6/EX6.4/6_4.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.4.sce', -1)
+
+Law of machines is P=0.06*W+10
+Maximum efficiency=66.67 percent
\ No newline at end of file diff --git a/3886/CH6/EX6.5/6_5.sce b/3886/CH6/EX6.5/6_5.sce new file mode 100644 index 000000000..d21562520 --- /dev/null +++ b/3886/CH6/EX6.5/6_5.sce @@ -0,0 +1,13 @@ +//Lifting machine
+VR=30
+W=5000 //N
+P=360 //N
+MA=W/P
+//efficiency
+eta=MA*100/VR //percent
+printf("Since the efficiency eta=%0.2d percent is less than 50 percent,it is a self locking machine",eta)
+//ideal load
+Wi=P*VR //N
+//Frictional resistance
+FR=Wi-W //N
+printf("\nIdeal load=%0.2d N\nFrictional resistance=%0.2d N",Wi,FR)
diff --git a/3886/CH6/EX6.5/6_5.txt b/3886/CH6/EX6.5/6_5.txt new file mode 100644 index 000000000..b26205998 --- /dev/null +++ b/3886/CH6/EX6.5/6_5.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.5.sce', -1)
+Since the efficiency eta=46 percent is less than 50 percent,it is a self locking machine
+Ideal load=10800 N
+Frictional resistance=5800 N
\ No newline at end of file diff --git a/3886/CH6/EX6.6/6_6.sce b/3886/CH6/EX6.6/6_6.sce new file mode 100644 index 000000000..5abfc27d7 --- /dev/null +++ b/3886/CH6/EX6.6/6_6.sce @@ -0,0 +1,16 @@ +//Finding the additional pulleys required
+//if n is number of movable pulleys then
+n=3
+VR=2^n
+eta=0.8
+MA=eta*VR
+P=6000/6.4 //N
+//In second case
+Effort=520 //N
+//efficiency eta=0.80-n1*0.05
+//n1=number of additional pulleys required=(n-3)
+//thus
+//W=P[0.8-(n-3)*0.05]*2^n
+//by using trial and erroe method
+nfinal=4
+printf("Number of movable pulleys required=%.0f",nfinal)
diff --git a/3886/CH6/EX6.6/6_6.txt b/3886/CH6/EX6.6/6_6.txt new file mode 100644 index 000000000..aee6fb38d --- /dev/null +++ b/3886/CH6/EX6.6/6_6.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.6.sce', -1)
+Number of movable pulleys required=4
\ No newline at end of file diff --git a/3886/CH6/EX6.7/6_7.sce b/3886/CH6/EX6.7/6_7.sce new file mode 100644 index 000000000..85c5d0c8a --- /dev/null +++ b/3886/CH6/EX6.7/6_7.sce @@ -0,0 +1,9 @@ +//calculate force required
+//refer fig.6.15
+//VR=2*number of movable pulleys
+VR=2*3
+eta=0.85
+MA=eta*VR
+W=12000
+P=W/5.1
+printf("The required force is P=%.2f N",P)
diff --git a/3886/CH6/EX6.7/6_7.txt b/3886/CH6/EX6.7/6_7.txt new file mode 100644 index 000000000..8e6ae6399 --- /dev/null +++ b/3886/CH6/EX6.7/6_7.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.7.sce', -1)
+The required force is P=2352.94 N
\ No newline at end of file diff --git a/3886/CH6/EX6.8/6_8.sce b/3886/CH6/EX6.8/6_8.sce new file mode 100644 index 000000000..6bac961d9 --- /dev/null +++ b/3886/CH6/EX6.8/6_8.sce @@ -0,0 +1,13 @@ +//Pull required
+//Refer fig.6.16
+//Let weight W be lifted by a distance x
+//Consider first order system
+//VR=2^2
+//Consider second order system
+VR=8
+eta=0.78
+MA=eta*VR
+W=12000
+P=W/6.24 //N
+printf("Required pull P=%0.2f N",P)
+
diff --git a/3886/CH6/EX6.8/6_8.txt b/3886/CH6/EX6.8/6_8.txt new file mode 100644 index 000000000..06a8c5c6e --- /dev/null +++ b/3886/CH6/EX6.8/6_8.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.8.sce', -1)
+Required pull P=1923.08 N
\ No newline at end of file diff --git a/3886/CH6/EX6.9/6_9.sce b/3886/CH6/EX6.9/6_9.sce new file mode 100644 index 000000000..e231e7bed --- /dev/null +++ b/3886/CH6/EX6.9/6_9.sce @@ -0,0 +1,13 @@ +//machine efficiency and effort lost in friction
+//For third order system of pulleys
+W=1000
+VR=(2^3)-1
+eta=(1000*100)/(180*7) //percent
+//ideal effort
+Pi=(W)/(VR) //N
+P=180 //N
+//effort lost in friction
+Pl=P-Pi //N
+printf("Efficiency=%.2f percent\nEffort lost in friction=%.2f N",eta,Pl)
+
+
diff --git a/3886/CH6/EX6.9/6_9.txt b/3886/CH6/EX6.9/6_9.txt new file mode 100644 index 000000000..320e971d9 --- /dev/null +++ b/3886/CH6/EX6.9/6_9.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\6. Lifting machines\6.9.sce', -1)
+Efficiency=79.37 percent
+Effort lost in friction=37.14 N
\ No newline at end of file diff --git a/3886/CH7/EX7.1/7_1.sce b/3886/CH7/EX7.1/7_1.sce new file mode 100644 index 000000000..2c43d0f11 --- /dev/null +++ b/3886/CH7/EX7.1/7_1.sce @@ -0,0 +1,14 @@ +//Diameter of wheels
+//refer fig.7.2
+d1=240 //mm
+N1=250 //rpm
+N2=100 //rpm
+d2=N1*d1/N2 //mm
+ans1=d1+d2 //mm
+d3=840/3 //mm
+d4=840-d3 //mm
+d5=(840)/(1+(5/3)) //mm
+d6=840-d5 //mm
+printf("\nd2=%.2f mm\nd3=%.2f mm\nd4=%.2f mm\nd5=%.2f mm\nd6=%.2f mm",d2,d3,d4,d5,d6)
+
+
diff --git a/3886/CH7/EX7.1/7_1.txt b/3886/CH7/EX7.1/7_1.txt new file mode 100644 index 000000000..93bd62bd2 --- /dev/null +++ b/3886/CH7/EX7.1/7_1.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\7. Transmission of power\7.1.sce', -1)
+
+d2=600.00 mm
+d3=280.00 mm
+d4=560.00 mm
+d5=315.00 mm
+d6=525.00 mm
\ No newline at end of file diff --git a/3886/CH7/EX7.10/7_10.sce b/3886/CH7/EX7.10/7_10.sce new file mode 100644 index 000000000..b6b55df36 --- /dev/null +++ b/3886/CH7/EX7.10/7_10.sce @@ -0,0 +1,13 @@ +//speed of pinion
+P=480 //N
+Pw=1800 //W
+v=Pw/P //m/sec
+//module
+m=8
+d=25*m //mm
+r=d/2000 //m
+omega=v/r //rad/sec
+N=(60*omega)/(2*%pi) //rpm
+//rest is theory
+printf("Speed of the pinion N=%.1f rpm",N)
+
diff --git a/3886/CH7/EX7.10/7_10.txt b/3886/CH7/EX7.10/7_10.txt new file mode 100644 index 000000000..c28bbd54b --- /dev/null +++ b/3886/CH7/EX7.10/7_10.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\7. Transmission of power\7.10.sce', -1)
+Speed of the pinion N=358.1 rpm
\ No newline at end of file diff --git a/3886/CH7/EX7.11/7_11.sce b/3886/CH7/EX7.11/7_11.sce new file mode 100644 index 000000000..728a0fa51 --- /dev/null +++ b/3886/CH7/EX7.11/7_11.sce @@ -0,0 +1,9 @@ +//speed of driven shaft
+//a-If the intermediate gears are on different shaft
+TA=25
+TD=100
+NA=160 //rpm
+aND=TA*NA/TD //rpm
+//b-If the intermediate gears are on the same shaft
+bND=(75*25*160)/(50*100) //rpm
+printf("\ncase a\nND=%.2f rpm\ncase b\nND=%.2f rpm",aND,bND)
diff --git a/3886/CH7/EX7.11/7_11.txt b/3886/CH7/EX7.11/7_11.txt new file mode 100644 index 000000000..2a156a29e --- /dev/null +++ b/3886/CH7/EX7.11/7_11.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\7. Transmission of power\7.11.sce', -1)
+
+case a
+ND=40.00 rpm
+case b
+ND=60.00 rpm
