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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": {}

-  }

- ]

-}
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