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author | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
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committer | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
commit | 7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch) | |
tree | 449d555969bfd7befe906877abab098c6e63a0e8 /3886/CH15 | |
parent | d1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff) | |
download | Scilab-TBC-Uploads-7bc77cb1ed33745c720952c92b3b2747c5cbf2df.tar.gz Scilab-TBC-Uploads-7bc77cb1ed33745c720952c92b3b2747c5cbf2df.tar.bz2 Scilab-TBC-Uploads-7bc77cb1ed33745c720952c92b3b2747c5cbf2df.zip |
Diffstat (limited to '3886/CH15')
-rw-r--r-- | 3886/CH15/EX15.10/15_10.sce | 11 | ||||
-rw-r--r-- | 3886/CH15/EX15.10/15_10.txt | 5 | ||||
-rw-r--r-- | 3886/CH15/EX15.11/15_11.sce | 11 | ||||
-rw-r--r-- | 3886/CH15/EX15.11/15_11.txt | 5 | ||||
-rw-r--r-- | 3886/CH15/EX15.12/15_12.sce | 33 | ||||
-rw-r--r-- | 3886/CH15/EX15.12/15_12.txt | 5 | ||||
-rw-r--r-- | 3886/CH15/EX15.2/15_2.sce | 13 | ||||
-rw-r--r-- | 3886/CH15/EX15.2/15_2.txt | 5 | ||||
-rw-r--r-- | 3886/CH15/EX15.3/15_3.sce | 13 | ||||
-rw-r--r-- | 3886/CH15/EX15.3/15_3.txt | 5 | ||||
-rw-r--r-- | 3886/CH15/EX15.4/15_4.sce | 12 | ||||
-rw-r--r-- | 3886/CH15/EX15.4/15_4.txt | 4 | ||||
-rw-r--r-- | 3886/CH15/EX15.5/15_5.sce | 12 | ||||
-rw-r--r-- | 3886/CH15/EX15.5/15_5.txt | 4 | ||||
-rw-r--r-- | 3886/CH15/EX15.6/15_6.sce | 20 | ||||
-rw-r--r-- | 3886/CH15/EX15.6/15_6.txt | 6 | ||||
-rw-r--r-- | 3886/CH15/EX15.7/15_7.sce | 21 | ||||
-rw-r--r-- | 3886/CH15/EX15.7/15_7.txt | 5 | ||||
-rw-r--r-- | 3886/CH15/EX15.8/15_8.sce | 23 | ||||
-rw-r--r-- | 3886/CH15/EX15.8/15_8.txt | 5 | ||||
-rw-r--r-- | 3886/CH15/EX15.9/15_9.sce | 28 | ||||
-rw-r--r-- | 3886/CH15/EX15.9/15_9.txt | 5 |
22 files changed, 251 insertions, 0 deletions
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 |