\ No newline at end of file diff --git a/3886/CH7/EX7.2/7_2.sce b/3886/CH7/EX7.2/7_2.sce new file mode 100644 index 000000000..3e1b1969a --- /dev/null +++ b/3886/CH7/EX7.2/7_2.sce @@ -0,0 +1,11 @@ +//speed of follower
+d1=600 //mm
+d2=400 //mm
+N1=160 //rpm
+N2=2 //rpm
+//if there is no slip
+aN2=(600*160)/(400) //rpm
+//when there is 2.5% slip
+p=2.5/100
+bN2=(N1*d1*(100-p))/(d2*100) //rpm
+printf("\nWhen there is no slip \nN2=%0.2f rpm\nWhen there is 2.5 percent slip \nN2=%0.2f rpm",aN2,bN2)
diff --git a/3886/CH7/EX7.2/7_2.txt b/3886/CH7/EX7.2/7_2.txt new file mode 100644 index 000000000..511e932fc --- /dev/null +++ b/3886/CH7/EX7.2/7_2.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\7. Transmission of power\7.2.sce', -1)
+
+When there is no slip
+N2=240.00 rpm
+When there is 2.5 percent slip
+N2=239.94 rpm
\ No newline at end of file diff --git a/3886/CH7/EX7.3/7_3.sce b/3886/CH7/EX7.3/7_3.sce new file mode 100644 index 000000000..5f9706855 --- /dev/null +++ b/3886/CH7/EX7.3/7_3.sce @@ -0,0 +1,11 @@ +//Length of belt
+N2=80
+N1=200
+d1=240
+d2=d1*N1/N2 //mm
+r1=120 //mm
+r2=300 //mm
+l=2500 //mm
+//length of crossbelt
+L=%pi*(r1+r2)+2*l+((r1+r2)^2)/l //mm
+printf("The length of crossbelt L=%.2f mm",L)
diff --git a/3886/CH7/EX7.3/7_3.txt b/3886/CH7/EX7.3/7_3.txt new file mode 100644 index 000000000..a629b93a3 --- /dev/null +++ b/3886/CH7/EX7.3/7_3.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\7. Transmission of power\7.3.sce', -1)
+The length of crossbelt L=6390.03 mm
\ No newline at end of file diff --git a/3886/CH7/EX7.4/7_4.sce b/3886/CH7/EX7.4/7_4.sce new file mode 100644 index 000000000..d8bd73ed6 --- /dev/null +++ b/3886/CH7/EX7.4/7_4.sce @@ -0,0 +1,17 @@ +//range of weight
+//refer fig. 7.5
+//angle of contact
+theta=1.25*2*%pi
+//case 1-Let impending motion of the weight be downward
+//T1=600 //N
+//T2=W
+//from law of rope friction
+//T2=T1*%e^mu*theta
+aW=600*(%e^(0.3*2.5*%pi)) //N
+//case-2 impending motion of W be upward
+//T1=W
+//T2=600 N
+//from law of rope friction
+bW=600/(%e^(0.75*%pi)) //N
+printf("The range of weight that can be supported is from %.2f N to %.2f N",aW,bW)
+
diff --git a/3886/CH7/EX7.4/7_4.txt b/3886/CH7/EX7.4/7_4.txt new file mode 100644 index 000000000..3e58193f3 --- /dev/null +++ b/3886/CH7/EX7.4/7_4.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\7. Transmission of power\7.4.sce', -1)
+The range of weight that can be supported is from 6330.43 N to 56.87 N
\ No newline at end of file diff --git a/3886/CH7/EX7.5/7_5.sce b/3886/CH7/EX7.5/7_5.sce new file mode 100644 index 000000000..da5c740eb --- /dev/null +++ b/3886/CH7/EX7.5/7_5.sce @@ -0,0 +1,15 @@ +//Maximum power that can be transmitted
+alpha=asin((500-300)/2*2500) //radians
+//angle of contact
+theta=%pi-2*alpha //radians
+f1=4
+b=100
+t=3
+//T2 may be allowed upto
+T2=f1*b*t //N
+mu=0.3
+T1=1200/2.505 //N
+r=500/2
+omega=(2*%pi*100)/60
+P=(T2-T1)*r*omega/1000000 //kW
+printf("The maximum power that can be transmitted=%.3f kW",P)
diff --git a/3886/CH7/EX7.5/7_5.txt b/3886/CH7/EX7.5/7_5.txt new file mode 100644 index 000000000..57e60aa08 --- /dev/null +++ b/3886/CH7/EX7.5/7_5.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\7. Transmission of power\7.5.sce', -1)
+The maximum power that can be transmitted=1.887 kW
\ No newline at end of file diff --git a/3886/CH7/EX7.6/7_6.sce b/3886/CH7/EX7.6/7_6.sce new file mode 100644 index 000000000..ac9ffc1ca --- /dev/null +++ b/3886/CH7/EX7.6/7_6.sce @@ -0,0 +1,11 @@ +//Force P required
+//refer fig. 7.6
+mu=0.3
+theta=(250*%pi)/180 //radians
+r=250 //mm
+T1=(300*10^3)/(250*(%e^mu*theta-1)) //N
+T2=(%e^mu*theta)*T1 //N
+//Considering equilibrium of lever arm
+P=(1644.06*50)/(300) //N
+printf("The required value is P=%.2f N",P)
+
diff --git a/3886/CH7/EX7.6/7_6.txt b/3886/CH7/EX7.6/7_6.txt new file mode 100644 index 000000000..0426b52d9 --- /dev/null +++ b/3886/CH7/EX7.6/7_6.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\7. Transmission of power\7.6.sce', -1)
+The required value is P=274.01 N
\ No newline at end of file diff --git a/3886/CH7/EX7.7/7_7.sce b/3886/CH7/EX7.7/7_7.sce new file mode 100644 index 000000000..80f5174e4 --- /dev/null +++ b/3886/CH7/EX7.7/7_7.sce @@ -0,0 +1,28 @@ +//maximum power that can be transmitted
+mu=0.3
+alpha=asin(((480/2)-(320/2))/2500) //radians
+theta=%pi-2*alpha
+//neglecting centrifugal tension
+f=3
+b=150
+t=8
+//maximum force permitted is
+T2=f*b*t //N
+T1=T2/%e^mu*theta
+r=480/2
+omega=(2*%pi*800)/60
+aP=((3600-1429.82)*480*2*800*%pi)/(2*60*(10^6)) //kW
+//If centrifugal tension is considered
+v=r*omega/1000 //m/sec
+m=1.32
+Tc=m*v^2 //N
+//maximum force that can be permitted on the belt
+//T=T2+Tc=f*b*t
+bT2=3600-533.62 //N
+bT1=3066.38/2.5178 //N
+//maximum torque that can be transferred
+bP=(bT2-bT1)*v/1000 //kW
+printf("\nneglecting centrifugal tension P=%0.3f kW\nConsidering centrifugal tension P=%0.3f kW",aP,bP)
+
+
+
diff --git a/3886/CH7/EX7.7/7_7.txt b/3886/CH7/EX7.7/7_7.txt new file mode 100644 index 000000000..e9d926892 --- /dev/null +++ b/3886/CH7/EX7.7/7_7.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\7. Transmission of power\7.7.sce', -1)
+
+neglecting centrifugal tension P=43.634 kW
+Considering centrifugal tension P=37.166 kW
\ No newline at end of file diff --git a/3886/CH7/EX7.8/7_8.sce b/3886/CH7/EX7.8/7_8.sce new file mode 100644 index 000000000..c5c42e147 --- /dev/null +++ b/3886/CH7/EX7.8/7_8.sce @@ -0,0 +1,13 @@ +//Maximum power transmitted
+m=0.9
+v=20
+Tc=m*v^2 //N
+f=1.5
+d=36
+T=(f*%pi*d^2)/(4) //N
+T2=T-Tc //N
+mu=0.3
+theta=220*%pi/180 //radians
+T1=T2/(%e^mu*theta*(1/sind(30))) //N
+P=(T2-T1)*v/1000 //kW
+printf("maximum power P=%.3f kW ",P)
diff --git a/3886/CH7/EX7.8/7_8.txt b/3886/CH7/EX7.8/7_8.txt new file mode 100644 index 000000000..26feb9aac --- /dev/null +++ b/3886/CH7/EX7.8/7_8.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\7. Transmission of power\7.8.sce', -1)
+maximum power P=21.085 kW
\ No newline at end of file diff --git a/3886/CH7/EX7.9/7_9.sce b/3886/CH7/EX7.9/7_9.sce new file mode 100644 index 000000000..8eb7c11b5 --- /dev/null +++ b/3886/CH7/EX7.9/7_9.sce @@ -0,0 +1,16 @@ +//parallel shafts connected with spur gearing
+l=540 //mm
+//d2=1080/4 //mm
+//d1=3*d2
+//for a module of 8
+//T1=810/8 but number of teeth is whole no.
