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   "source": [
    "# Chapter 5 General Case of Forces in a plane"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
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   },
   "outputs": [],
   "source": [
    "##Example 5.2 Equations of equilibrium"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The reaction at P is 5656.85424949238 N\n",
      "The reaction at Q is  4000.0 N\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "#Initialization of Variables\n",
    "W=2000 #N\n",
    "Lab=2 #m #length of the member from the vertical to the 1st load of 2000 N\n",
    "Lac=5 #m #length of the member from the vertical to the 2nd load of 2000 N\n",
    "Lpq=3.5 #m\n",
    "\n",
    "#Calculations\n",
    "Rq=((W*Lab)+(W*Lac))/Lpq #N #take moment abt. pt P\n",
    "Xp=Rq #N #sum Fx=0\n",
    "Yp=2*W #N #sum Fy=0\n",
    "Rp=math.sqrt(Xp**2+Yp**2) #N\n",
    "\n",
    "#Resuts\n",
    "print('The reaction at P is' ,Rp ,'N')\n",
    "print('The reaction at Q is ',Rq ,'N')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "##Example 5.3 Equations of equilibrium"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The reaction at A i.e Ra is  [[ 120.27406887]] N\n",
      "The reaction at B i.e Rb is  [[ 35.13703443]] N\n",
      "The required tension in the string is  [[ 40.57275258]] N\n"
     ]
    }
   ],
   "source": [
    "import math,numpy\n",
    "#Initilization of vaiables\n",
    "W=25 #N # self weight of the ladder\n",
    "M=75 #N # weight of the man standing o the ladder\n",
    "theta=63.43 #degree # angle which the ladder makes with the horizontal\n",
    "alpha=30 #degree # angle made by the string with the horizontal\n",
    "Loa=2 #m # spacing between the wall and the ladder\n",
    "Lob=4 #m #length from the horizontal to the top of the ladder touching the wall(vertical)\n",
    "\n",
    "#Calculations\n",
    "#Using matrix to solve the simultaneous eqn's 3 & 4\n",
    "A=numpy.matrix('2 -4; 1 -0.577')\n",
    "B=numpy.matrix('100;100')\n",
    "C=numpy.linalg.inv(A)*B\n",
    "\n",
    "#Results\n",
    "print('The reaction at A i.e Ra is ',C[0] ,'N')\n",
    "print('The reaction at B i.e Rb is ',C[1] ,'N')\n",
    "\n",
    "#Calculations\n",
    "T=C[1]/math.cos(math.radians(alpha)) #N # from (eqn 1)\n",
    "\n",
    "#Results\n",
    "print('The required tension in the string is ',T, 'N')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "##Example 5.4 Equations of Equilibrium"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The reaction at B i.e Rb is  25.0 N\n",
      "The horizontal reaction at A i.e Xa is  21.650635094610966 N\n",
      "The vertical reaction at A i.e Ya is  112.5 N\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "#Initilization of variables\n",
    "W=100 #N\n",
    "theta=60 #degree angle made by the ladder with the horizontal\n",
    "alpha=30 #degree angle made by the ladder with the vertical wall\n",
    "Lob=4 #m  length from the horizontal to the top of the ladder touching the wall(vertical)\n",
    "Lcd=2 #m  length from the horizontal to the centre of the ladder where the man stands\n",
    "\n",
    "#Calculations\n",
    "Lab=Lob*(1/math.cos(math.radians(alpha))) #m length of the ladder\n",
    "Lad=Lcd*math.tan(math.radians(alpha)) #m\n",
    "Rb=(W*Lad)/Lab #N take moment at A\n",
    "Xa=Rb*math.sin(math.radians(theta)) #N  From eq'n 1\n",
    "Ya=W+Rb*math.cos(math.radians(theta)) #N From eq'n 2\n",
    "\n",
    "#Results\n",
    "print('The reaction at B i.e Rb is ',Rb, 'N')\n",
    "print('The horizontal reaction at A i.e Xa is ',Xa, 'N')\n",
    "print('The vertical reaction at A i.e Ya is ',Ya,'N')\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "##Example 5.5 Equations of Equilibrium"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The horizontal reaction at A i.e Xa is  28.867513459481287 N\n",
      "The vertical reaction at A i.e Ya is  100 N\n",
      "The reaction at B i.e Rb is  28.867513459481287 N\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "#Initilization of variables\n",
    "W=100 #N self weight of the man\n",
    "alpha=30 #degree angle made by the ladder with the wall\n",
    "Lob=4 #m  length from the horizontal to the top of the ladder touching the wall(vertical)\n",
    "Lcd=2 #m\n",
    "\n",
    "#Calculations\n",
    "# using the equiblirium equations\n",
    "Ya=W #N  From eq'n 2\n",
    "Lad=Lcd*math.tan(math.radians(alpha)) #m Lad is the distance fom pt A to the point where the line from the cg intersects the horizontal\n",
    "Rb=(W*Lad)/Lob #N  Taking sum of moment abt A\n",
    "Xa=Rb #N From eq'n 1\n",
    "\n",
    "#Results\n",
    "print('The horizontal reaction at A i.e Xa is ',Xa, 'N')\n",
    "print('The vertical reaction at A i.e Ya is ',Ya,'N' )\n",
    "print('The reaction at B i.e Rb is ',Rb ,'N')\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "##Example 5.6 Equations of Equilibrium"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The horizontal reaction at A i.e Xa is  3.84 N\n",
      "The vertical reaction at A i.e Ya is  7.12 N\n",
      "Therefore the reaction at A i.e Ra is  8.089499366462674 N\n",
      "The reaction at D i.e Rd is  4.8 N\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "#Initilization of variables\n",
    "d=0.09 #m diametre of the right circular cylinder\n",
    "h=0.12 #m height of the cyinder\n",
    "W=10 #N  self weight of the bar\n",
    "l=0.24 #m length of the bar\n",
    "\n",
    "#Calculations\n",
    "theta=math.degrees(math.atan(h/d))  #angle which the bar makes with the horizontal\n",
    "Lad=math.sqrt(d**2+h**2) #m Lad is the length of the bar from point A to point B\n",
    "Rd=(W*h*(math.cos(theta*math.pi/180)))/Lad #N  Taking moment at A\n",
    "Xa=Rd*(math.sin(theta*math.pi/180)) #N  sum Fx=0.... From eq'n 1\n",
    "Ya=W-(Rd*(math.cos(theta*math.pi/180))) #N  sum Fy=0..... From eq'n 2\n",
    "Ra=math.sqrt(Xa**2+Ya**2) #resultant of Xa & Ya\n",
    "\n",
    "#Results\n",
    "print('The horizontal reaction at A i.e Xa is ',Xa, 'N')\n",
    "print('The vertical reaction at A i.e Ya is ',Ya, 'N')\n",
    "print('Therefore the reaction at A i.e Ra is ',Ra,'N')\n",
    "print('The reaction at D i.e Rd is ',Rd,'N')"
   ]
  }
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