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-{

- "metadata": {

-  "name": "chapter7.ipynb"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter 7: Application Of Friction"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 7.7-1,Page No:152"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "# Initilization of variables\n",

-      "\n",

-      "d1=24 # cm # diameter of larger pulley\n",

-      "d2=12 # cm # diameter of smaller pulley\n",

-      "d=30 #cm # seperation betweem 1st & the 2nd pulley\n",

-      "pie=3.14\n",

-      "\n",

-      "# Calcuations\n",

-      "\n",

-      "r1=d1*0.5 #cm # radius of 1st pulley #1/2=0.5\n",

-      "r2=d2*0.5 #cm # radius of 2nd pulley #1/2=0.5\n",

-      "theta=(arcsin((r1-r2)/d))*(180/pi) #degrees \n",

-      "\n",

-      "# Angle of lap\n",

-      "beta_1=180+(2*theta) #degree # for larger pulley\n",

-      "beta_2=180-(2*theta) #degree # for smaller pulley\n",

-      "L=pie*(r1+r2)+(2*d)+((r1-r2)**(2/d)) #cm # Length of the belt\n",

-      "\n",

-      "# Results\n",

-      "\n",

-      "print\"The angle of lap for the larger pulley is \",round(beta_1,2),\"degree\"\n",

-      "print\"he angle of lap for the smaller pulley is \",round(beta_2,2),\"degree\"\n",

-      "print\"he length of pulley required is \",round(L,2),\"cm\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The angle of lap for the larger pulley is  203.07 degree\n",

-        "he angle of lap for the smaller pulley is  156.93 degree\n",

-        "he length of pulley required is  117.52 cm\n"

-       ]

-      }

-     ],

-     "prompt_number": 22

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 7.7-2,Page No:153"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "# Initilization of variables\n",

-      "\n",

-      "d1=0.6 #m # diameter of larger pulley\n",

-      "d2=0.3 #m diameter of smaller pulley\n",

-      "d=3.5 #m #separation between the pulleys\n",

-      "pie=3.14\n",

-      "\n",

-      "# Calculations\n",

-      "\n",

-      "r1=d1*0.5 #m # radius of larger pulley #1/2=0.5\n",

-      "r2=d2*0.5 #m # radius of smaller pulley #1/2=0.5\n",

-      "theta=arcsin((r1+r2)/d)*(180/pi) #degree\n",

-      "\n",

-      "# Angle of lap for both the pulleys is same, i.e\n",

-      "\n",

-      "beta=180+(2*theta) # degree\n",

-      "L=((pie*(r1+r2))+(2*d)+(((r1+r2)**2)/d)) #cm # Length of the belt\n",

-      "\n",

-      "# Results\n",

-      "\n",

-      "print\"The angle of lap for the pulley is \",round(beta,1),\"degree\"\n",

-      "print\"The length of pulley required is \",round(L,3),\"m\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The angle of lap for the pulley is  194.8 degree\n",

-        "The length of pulley required is  8.471 m\n"

-       ]

-      }

-     ],

-     "prompt_number": 29

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 7.7-4,Page No:161"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "# Initilization of variables\n",

-      "\n",

-      "W1=1000 #N\n",

-      "mu=0.25 #coefficient of friction\n",

-      "pie=3.14\n",

-      "beta=pie\n",

-      "T1=W1 # Tension in the 1st belt carrying W1\n",

-      "e=2.718 #constant\n",

-      "\n",

-      "# Calculations\n",

-      "\n",

-      "T2=T1/(e**(mu*beta)) #N # Tension in the 2nd belt\n",

-      "W2=T2 #N\n",

-      "\n",

-      "# Results\n",

-      "\n",

-      "print\"The minimum weight W2 to keep W1 in equilibrium is \",round(W2),\"N\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The minimum weight W2 to keep W1 in equilibrium is  456.0 N\n"

-       ]

-      }

-     ],

-     "prompt_number": 31

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 7.7-5,Page No:162"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "# Initilization of variables\n",

-      "\n",

-      "mu=0.5 # coefficient of friction between the belt and the wheel\n",

-      "W=100 #N\n",

-      "theta=45 #degree\n",

-      "e=2.718\n",

-      "Lac=0.75 #m # ength of the lever\n",

-      "Lab=0.25 #m\n",

-      "Lbc=0.50 #m\n",

-      "r=0.25 #m\n",

-      "pie=3.14 # constant\n",

-      "\n",

-      "# Calculations\n",

-      "\n",

-      "beta=((180+theta)*pie)/180 # radian # angle of lap\n",

-      "\n",

-      "# from eq'n 2\n",

-      "T1=(W*Lbc)/Lab #N \n",

-      "T2=T1/(e**(mu*beta)) #N # from eq'n 1\n",

-      "\n",

-      "# consider the F.B.D of the pulley and take moment about its center, we get Braking Moment (M)\n",

-      "M=r*(T1-T2) #N-m\n",

-      "\n",

-      "# Results\n",

-      "\n",

-      "print\"The braking moment (M) exerted by the vertical weight W is \",round(M,2),\"N-m\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The braking moment (M) exerted by the vertical weight W is  42.97 N-m\n"