+T1=102
+//T2=270/8 but number of teeth is whole no.
+T2=34
+//pitch circle diameter are
+d1=102*8 //mm
+d2=34*8 //mm
+//The exact distance between the shafts
+l=(d1+d2)/2 //mm
+printf("\nT1=%.0d \nT2=%.0d \nl=%.0d mm",T1,T2,l)
+
diff --git a/3886/CH7/EX7.9/7_9.txt b/3886/CH7/EX7.9/7_9.txt new file mode 100644 index 000000000..a72f4435a --- /dev/null +++ b/3886/CH7/EX7.9/7_9.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\7. Transmission of power\7.9.sce', -1)
+
+T1=102
+T2=34
+l=544 mm
\ No newline at end of file diff --git a/3886/CH8/EX8.1/8_1.sce b/3886/CH8/EX8.1/8_1.sce new file mode 100644 index 000000000..f70b4fc26 --- /dev/null +++ b/3886/CH8/EX8.1/8_1.sce @@ -0,0 +1,12 @@ +//Resultant of system
+//Refer fig. 8.5
+//Let E be the equilibriant
+//using virtual work principle
+//-E*cosd(theta)=50*cosd(45)+80*cosd(25)+70*cosd(50)=152.86 N
+//-E*sind(theta)=-50*sind(45)+80*sind(25)+70*sind(50)=52.07*sind(25)
+//Thus
+a=152.86 //N (R*cosd(theta))
+b=52.07 //N (R*sind(theta))
+R=sqrt(a^2+b^2) //N
+theta=atand(b/a) //degree
+printf("Resultant R=%.2f N inclined at theta=%.2f degree w.r.t positive x-axis",R,theta)
diff --git a/3886/CH8/EX8.1/8_1.txt b/3886/CH8/EX8.1/8_1.txt new file mode 100644 index 000000000..5cf1ab4fe --- /dev/null +++ b/3886/CH8/EX8.1/8_1.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.1.sce', -1)
+Resultant R=161.49 N inclined at theta=18.81 degree w.r.t positive x-axis
\ No newline at end of file diff --git a/3886/CH8/EX8.10/8_10.sce b/3886/CH8/EX8.10/8_10.sce new file mode 100644 index 000000000..5f8e97bd5 --- /dev/null +++ b/3886/CH8/EX8.10/8_10.sce @@ -0,0 +1,6 @@ +//Horizontal force P
+//refer fig.8.14
+//Applying virtual work principle
+//-1500*(delta(s)*sind(30))+P*delta(s)*cosd(30)+0=0
+P=1500*tand(30) //N
+printf("Magnitude of P=%.2f N",P)
diff --git a/3886/CH8/EX8.10/8_10.txt b/3886/CH8/EX8.10/8_10.txt new file mode 100644 index 000000000..0ebbbf703 --- /dev/null +++ b/3886/CH8/EX8.10/8_10.txt @@ -0,0 +1,2 @@ +--> exec('E:\My program EM\8. Virtual work\8.10.sce', -1)
+Magnitude of P=866.03 N
\ No newline at end of file diff --git a/3886/CH8/EX8.11/8_11.sce b/3886/CH8/EX8.11/8_11.sce new file mode 100644 index 000000000..80f80af83 --- /dev/null +++ b/3886/CH8/EX8.11/8_11.sce @@ -0,0 +1,13 @@ +//Range of force P
+//refer fig. 8.15
+//a-when the motion is impending up the plane
+mu=0.3
+N=1000*cosd(70)
+F=mu*N //N
+//Applying virtual work principle
+aP=1000*sind(70)+300*cosd(70) //N
+//b-when the motion is impending down the plane
+//Applying virtual work principle
+bP=1000*sind(70)-300*cosd(70) //N
+printf("Block is in equilibrium for P=%.2f N to %.2f N",bP,aP)
+
diff --git a/3886/CH8/EX8.11/8_11.txt b/3886/CH8/EX8.11/8_11.txt new file mode 100644 index 000000000..1b8e2919b --- /dev/null +++ b/3886/CH8/EX8.11/8_11.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.11.sce', -1)
+Block is in equilibrium for P=837.09 N to 1042.30 N
\ No newline at end of file diff --git a/3886/CH8/EX8.12/8_12.sce b/3886/CH8/EX8.12/8_12.sce new file mode 100644 index 000000000..bea441e90 --- /dev/null +++ b/3886/CH8/EX8.12/8_12.sce @@ -0,0 +1,7 @@ +//Position of the balls
+//refer fig. 8.16
+//Let a virtual displacement be given to the system of balls as shown
+//Applying virtual work principle
+//200*sind(30)*delta(DB)+150*sind(60)*delta(EB)=0
+theta=atand((150*sind(60))/(200*sind(30))) //degree
+printf("Thus theta=%.2f degree",theta)
diff --git a/3886/CH8/EX8.12/8_12.txt b/3886/CH8/EX8.12/8_12.txt new file mode 100644 index 000000000..b0ec00b10 --- /dev/null +++ b/3886/CH8/EX8.12/8_12.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.12.sce', -1)
+Thus theta=52.41 degree
\ No newline at end of file diff --git a/3886/CH8/EX8.13/8_13.sce b/3886/CH8/EX8.13/8_13.sce new file mode 100644 index 000000000..de1cbf0aa --- /dev/null +++ b/3886/CH8/EX8.13/8_13.sce @@ -0,0 +1,14 @@ +//Force P to make motion impending to the left
+//Refer fig.8.17
+N1=250
+N2=1000*cosd(45)
+N3=500
+F1=0.25*N1 //N
+F2=0.25*N2 //N
+F3=0.25*N3 //N
+//let us give virtual displacement towards left
+//Applying virtual work principle
+//(P-F1-1000*sind(45)-F2-F3)*delta(s)=0
+P=F1+1000*sind(45)+F2+F3 //N
+printf("The required force is P=%.2f N",P)
+
diff --git a/3886/CH8/EX8.13/8_13.txt b/3886/CH8/EX8.13/8_13.txt new file mode 100644 index 000000000..e0186396a --- /dev/null +++ b/3886/CH8/EX8.13/8_13.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.13.sce', -1)
+The required force is P=1071.38 N
\ No newline at end of file diff --git a/3886/CH8/EX8.14/8_14.sce b/3886/CH8/EX8.14/8_14.sce new file mode 100644 index 000000000..a34f24830 --- /dev/null +++ b/3886/CH8/EX8.14/8_14.sce @@ -0,0 +1,12 @@ +//Determine WB
+//refer fig.8.18
+theta=atand(3/4)
+mu=0.3
+WA=200
+F=mu*WA*cosd(theta) //N
+//Let us give virtual displacement of delta(s) up the plane to block A
+//Applying virtual work principle
+//(-200*sind(theta)-F+WB/2)*delta(s)=0
+WB=2*(200*sind(theta)+F) //N
+printf("Required value of WB=%.2f N",WB)
+
