summaryrefslogtreecommitdiff
path: root/3886/CH5
diff options
context:
space:
mode:
Diffstat (limited to '3886/CH5')
-rw-r--r--3886/CH5/EX5.1/5_1.sce19
-rw-r--r--3886/CH5/EX5.1/5_1.txt5
-rw-r--r--3886/CH5/EX5.10/5_10.sce12
-rw-r--r--3886/CH5/EX5.10/5_10.txt3
-rw-r--r--3886/CH5/EX5.11/5_11.sce14
-rw-r--r--3886/CH5/EX5.11/5_11.txt2
-rw-r--r--3886/CH5/EX5.12/5_12.sce15
-rw-r--r--3886/CH5/EX5.12/5_12.txt8
-rw-r--r--3886/CH5/EX5.15/5_15.sce25
-rw-r--r--3886/CH5/EX5.15/5_15.txt4
-rw-r--r--3886/CH5/EX5.16/5_16.sce17
-rw-r--r--3886/CH5/EX5.16/5_16.txt4
-rw-r--r--3886/CH5/EX5.2/5_2.sce13
-rw-r--r--3886/CH5/EX5.2/5_2.txt4
-rw-r--r--3886/CH5/EX5.3/5_3.sce18
-rw-r--r--3886/CH5/EX5.3/5_3.txt5
-rw-r--r--3886/CH5/EX5.4/5_4.sce15
-rw-r--r--3886/CH5/EX5.4/5_4.txt4
-rw-r--r--3886/CH5/EX5.5/5_5.sce17
-rw-r--r--3886/CH5/EX5.5/5_5.txt4
-rw-r--r--3886/CH5/EX5.6/5_6.sce19
-rw-r--r--3886/CH5/EX5.6/5_6.txt4
-rw-r--r--3886/CH5/EX5.7/5_7.sce21
-rw-r--r--3886/CH5/EX5.7/5_7.txt5
-rw-r--r--3886/CH5/EX5.8/5_8.sce17
-rw-r--r--3886/CH5/EX5.8/5_8.txt3
-rw-r--r--3886/CH5/EX5.9/5_9.sce8
-rw-r--r--3886/CH5/EX5.9/5_9.txt3
28 files changed, 288 insertions, 0 deletions
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
generated by cgit (git 2.34.1) at 2025-03-30 21:04:52 +0000