-       ]

-      }

-     ],

-     "prompt_number": 32

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 7.7-6,Page No:163"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "# Initiization of variables\n",

-      "\n",

-      "W= 1000 #N # or 1kN\n",

-      "mu=0.3 # coefficient of friction between the rope and the cylinder\n",

-      "e=2.718 # constant\n",

-      "pie=3.14 # constant\n",

-      "alpha=90 # degree # since 2*alpha=180 egree\n",

-      "\n",

-      "# Calculations\n",

-      "\n",

-      "beta=2*pie*3 # radian # for 3 turn of the rope\n",

-      "\n",

-      "# Here T1 is the larger tension in that section of the rope which is about to slip\n",

-      "T1=W #N\n",

-      "F=W/e**(mu*(1/(sin(alpha*(pi/180))))*(beta)) #N # Here T2=F\n",

-      "\n",

-      "# Results\n",

-      "\n",

-      "print\"The  force required to suport the weight of 1000 N i.e 1kN is \",round(F,1),\"N\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The  force required to suport the weight of 1000 N i.e 1kN is  3.5 N\n"

-       ]

-      }

-     ],

-     "prompt_number": 34

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 7.7-7,Page No:163"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "# Initilization of variables\n",

-      "\n",

-      "Pw=50 #kW\n",

-      "T_max=1500 #N\n",

-      "v=10 # m/s # velocity of rope\n",

-      "w=4 # N/m # weight of rope\n",

-      "mu=0.2 # coefficient of friction \n",

-      "g=9.81 # m/s^2 # acceleration due to gravity\n",

-      "e=2.718 # constant\n",

-      "pie=3.14 # constant\n",

-      "alpha=30 # degree # since 2*alpha=60 \n",

-      "\n",

-      "# Calcuations\n",

-      "\n",

-      "beta=pie # radian # angle of lap\n",

-      "T_e=(w*v**2)/g # N # where T_e is the centrifugal tension\n",

-      "T1=(T_max)-(T_e) #N\n",

-      "T2=T1/(e**(mu*(1/sin(alpha*(pi/180)))*(beta))) #N # From eq'n T1/T2=e^(mu*cosec(alpha)*beta)\n",

-      "P=(T1-T2)*v*(10**-3) #kW # power transmitted by a single rope\n",

-      "N=Pw/P # Number of ropes required\n",

-      "\n",

-      "# Results\n",

-      "\n",

-      "print\"The number of ropes required to transmit 50 kW is \",round(N),\"Nos\"\n",

-      "# approx no of ropes is 5\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The number of ropes required to transmit 50 kW is  5.0 Nos\n"

-       ]

-      }

-     ],

-     "prompt_number": 36

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 7.7-8,Page No:164"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "# Initilization of variables\n",

-      "\n",

-      "d1=0.45 #m # diameter of larger pulley\n",

-      "d2=0.20 #m # diameter of smaller pulley\n",

-      "d=1.95 #m # separation between the pulley's\n",

-      "T_max=1000 #N # or 1kN which is the maximum permissible tension\n",

-      "mu=0.20 # coefficient of friction\n",

-      "N=100 # r.p.m # speed of larger pulley\n",

-      "e=2.718 # constant\n",

-      "pie=3.14 # constant\n",

-      "T_e=0 #N # as the data for calculating T_e is not given we assume T_e=0\n",

-      "\n",

-      "# Calculations\n",

-      "\n",

-      "r1=d1*0.5 #m # radius of larger pulley #1/2=0.5\n",

-      "r2=d2*0.5 #m # radius of smaller pulley #1/2=0.5\n",

-      "theta=arcsin((r1+r2)/d)*(180/pi) # degree\n",

-      "\n",

-      "# for cross drive the angle of lap for both the pulleys is same\n",

-      "beta=(180+(2*(theta)))*(pi/180) #radian\n",

-      "T1=T_max-T_e #N\n",

-      "T2=T1/(e**(mu*(beta))) #N # from formulae, T1/T2=e^(mu*beta)\n",

-      "v=((2*pie)*N*r1)/60 # m/s # where v=velocity of belt which is given as, v=wr=2*pie*N*r/60\n",

-      "P=(T1-T2)*v*(10**-3) #kW # Power\n",

-      "\n",

-      "# Results\n",

-      "\n",

-      "print\"The power transmitted by the cross belt drive is \",round(P,1),\"kW\"\n",

-      "# the approx answer is 1.3 kW The answer given in the book (i.e 1.81kW) is wrong.\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The power transmitted by the cross belt drive is  1.2 kW\n"

-       ]

-      }

-     ],

-     "prompt_number": 48

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 7.7-9,Page No:165"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "# Initilization of variabes\n",