diff --git a/3886/CH8/EX8.14/8_14.txt b/3886/CH8/EX8.14/8_14.txt new file mode 100644 index 000000000..eb621d93c --- /dev/null +++ b/3886/CH8/EX8.14/8_14.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.14.sce', -1)
+Required value of WB=336.00 N
\ No newline at end of file diff --git a/3886/CH8/EX8.15/8_15.sce b/3886/CH8/EX8.15/8_15.sce new file mode 100644 index 000000000..3d7112b13 --- /dev/null +++ b/3886/CH8/EX8.15/8_15.sce @@ -0,0 +1,13 @@ +//Determine force P
+//refer fig.8.19
+//(a) P at floor level
+//Applying virtual work principle
+//P*delta(x)-200*delta(y)=0
+P=100*tand(30) //N
+//(b) If the rope is used instead of force P
+//refer fig.8.20
+//Taking C as origin
+//Applying virtual work principle
+//(-3*T*cosd(theta)+400*sind(theta))*delta(theta)=0
+T=(400*tand(30))/3 //N
+printf("\nP=%.2f N\nT=%.2f N",P,T)
diff --git a/3886/CH8/EX8.15/8_15.txt b/3886/CH8/EX8.15/8_15.txt new file mode 100644 index 000000000..b0fe2b738 --- /dev/null +++ b/3886/CH8/EX8.15/8_15.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.15.sce', -1)
+
+P=57.74 N
+T=76.98 N
\ No newline at end of file diff --git a/3886/CH8/EX8.16/8_16.sce b/3886/CH8/EX8.16/8_16.sce new file mode 100644 index 000000000..bb8ccec03 --- /dev/null +++ b/3886/CH8/EX8.16/8_16.sce @@ -0,0 +1,9 @@ +//Least value of theta to avoid slipping of ladder
+//refer fig.8.21
+//Applying virtual work principle
+//-0.4*NA*6*cosd(theta)*delta(theta)-200*(-3*sind(theta)*delta(theta))-900*(-5*sind(theta)*delta(theta))+0.25*NB(-6*sind(theta)*delta(theta))=0
+NA=1100/(1+0.25*0.4) //N
+NB=0.4*1000 //N
+theta=atand(2400/4500) //degree
+printf("Thus theta=%.2f degree",theta)
+
diff --git a/3886/CH8/EX8.16/8_16.txt b/3886/CH8/EX8.16/8_16.txt new file mode 100644 index 000000000..318209712 --- /dev/null +++ b/3886/CH8/EX8.16/8_16.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.16.sce', -1)
+Thus theta=28.07 degree
\ No newline at end of file diff --git a/3886/CH8/EX8.17/8_17.sce b/3886/CH8/EX8.17/8_17.sce new file mode 100644 index 000000000..5e8339399 --- /dev/null +++ b/3886/CH8/EX8.17/8_17.sce @@ -0,0 +1,7 @@ +//Force in the member FH
+//refer fig.8.22
+//use symmetry
+//Apply virtual work principle
+//RA*12*delta(theta)-10*10*delta(theta)-10*6*delta(theta)-10*2*delta(theta)+FFH*2*tand(60)=0
+FFH=-240/(2*tand(60)) //kN
+printf("FFH is a compressive force of %.2f kN",-FFH)
diff --git a/3886/CH8/EX8.17/8_17.txt b/3886/CH8/EX8.17/8_17.txt new file mode 100644 index 000000000..a80319a14 --- /dev/null +++ b/3886/CH8/EX8.17/8_17.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.17.sce', -1)
+FFH is a compressive force of 69.28 kN
\ No newline at end of file diff --git a/3886/CH8/EX8.18/8_18.sce b/3886/CH8/EX8.18/8_18.sce new file mode 100644 index 000000000..0cd98b2ea --- /dev/null +++ b/3886/CH8/EX8.18/8_18.sce @@ -0,0 +1,6 @@ +//Force developed in member DF
+//refer fig. 8.23
+//Applying virtual work principle
+//-100*10*delta(theta)-100*5*delta(theta)+FDF*5*delta(theta)*cosd(45)=0
+FDF=1500/(5*cosd(45)) //kN (tensile)
+printf("Thus force developed in member DF is FDF=%.2f kN (Tensile)",FDF)
diff --git a/3886/CH8/EX8.18/8_18.txt b/3886/CH8/EX8.18/8_18.txt new file mode 100644 index 000000000..04c06d618 --- /dev/null +++ b/3886/CH8/EX8.18/8_18.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.18.sce', -1)
+Thus force developed in member DF is FDF=424.26 kN (Tensile)
\ No newline at end of file diff --git a/3886/CH8/EX8.2/8_2.sce b/3886/CH8/EX8.2/8_2.sce new file mode 100644 index 000000000..c1d5c8f5c --- /dev/null +++ b/3886/CH8/EX8.2/8_2.sce @@ -0,0 +1,7 @@ +//Reactions developed in beam
+//refer fig 8.6
+//Let RA and RB be the reactions at supports A and B
+//applying virtual work principle
+RB=(20/3)+(80/3) //kN
+RA=(40/3)+(40/3) //kN
+printf("Reactions are-\nRA=%.2f kN\nRB=%.2f kN",RA,RB)
diff --git a/3886/CH8/EX8.2/8_2.txt b/3886/CH8/EX8.2/8_2.txt new file mode 100644 index 000000000..4ef7f3fa2 --- /dev/null +++ b/3886/CH8/EX8.2/8_2.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.2.sce', -1)
+Reactions are-
+RA=26.67 kN
+RB=33.33 kN
\ No newline at end of file diff --git a/3886/CH8/EX8.3/8_3.sce b/3886/CH8/EX8.3/8_3.sce new file mode 100644 index 000000000..46e19ea52 --- /dev/null +++ b/3886/CH8/EX8.3/8_3.sce @@ -0,0 +1,6 @@ +//Reactions of overhanging beam
+//refer fig. 8.7
+//Applying virtual work principle for beam in equilibrium
+RB=(8/6)*(30+60*4/8-20/8) //kN
+RA=(7/6)*(20+60*2/7-30*2/7) //kN
+printf("The reactions are-\nRA=%0.2f kN\nRB=%.2f kN",RA,RB)
diff --git a/3886/CH8/EX8.3/8_3.txt b/3886/CH8/EX8.3/8_3.txt new file mode 100644 index 000000000..281ed61d3 --- /dev/null +++ b/3886/CH8/EX8.3/8_3.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.3.sce', -1)
+The reactions are-
+RA=33.33 kN
+RB=76.67 kN
\ No newline at end of file diff --git a/3886/CH8/EX8.4/8_4.sce b/3886/CH8/EX8.4/8_4.sce new file mode 100644 index 000000000..6abb120a4 --- /dev/null +++ b/3886/CH8/EX8.4/8_4.sce @@ -0,0 +1,7 @@ +//Reactions at A
+//refer fig. 8.8
+//applying virtual work principle
+//(RA-60)delta(y)=0
+//thus
+RA=60 //kN
+printf("\n Reaction is RA=%.2f kN",RA)
