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{
"metadata": {
"name": "chapter6.ipynb"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 6: Friction"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6-1,Page No:126"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"import numpy as np\n",
"\n",
"# Initilization of variables\n",
"\n",
"m=5 #kg # mass of the bock\n",
"g=9.81 # m/s^2 # acceleration due to gravity\n",
"theta=15 # degree # angle made by the forces (P1 & P2) with the horizontal of the block\n",
"mu=0.4 #coefficient of static friction\n",
"\n",
"#Calculations\n",
"\n",
"# Case 1. Where P1 is the force required to just pull the bock\n",
"\n",
"# Solving eqn's 1 & 2 using matrix\n",
"A=np.array([[cos(theta*(pi/180)), -mu],[sin(theta*(pi/180)), 1]])\n",
"B=np.array([0,(m*g)])\n",
"C=np.linalg.solve(A,B)\n",
"\n",
"# Calculations \n",
"\n",
"# Case 2. Where P2 is the force required to push the block\n",
"\n",
"# Solving eqn's 1 & 2 using matrix\n",
"P=np.array([[-cos(theta*(pi/180)), mu],[-sin(theta*(pi/180)) ,1]])\n",
"Q=np.array([0,(m*g)])\n",
"R=np.linalg.solve(P,Q)\n",
"\n",
"# Results\n",
"\n",
"print\"The required pull force P1 is \",round(C[0],2),\"N\" #answer in textbook is wrong\n",
"print\"The required push force P2 is \",round(R[0]),\"N\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required pull force P1 is 18.35 N\n",
"The required push force P2 is 23.0 N\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6-4,Page No:129"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Initilization of variables\n",
"\n",
"W1=50 # N # weight of the first block\n",
"W2=50 # N # weight of the second block\n",
"mu_1=0.3 # coefficient of friction between the inclined plane and W1\n",
"mu_2=0.2 # coefficient of friction between the inclined plane and W2\n",
"\n",
"# Calculations\n",
"\n",
"# On adding eq'ns 1&3 and substuting the values of N1 & N2 from eqn's 2&4 in this and on solving for alpha we get,\n",
"\n",
"alpha=(arctan(((mu_1*W1)+(mu_2*W2))/(W1+W2)))*(180/pi) # degrees\n",
"\n",
"# Results\n",
"\n",
"print\"The inclination of the plane is \",round(alpha),\"degree\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The inclination of the plane is 14.0 degree\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6-7,Page No:133"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Initilization of variables\n",
"\n",
"M=2000 # kg # mass of the car\n",
"mu=0.3 # coefficient of static friction between the tyre and the road\n",
"g=9.81 # m/s^2 # acc. due to gravity\n",
"\n",
"# Calculations\n",
"\n",
"# Divide eqn 1 by eqn 2, We get\n",
"theta=arctan(mu)*(180/pi) #degree\n",
"\n",
"# Results\n",
"\n",
"print\"The angle of inclination is \",round(theta,1),\"degree\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The angle of inclination is 16.7 degree\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6-9,Page No:135"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"import numpy as np\n",
"\n",
"# Initilization of variabes\n",
"\n",
"Wa=1000 #N # weight of block A\n",
"Wb=500 #N # weight of block B\n",
"theta=15 # degree # angle of the wedge\n",
"mu=0.2 # coefficient of friction between the surfaces in contact\n",
"phi=7.5 # degrees # used in case 2\n",
"\n",
"# Caculations \n",
"\n",
"# CASE (a)\n",
"\n",
"# consider the equilibrium of upper block A\n",
"# rearranging eq'ns 1 &2 and solving them using matrix for N1 & N2\n",
"A=np.array([[1 ,-0.4522],[-0.2 ,0.914]])\n",
"B=np.array([0,1000])\n",
"C=np.linalg.solve(A,B)\n",
"\n",
"# Now consider the equilibrium of lower block B\n",
"# From eq'n 4\n",
"N3=Wb+(C[1]*cos(theta*(pi/180)))-(mu*C[1]*sin(theta*(pi/180))) #N\n",
"# Now from eq'n 3\n",
"P=(mu*N3)+(mu*C[1]*cos(theta*(pi/180)))+(C[1]*sin(theta*(pi/180))) # N\n",
"\n",
"# CASE (b)\n",
"\n",
"# The eq'n for required coefficient for the wedge to be self locking is,\n",
"mu_req=(theta*pi)/360 # multiplying with (pi/180) to convert it into radians\n",
"\n",
"# Results\n",
"\n",
"print\"The minimum horizontal force (P) which should be applied to raise the block is \",round(P),\"N\"\n",
"print\"The required coefficient for the wedge to be self locking is \",round(mu_req,4) #answer in textbook is wrong\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minimum horizontal force (P) which should be applied to raise the block is 871.0 N\n",
"The required coefficient for the wedge to be self locking is 0.1309\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6-13.Page No:141"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"import numpy as np\n",
"\n",
"# Initilization of variables\n",
"\n",
"P=100 #N # force acting at 0.2 m from A\n",
"Q=200 #N # force acting at any distance x from B\n",
"l=1 #m # length of the bar\n",
"theta=45 #degree #angle made by the normal reaction at A&B with horizontal\n",
"\n",
"# Calculations\n",
"\n",
"#solving eqn's 1 & 2 using matrix for Ra & Rb,\n",
"A=np.array([[1, -1],[sin(theta*(pi/180)) ,sin(theta*(pi/180))]])\n",
"B=np.array([0,(P+Q)])\n",
"C=np.linalg.solve(A,B)\n",
"\n",
"# Now take moment about B\n",
"x=((C[0]*l*sin(theta*(pi/180)))-(P*(l-0.2)))/200 #m # here 0.2 is the distance where 100 N load lies from A\n",
"\n",
"# Results\n",
"\n",
"print\"The minimum value of x at which the load Q=200 N may be applied before slipping impends is \",round(x,2),\"m\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minimum value of x at which the load Q=200 N may be applied before slipping impends is 0.35 m\n"
]
}
],
"prompt_number": 20
}
],
"metadata": {}
}
]
}
|