-      "\n",

-      "b=0.1 #m #width of the belt\n",

-      "t=0.008 #m #thickness of the belt\n",

-      "v=26.67 # m/s # belt speed\n",

-      "pie=3.14 # constant\n",

-      "beta=165 # radian # angle of lap for the smaller belt\n",

-      "mu=0.3 # coefficient of friction\n",

-      "sigma_max=2 # MN/m^2 # maximum permissible stress in the belt\n",

-      "m=0.9 # kg/m # mass of the belt\n",

-      "g=9.81 # m/s^2\n",

-      "e=2.718 # constant\n",

-      "\n",

-      "# Calculations\n",

-      "\n",

-      "A=b*t # m^2 # cross-sectional area of the belt\n",

-      "T_e=m*v**2 # N # where T_e is the Centrifugal tension\n",

-      "T_max=(sigma_max)*(A)*(10**6) # N # maximum tension in the belt\n",

-      "T1=(T_max)-(T_e) # N \n",

-      "T2=T1/(e**((mu*pie*beta)/180)) #N # from formulae T1/T2=e^(mu*beta)\n",

-      "P=(T1-T2)*v*(10**-3) #kW # Power transmitted\n",

-      "T_o=(T1+T2)/2 # N # Initial tension\n",

-      "\n",

-      "# Now calculations to transmit maximum power\n",

-      "\n",

-      "Te=T_max/3 # N # max tension\n",

-      "u=(T_max/(3*m))**0.5 # m/s # belt speed for max power\n",

-      "T_1=T_max-Te # N # T1 for case 2\n",

-      "T_2=T_1/(e**((mu*pie*beta)/180)) # N \n",

-      "P_max=(T_1-T_2)*u*(10**-3) # kW # Max power transmitted\n",

-      "\n",

-      "# Results\n",

-      "\n",

-      "print\"The initial power transmitted is \",round(P,2),\"kW\"\n",

-      "print\"The initial tension in the belt is \",round(T_o,1),\"N\"\n",

-      "print\"The maximum power that can be transmitted is \",round(P_max,2),\"kW\" # the answer is approx 15.017kW wheres the answer 14.99kW given in book is wrong      \n",

-      "print\"The maximum power is transmitted at a belt speed of \",round(u,2),\"m/s\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The initial power transmitted is  14.8 kW\n",

-        "The initial tension in the belt is  682.3 N\n",

-        "The maximum power that can be transmitted is  15.02 kW\n",

-        "The maximum power is transmitted at a belt speed of  24.34 m/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 54

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 7.7-10,Page No:169"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "# Initilization of variables\n",

-      "\n",

-      "p=0.0125 # m # pitch of screw\n",

-      "d=0.1 #m # diameter of the screw\n",

-      "r=0.05 #m # radius of the screw\n",

-      "l=0.5 #m # length of the lever\n",

-      "W=50 #kN # load on the lever\n",

-      "mu=0.20 # coefficient of friction \n",

-      "pie=3.14 #constant\n",

-      "\n",

-      "# Calculations\n",

-      "\n",

-      "theta=arctan(p/(2*pie*r))*(180/pi) #degree # theta is the Helix angle\n",

-      "phi=arctan(mu)*(180/pi) # degree # phi is the angle of friction\n",

-      "\n",

-      "# Taking the leverage due to handle into account,force F1 required is,\n",

-      "F1=(W*(tan(theta*(pi/180)+phi*(pi/180))))*(r/l) #kN\n",

-      "\n",

-      "# To lower the load\n",

-      "F2=(W*(tan(theta*(pi/180)-phi*(pi/180))))*(r/l) #kN # -ve sign of F2 indicates force required is in opposite sense\n",

-      "E=(tan(theta*(pi/180))/tan(theta*(pi/180)+phi*(pi/180)))*100 # % # here E=eata=efficiency in %\n",

-      "\n",

-      "# Results\n",

-      "\n",

-      "print\"The force required (i.e F1) to raise the weight is \",round(F1,3),\"kN\" #due to decimal variance answer varies by 0.004kN \n",

-      "print\"The force required (i.e F2) to lower the weight is \",round(F2,3),\"kN\"\n",

-      "print\"The efficiency of the jack is \",round(E,2),\"percent\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The force required (i.e F1) to raise the weight is  1.209 kN\n",

-        "The force required (i.e F2) to lower the weight is  -0.795 kN\n",

-        "The efficiency of the jack is  16.47 percent\n"

-       ]

-      }

-     ],

-     "prompt_number": 58

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 7.7-11,Page No:172"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "# Initilization of variabes\n",

-      "\n",

-      "P=20000 #N #Weight of the shaft\n",

-      "D=0.30 #m #diameter of the shaft\n",

-      "R=0.15 #m #radius of the shaft\n",

-      "mu=0.12 # coefficient of friction\n",

-      "\n",

-      "# Calculations\n",

-      "\n",

-      "# Friction torque T is given by formulae,\n",

-      "T=(0.666)*P*R*mu #N-m #2/3=0.666\n",

-      "M=T #N-m\n",

-      "\n",

-      "# Results\n",

-      "\n",

-      "print\"The frictional torque is \",round(M),\"N-m\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The frictional torque is  240.0 N-m\n"

-       ]

-      }

-     ],

-     "prompt_number": 61

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file