diff --git a/3886/CH8/EX8.4/8_4.txt b/3886/CH8/EX8.4/8_4.txt new file mode 100644 index 000000000..18ccdf0ae --- /dev/null +++ b/3886/CH8/EX8.4/8_4.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.4.sce', -1)
+
+ Reaction is RA=60.00 kN
\ No newline at end of file diff --git a/3886/CH8/EX8.5/8_5.sce b/3886/CH8/EX8.5/8_5.sce new file mode 100644 index 000000000..4bd95c5b9 --- /dev/null +++ b/3886/CH8/EX8.5/8_5.sce @@ -0,0 +1,11 @@ +//Reaction at A in overhanging beam
+//refer fig. 8.9
+//Applying virtual work principle
+//(-2*RA+180-20)*delta(y)=0
+//Thus
+RA=(180-20)/2 //kN
+printf("\n Reaction is RA=%.2f kN",RA)
+
+
+
+
diff --git a/3886/CH8/EX8.5/8_5.txt b/3886/CH8/EX8.5/8_5.txt new file mode 100644 index 000000000..a936f647e --- /dev/null +++ b/3886/CH8/EX8.5/8_5.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.5.sce', -1)
+
+ Reaction is RA=80.00 kN
\ No newline at end of file diff --git a/3886/CH8/EX8.6/8_6.sce b/3886/CH8/EX8.6/8_6.sce new file mode 100644 index 000000000..e0bbe045f --- /dev/null +++ b/3886/CH8/EX8.6/8_6.sce @@ -0,0 +1,7 @@ +//Reaction at B
+//refer fig. 8.10
+//Applying virtual work principle
+//(-4-6+RB)*delta(y)=0
+//Thus
+RB=6+4 //kN
+printf("Reaction at B is RB=%.2f kN",RB)
diff --git a/3886/CH8/EX8.6/8_6.txt b/3886/CH8/EX8.6/8_6.txt new file mode 100644 index 000000000..e2dc2ec61 --- /dev/null +++ b/3886/CH8/EX8.6/8_6.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.6.sce', -1)
+Reaction at B is RB=10.00 kN
\ No newline at end of file diff --git a/3886/CH8/EX8.7/8_7.sce b/3886/CH8/EX8.7/8_7.sce new file mode 100644 index 000000000..3050647db --- /dev/null +++ b/3886/CH8/EX8.7/8_7.sce @@ -0,0 +1,8 @@ +//Determine reaction RA
+//refer fig. 8.11
+//give virtual displacement at A
+//Applying virtual work principle
+//(RA-32+8+0+4)*delta(y)=0
+RA=32-8-4 //kN
+printf("Reaction at A is RA=%.2f kN",RA)
+
diff --git a/3886/CH8/EX8.7/8_7.txt b/3886/CH8/EX8.7/8_7.txt new file mode 100644 index 000000000..6e57c581d --- /dev/null +++ b/3886/CH8/EX8.7/8_7.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.7.sce', -1)
+Reaction at A is RA=20.00 kN
\ No newline at end of file diff --git a/3886/CH8/EX8.8/8_8.sce b/3886/CH8/EX8.8/8_8.sce new file mode 100644 index 000000000..0c9cacbb7 --- /dev/null +++ b/3886/CH8/EX8.8/8_8.sce @@ -0,0 +1,8 @@ +//Determine reaction RA
+//refer fig.8.12
+//Applying virtual work principle
+//give virtual displacement at A
+//Virtual work equation is
+//(RA-20+8)*delta(y)=0
+RA=20-8 //kN
+printf("The reaction RA=%.2d kN",RA)
\ No newline at end of file diff --git a/3886/CH8/EX8.8/8_8.txt b/3886/CH8/EX8.8/8_8.txt new file mode 100644 index 000000000..805fc6b8f --- /dev/null +++ b/3886/CH8/EX8.8/8_8.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\8. Virtual work\8.8.sce', -1)
+The reaction RA=12 kN
\ No newline at end of file diff --git a/3886/CH9/EX9.1/9_1.sce b/3886/CH9/EX9.1/9_1.sce new file mode 100644 index 000000000..81138e071 --- /dev/null +++ b/3886/CH9/EX9.1/9_1.sce @@ -0,0 +1,6 @@ +//Locating the centroid of T section
+//Refer fig. 9.10
+//due to symmetry the centroid lies on y axis
+//distance of centroid from top is
+ybar=(100*20*10+20*100*70)/(100*20+20*100)
+printf("Centroid of T-section is on the symmetric axis at a distance of %.2f mm from the top",ybar)
diff --git a/3886/CH9/EX9.1/9_1.txt b/3886/CH9/EX9.1/9_1.txt new file mode 100644 index 000000000..b227287c1 --- /dev/null +++ b/3886/CH9/EX9.1/9_1.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.1.sce', -1)
+Centroid of T-section is on the symmetric axis at a distance of 40.00 mm from the top
\ No newline at end of file diff --git a/3886/CH9/EX9.10/9_10.sce b/3886/CH9/EX9.10/9_10.sce new file mode 100644 index 000000000..af8e232a5 --- /dev/null +++ b/3886/CH9/EX9.10/9_10.sce @@ -0,0 +1,13 @@ +//Moment of inertia of L-section
+//Divide the section into two rectangles A1 and A2
+A1=125*10 //mm^2
+A2=75*10 //mm^2
+A=A1+A2 //mm^2
+xbar=((1250*5)+750*(10+75/2))/A //mm
+ybar=((1250*125/2)+(750*5))/A //mm
+Ixx=((10*125^3)/12)+(1250*21.56^2)+((75/12)*10^3)+(750*39.94^2) //mm^4
+Iyy=((125*10^3)/12)+(1250*15.94^2)+((10*75^3)/12)+(750*26.56^2) //mm^4
+Izz=Ixx+Iyy //mm^4
+printf("\nIxx=%.2f mm^4\nIyy=%.2f mm^4\nIzz=%.2f mm^4",Ixx,Iyy,Izz)
+
+
diff --git a/3886/CH9/EX9.10/9_10.txt b/3886/CH9/EX9.10/9_10.txt new file mode 100644 index 000000000..78517996d --- /dev/null +++ b/3886/CH9/EX9.10/9_10.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.10.sce', -1)
+
+Ixx=3411298.87 mm^4
+Iyy=1208658.87 mm^4
+Izz=4619957.73 mm^4
\ No newline at end of file diff --git a/3886/CH9/EX9.11/9_11.sce b/3886/CH9/EX9.11/9_11.sce new file mode 100644 index 000000000..7ecc83b92 --- /dev/null +++ b/3886/CH9/EX9.11/9_11.sce @@ -0,0 +1,11 @@ +//Moment of inertia of I section
+//Refer fig. 9.36
+A1=200*9 //mm^2
+A2=(250-9*2)*6.7 //mm^2
+A3=200*9 //mm^2
+A=A1+A2+A3 //mm^2
+Ixx=((200*9^3)/12)+(1800*120.5^2)+((6.7*232^3)/12)+(0)+((200*9^3)/12)+(1800*120.5^2) //mm^4
+Iyy=((9*200^3)/12)+((232*6.7^3)/12)+((9*200^3)/12) //mm^4
+Izz=Ixx+Iyy //mm^4
+printf("\nIxx=%.2f mm^4\nIyy=%.2f mm^4\nIzz=%.2f mm^4",Ixx,Iyy,Izz)
+//The answers vary due to round off error
diff --git a/3886/CH9/EX9.11/9_11.txt b/3886/CH9/EX9.11/9_11.txt new file mode 100644 index 000000000..7646a75cc --- /dev/null +++ b/3886/CH9/EX9.11/9_11.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.11.sce', -1)
+
+Ixx=59269202.13 mm^4
+Iyy=12005814.75 mm^4
+Izz=71275016.88 mm^4
\ No newline at end of file diff --git a/3886/CH9/EX9.12/9_12.sce b/3886/CH9/EX9.12/9_12.sce new file mode 100644 index 000000000..a4fa20ad3 --- /dev/null +++ b/3886/CH9/EX9.12/9_12.sce @@ -0,0 +1,9 @@ +//Second moment of Inertia
+//refer fig. 9.37
+A1=100*13.5 //mm^2
+A2=(400-27)*8.1 //mm^2
+A3=100*13.5 //mm^2
+A=A1+A2+A3 //mm^2
+Ixx=((100*13.5^3)/12)+(1350*193.25^2)+((8.1*373^3)/12)+((100*13.5^3)/12)+(1350*193.25^2) //mm^4
+Iyy=((13.5*100^3)/12)+(1350*24.27^2)+((373*8.1^3)/12)+(3021.3*21.68^2)+((13.5*100^3)/12)+(1350*24.27^2)
+printf("\nIxx=%.2d mm^4\nIyy=%.2d mm^4",Ixx,Iyy)
diff --git a/3886/CH9/EX9.12/9_12.txt b/3886/CH9/EX9.12/9_12.txt new file mode 100644 index 000000000..b4cad0683 --- /dev/null +++ b/3886/CH9/EX9.12/9_12.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.12.sce', -1)
+
+Ixx=135903228 mm^4
+Iyy=5276986 mm^4
\ No newline at end of file diff --git a/3886/CH9/EX9.13/9_13.sce b/3886/CH9/EX9.13/9_13.sce new file mode 100644 index 000000000..ad31e5007 --- /dev/null +++ b/3886/CH9/EX9.13/9_13.sce @@ -0,0 +1,19 @@ +//Polar moment of inertia
+//Refer fig.9.38
+//section can be divided into three triangles
+A1=80*12 //mm^2
+A2=(150-22)*12 //mm^2
+A3=120*10 //mm^2
+A=A1+A2+A3 //mm^2
+Ixx=((80*12^3)/12)+(960*74.22^2)+((12*128^3)/12)+(1536*4.22^2)+((120*10^3)/12)+(1200*64.78^2) //mm^4
+Iyy=((12*80^3)/12)+((128*12^3)/12)+((10*120^3)/12) //mm^4
+//Polar moment of Inertia (Izz)
+Izz=Ixx+Iyy //mm^4
+kxx=sqrt(Ixx/A) //mm
+Kyy=sqrt(Iyy/A) //mm The answer provided in the textbook is wrong
+printf("\nIzz=%.2d mm^4\nkxx=%.2f mm\nkyy=%.2f mm",Izz,kxx,kyy)
+
+
+
+
+
diff --git a/3886/CH9/EX9.13/9_13.txt b/3886/CH9/EX9.13/9_13.txt new file mode 100644 index 000000000..553fb7d87 --- /dev/null +++ b/3886/CH9/EX9.13/9_13.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.13.sce', -1)
+
+Izz=14440459 mm^4
+kxx=58.09 mm
+kyy=31.21 mm
\ No newline at end of file diff --git a/3886/CH9/EX9.14/9_14.sce b/3886/CH9/EX9.14/9_14.sce new file mode 100644 index 000000000..f1a6022b9 --- /dev/null +++ b/3886/CH9/EX9.14/9_14.sce @@ -0,0 +1,13 @@ +//Determine moment of inertia
+//refer fig.9.39
+A1=100*30 //mm^2
+A2=100*25 //mm^2
+A3=200*20 //mm^2
+A4=87.5*20/2 //mm^2
+A5=87.5*20/2 //mm^2
+A=A1+A2+A3+A4+A5 //mm^2
+ybar=(3000*135+2500*70+4000*10+875*(20/3+20)*2)/A //mm
+Ixx=((100*30*30*30)/(12))+(3000*(75.74^2))+((25*(100^3))/(12))+(2500*(10.74^2))+((200*(20^3))/(12))+(4000*(49.26^2))+((87.5*(20^3))/(36))+(875*(32.59^2))+((87.5*(20^3))/(36))+(875*(32.59^2)) //mm^4
+Iyy=((30*(100^3))/(12))+((100*(25^3))/(12))+((20*(200^3))/(12))+((20*(87.5^3))/(36))+(875*(41.66^2))+((20*(87.5^3))/(36))+(875*(41.66^2)) //mm^4
+printf("\nIxx=%.2d mm^4\nIyy=%.2d mm^4",Ixx,Iyy)
+//The answers vary due to round off error
diff --git a/3886/CH9/EX9.14/9_14.txt b/3886/CH9/EX9.14/9_14.txt new file mode 100644 index 000000000..8a4ef1863 --- /dev/null +++ b/3886/CH9/EX9.14/9_14.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.14.sce', -1)
+
+Ixx=31543446 mm^4
+Iyy=19745121 mm^4
\ No newline at end of file diff --git a/3886/CH9/EX9.15/9_15.sce b/3886/CH9/EX9.15/9_15.sce new file mode 100644 index 000000000..c4a5b88a8 --- /dev/null +++ b/3886/CH9/EX9.15/9_15.sce @@ -0,0 +1,4 @@ +//Moment of inertia of section
+//refer fig. 9.40
+IAB=((400*20^3)/12)+(400*20*10^2)+(((100*10^3)/12)+100*10*(20+5)^2)*2+(((10*380^3)/12)+10*380*(30+190)^2)*2+(((100*10^3)/12)+100*10*(20+10+380+5)^2)*2 //mm^4
+printf("IAB=%.2d mm^4",IAB)
diff --git a/3886/CH9/EX9.15/9_15.txt b/3886/CH9/EX9.15/9_15.txt new file mode 100644 index 000000000..3aa3569e1 --- /dev/null +++ b/3886/CH9/EX9.15/9_15.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.15.sce', -1)
+IAB=806093333 mm^4
\ No newline at end of file diff --git a/3886/CH9/EX9.16/9_16.sce b/3886/CH9/EX9.16/9_16.sce new file mode 100644 index 000000000..eefe5eedd --- /dev/null +++ b/3886/CH9/EX9.16/9_16.sce @@ -0,0 +1,7 @@ +//Moment of inertia
+//refer fig.9.41
+sumAiyi=250*10*5+2*40*10*(10+20)+40*10*(10+5)+40*10*255+250*10*(10+125) //mm^3
+A=2*250*10+40*10*4 //mm^2
+ybar=sumAiyi/A
+Ixx=((250*10^3)/12)+(250*10*(73.03-5)^2)+(((10*40^3)/12)+40*10*(73.03-30)^2)*2+((40*10^3)/12)+40*10*(73.03-15)^2+((10*250^3)/12)+250*10*(73.03-135)^2+((40*10^3)/12)+40*10*(73.03-255)^2 //mm^4
+printf("Ixx=%.2d mm^4",Ixx)
diff --git a/3886/CH9/EX9.16/9_16.txt b/3886/CH9/EX9.16/9_16.txt new file mode 100644 index 000000000..60165db26 --- /dev/null +++ b/3886/CH9/EX9.16/9_16.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.16.sce', -1)
+Ixx=50399393 mm^4
\ No newline at end of file diff --git a/3886/CH9/EX9.17/9_17.sce b/3886/CH9/EX9.17/9_17.sce new file mode 100644 index 000000000..c396c0e98 --- /dev/null +++ b/3886/CH9/EX9.17/9_17.sce @@ -0,0 +1,9 @@ +//moment of inertia
+//refer fig.9.42
+sumAiyi=(600*15)*((600/2)+20)+140*10*2*(70+30)+150*10*2*(5+20)+400*20*10 //mm^3
+A=600*15+140*10*2+150*10*2+400*20 //mm^2
+ybar=sumAiyi/A //mm
+Ixx=((15*(600^3))/(12))+(600*15*((145.39-320)^2))+((10*2*(140^3))/(12))+(1400*2*((145.39-100)^2))+((150*2*(10^3))/(12))+(1500*2*((145.39-15)^2))+((400*(20^3))/(12))+(400*20*((145.39-10)^2))
+printf("Ixx=%.2f mm^4",Ixx)
+//The answer provided in the textbook is wrong
+
diff --git a/3886/CH9/EX9.17/9_17.txt b/3886/CH9/EX9.17/9_17.txt new file mode 100644 index 000000000..b9b667f95 --- /dev/null +++ b/3886/CH9/EX9.17/9_17.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.17.sce', -1)
+Ixx=752679847.88 mm^4
\ No newline at end of file diff --git a/3886/CH9/EX9.18/9_18.sce b/3886/CH9/EX9.18/9_18.sce new file mode 100644 index 000000000..e4c6b47e3 --- /dev/null +++ b/3886/CH9/EX9.18/9_18.sce @@ -0,0 +1,4 @@ +//Compute moment of inertia
+//refer fig. 9.43
+Ixx=((125*60^3)/36)+(125*(60/2)*(60+60/3)^2)+((125*60^3)/36)+(125*(60/2)*(2*60/3)^2)+((125*60^3)/36)+(125*(60/2)*(60/3)^2)+((125*60^3)/36)+(125*(60/2)*(60/3)^2) //mm^4
+printf("Ixx=%.2d mm^4",Ixx)
diff --git a/3886/CH9/EX9.18/9_18.txt b/3886/CH9/EX9.18/9_18.txt new file mode 100644 index 000000000..89bd66c2c --- /dev/null +++ b/3886/CH9/EX9.18/9_18.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.18.sce', -1)
+Ixx=36000000 mm^4
\ No newline at end of file diff --git a/3886/CH9/EX9.19/9_19.sce b/3886/CH9/EX9.19/9_19.sce new file mode 100644 index 000000000..04ea5aa68 --- /dev/null +++ b/3886/CH9/EX9.19/9_19.sce @@ -0,0 +1,5 @@ +//moment of inertia of shaded region
+//refer fig. 9.44
+//The figure is divided into simple geometry
+IAB=((80*80^3)/12)+((%pi*80^4)/128)-((%pi*40^4)/64) //mm^4
+printf("IAB=%.2d mm^4",IAB)
diff --git a/3886/CH9/EX9.19/9_19.txt b/3886/CH9/EX9.19/9_19.txt new file mode 100644 index 000000000..1e51c7115 --- /dev/null +++ b/3886/CH9/EX9.19/9_19.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.19.sce', -1)
+IAB=4292979 mm^4
\ No newline at end of file diff --git a/3886/CH9/EX9.2/9_2.sce b/3886/CH9/EX9.2/9_2.sce new file mode 100644 index 000000000..9f5a5dd70 --- /dev/null +++ b/3886/CH9/EX9.2/9_2.sce @@ -0,0 +1,11 @@ +//Centroid of angle
+//refer fig. 9.11
+//the given figure can be divided into two rectangles
+A1=150*12 //mm^2
+A2=(200-12)*12 //mm^2
+//total area
+A=A1+A2 //mm^2
+xbar=(1800*75+2256*6)/(4056) //mm
+ybar=(1800*6+2256*106)/(4056) //mm
+printf("The centroid is at \nxbar=%.2f mm\nybar=%.2f mm ",xbar,ybar)
+
diff --git a/3886/CH9/EX9.2/9_2.txt b/3886/CH9/EX9.2/9_2.txt new file mode 100644 index 000000000..e16ddcce8 --- /dev/null +++ b/3886/CH9/EX9.2/9_2.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.2.sce', -1)
+The centroid is at
+xbar=36.62 mm
+ybar=61.62 mm
\ No newline at end of file diff --git a/3886/CH9/EX9.20/9_20.sce b/3886/CH9/EX9.20/9_20.sce new file mode 100644 index 000000000..8acc440c9 --- /dev/null +++ b/3886/CH9/EX9.20/9_20.sce @@ -0,0 +1,8 @@ +//Find second moment of inertia
+ybar=28.47 //mm
+xbar=39.21 //mm
+Ixx=((80*20^3)/36)+(80*(20/2)*(60-(2*20/3)-28.47)^2)+((80*40^3)/12)+(80*40*(28.47-20)^2)-((0.0068598*20^4)+(20^2)*(%pi/2)*(28.47-((4*20)/(3*%pi)))^2) //mm^4
+Iyy=((20*30^3)/36)+((20/2)*30*(39.21-(2*30/3))^2)+((20*50^3)/36)+(20/2)*50*(39.21-(30+50/3))^2+((40*80^3)/12)+(40*80*(39.21-40)^2)-(%pi*40^4)/(2*64)-((%pi)*(40^2)*(40-39.21)^2)/(4*2) //mm^4
+printf("\nIxx=%.2d mm^4\nIyy=%.2d mm^4",Ixx,Iyy)
+
+
diff --git a/3886/CH9/EX9.20/9_20.txt b/3886/CH9/EX9.20/9_20.txt new file mode 100644 index 000000000..ba2e7ee4b --- /dev/null +++ b/3886/CH9/EX9.20/9_20.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.20.sce', -1)
+
+Ixx=686944 mm^4
+Iyy=1868392 mm^4
\ No newline at end of file diff --git a/3886/CH9/EX9.3/9_3.sce b/3886/CH9/EX9.3/9_3.sce new file mode 100644 index 000000000..928930bcc --- /dev/null +++ b/3886/CH9/EX9.3/9_3.sce @@ -0,0 +1,16 @@ +//locating centroid
+//refer fig. 9.12
+//due to symmetry centroid must lie on y-axis
+xbar=0
+A1=100*20 //mm^2
+//for A1
+y1=30+100+20/2 //mm
+//for A2
+A2=100*20 //mm^2
+y2=30+100/2 //mm
+A3=150*30 //mm^2
+y3=30/2 //mm
+A=2000+2000+4500
+ybar=(A1*y1+A2*y2+A3*y3)/A //mm
+printf("The centroid is on the symmetric axis at a distance of %.2f mm from the bottom as shown in figure 9.12",ybar)
+
diff --git a/3886/CH9/EX9.3/9_3.txt b/3886/CH9/EX9.3/9_3.txt new file mode 100644 index 000000000..6088ee43f --- /dev/null +++ b/3886/CH9/EX9.3/9_3.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.3.sce', -1)
+The centroid is on the symmetric axis at a distance of 59.71 mm from the bottom as shown in figure 9.12
\ No newline at end of file diff --git a/3886/CH9/EX9.4/9_4.sce b/3886/CH9/EX9.4/9_4.sce new file mode 100644 index 000000000..114872319 --- /dev/null +++ b/3886/CH9/EX9.4/9_4.sce @@ -0,0 +1,20 @@ +//Centroid of dam
+//refer fig. 9.13 and select axis accordingly
+A1=2*6/2 //m^2
+A2=2*7.5 //m^2
+A3=3*5/2 //m^2
+A4=1*4 //m^2
+A=A1+A2+A3+A4 //m^2
+//centroid of simple geometries are
+x1=2*2/3 //m
+y1=6/3 //m
+x2=2+1 //m
+y2=7.5/2 //m
+x3=2+2+3/3 //m
+y3=1+5/3 //m
+x4=4+4/2 //m
+y4=0.5 //m
+xbar=(A1*x1+A2*x2+A3*x3+A4*x4)/(A) //m
+ybar=(A1*y1+A2*y2+A3*y3+A4*y4)/(A) //m
+printf("centroid is at\nxbar=%.3f mm\nybar=%.3f mm",xbar,ybar)
+
diff --git a/3886/CH9/EX9.4/9_4.txt b/3886/CH9/EX9.4/9_4.txt new file mode 100644 index 000000000..16dfa7458 --- /dev/null +++ b/3886/CH9/EX9.4/9_4.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.4.sce', -1)
+centroid is at
+xbar=3.523 mm
+ybar=2.777 mm
\ No newline at end of file diff --git a/3886/CH9/EX9.5/9_5.sce b/3886/CH9/EX9.5/9_5.sce new file mode 100644 index 000000000..c341eed72 --- /dev/null +++ b/3886/CH9/EX9.5/9_5.sce @@ -0,0 +1,21 @@ +//Determine centroid
+//refer fig. 9.14
+//This figure is divided into three simple figures
+A1=3*4/2 //m^2
+A2=6*4 //m^2
+A3=%pi*(1/2)*2^2 //m^2
+A=A1+A2+A3 //m^2
+//Co-ordinates of centroid
+x1=6+3/3 //m
+ya=4/3 //m
+x2=3 //m
+y2=2 //m
+R=2
+x3=(-4*R)/(3*%pi) //m
+y3=2 //m
+xbar=(A1*x1+A2*x2+A3*x3)/(A) //m
+ybar=(A1*ya+A2*y2+A3*y3)/(A) //m
+printf("\nxbar=%.3f m\nybar=%.3f m",xbar,ybar)
+
+
+
diff --git a/3886/CH9/EX9.5/9_5.txt b/3886/CH9/EX9.5/9_5.txt new file mode 100644 index 000000000..b987a37fd --- /dev/null +++ b/3886/CH9/EX9.5/9_5.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.5.sce', -1)
+
+xbar=2.995 m
+ybar=1.890 m
\ No newline at end of file diff --git a/3886/CH9/EX9.6/9_6.sce b/3886/CH9/EX9.6/9_6.sce new file mode 100644 index 000000000..d097410de --- /dev/null +++ b/3886/CH9/EX9.6/9_6.sce @@ -0,0 +1,43 @@ +//Centroid of the gusset plate
+//refer fig. 9.15
+//The composite area is divided into algebraic sum and differences of simple geometries
+//for rectangle
+A1=160*280 //mm^2
+x1=140 //mm
+y1=80 //mm
+//for triangle
+A2=280*40/2 //mm^2
+x2=2*280/3 //mm
+y2=160+40/3 //mm
+//1st hole
+A3=(-%pi*21.5^2)/(4) //mm^2
+x3=70 //mm
+y3=50 //mm
+//second hole
+A4=-363.05 //mm^2
+x4=140 //mm
+y4=50 //mm
+//third hole
+A5=-363.05 //mm^2
+x5=210 //mm
+y5=50 //mm
+//fourth hole
+A6=-363.05 //mm^2
+x6=70 //mm
+y6=120 //mm
+//fifth hole
+A7=-363.05 //mm^2
+x7=140 //mm
+y7=130 //mm
+//sixth hole
+A8=-363.05 //mm^2
+x8=210 //mm
+y8=140 //mm
+A=A1+A2+A3+A4+A5+A6+A7+A8 //mm^2
+sumAixi=A1*x1+A2*x2+A3*x3+A4*x4+A5*x5+A6*x6+A7*x7+A8*x8 //mm^3
+xbar=sumAixi/A //mm
+sumAiyi=A1*y1+A2*y2+A3*y3+A4*y4+A5*y5+A6*y6+A7*y7+A8*y8
+ybar=sumAiyi/A //mm
+printf("\xbar=%.3f mm \nybar=%.3f mm",xbar,ybar)
+
+
diff --git a/3886/CH9/EX9.6/9_6.txt b/3886/CH9/EX9.6/9_6.txt new file mode 100644 index 000000000..9fad73cab --- /dev/null +++ b/3886/CH9/EX9.6/9_6.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.6.sce', -1)
+xbar=145.419 mm
+ybar=90.387 mm
\ No newline at end of file diff --git a/3886/CH9/EX9.7/9_7.sce b/3886/CH9/EX9.7/9_7.sce new file mode 100644 index 000000000..47a4f987a --- /dev/null +++ b/3886/CH9/EX9.7/9_7.sce @@ -0,0 +1,8 @@ +//Determine co-ordinates
+//total area
+A=200*150-(100*75/2)-((%pi*100^2)/(4)) //mm^2
+xc=2375000/26250 //mm
+yc=1781250/26250 //mm
+printf("\xc=%.3f mm\nyc=%.3f mm",xc,yc)
+
+
diff --git a/3886/CH9/EX9.7/9_7.txt b/3886/CH9/EX9.7/9_7.txt new file mode 100644 index 000000000..c4bac107a --- /dev/null +++ b/3886/CH9/EX9.7/9_7.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.7.sce', -1)
+xc=90.476 mm
+yc=67.857 mm
\ No newline at end of file diff --git a/3886/CH9/EX9.8/9_8.sce b/3886/CH9/EX9.8/9_8.sce new file mode 100644 index 000000000..f8406efd5 --- /dev/null +++ b/3886/CH9/EX9.8/9_8.sce @@ -0,0 +1,25 @@ +//Locate centroid
+//refer fig.9.17
+x=40 //mm
+A1=168*x^2
+A2=12*x^2
+A3=-16*x^2
+A4=-8*%pi*x^2
+A5=-4*%pi*x^2
+x1=7*x
+x2=16*x
+x3=2*x
+x4=6*x
+x5=12.3023*x
+y1=6*x
+y2=4*x/3
+y3=10*x
+y4=(16*x/(3*%pi))
+y5=10.3023*x
+A=126.3009*x^2
+sumAixi=1030.6083*x^3
+sumAiyi=691.8708*x^3
+xbar=1030.6083*x/126.3009 //mm
+ybar=691.8708*x/126.3009 //mm
+printf("centroid is at (%.2f, %.2f)",xbar,ybar)
+
diff --git a/3886/CH9/EX9.8/9_8.txt b/3886/CH9/EX9.8/9_8.txt new file mode 100644 index 000000000..b795417e5 --- /dev/null +++ b/3886/CH9/EX9.8/9_8.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.8.sce', -1)
+centroid is at (326.40, 219.12)
\ No newline at end of file diff --git a/3886/CH9/EX9.9/9_9.sce b/3886/CH9/EX9.9/9_9.sce new file mode 100644 index 000000000..b50f36305 --- /dev/null +++ b/3886/CH9/EX9.9/9_9.sce @@ -0,0 +1,13 @@ +//Determine moment of inertia
+//refer fig.9.34
+//composite section can be divided into simple ones
+A1=150*10 //mm^2
+A2=140*10 //mm^2
+A=A1+A2 //mm^2
+//due to symmetry centroid lies on y-axis
+ybar=(1500*5+1400*(10+70))/(2900) //mm
+Ixx=((150/12)*10^3)+(1500*36.21^2)+((10/12)*140^3)+(1400*38.79^2) //mm^4
+Iyy=((10*150^3)/(12))+((140*10^3)/(12)) //mm^4
+kxx=sqrt(Ixx/A) //mm
+kyy=sqrt(Iyy/A) //mm
+printf("\nIxx=%.2f mm^4\nIyy=%.2f mm^4\nkxx=%.2f mm\nkyy=%.2f mm",Ixx,Iyy,kxx,kyy)
diff --git a/3886/CH9/EX9.9/9_9.txt b/3886/CH9/EX9.9/9_9.txt new file mode 100644 index 000000000..fbb8c2e0d --- /dev/null +++ b/3886/CH9/EX9.9/9_9.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\9. Centroid and moment of inertia\9.9.sce', -1)
+
+Ixx=6372442.56 mm^4
+Iyy=2824166.67 mm^4
+kxx=46.88 mm
+kyy=31.21 mm
\ No newline at end of file |