{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Exercises in the IPy Notebook" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "In the following exercises, the \"examples\" are completed for you and the \"problems\" are for you to try. Each problem can be solved by a method very similar to the associated example. " ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "from numpy import *" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## example 1 (simultaneous equations)\n", "\n", "Solve the following equations for $x$ and $y$:\n", "\n", "$5x+4y=-1$\n", "\n", "$-7x-2y=-13$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### solution:\n", "\n", "There are several ways to solve simultaneous equatiions. Here we use the matrix method for solving linear systems of equations. It amounts to writing the two equations as a $2\\times 2$ matrix of coefficients times a $2\\times1$ vector of variables.\n", "\n", "So\n", "\n", "$$Ax + By = E\\\\\n", " Cx + Dy = F $$\n", " \n", "is written as\n", "\n", "$$\\begin{bmatrix}A & B \\\\ C & D \\end{bmatrix}\n", "\\begin{bmatrix} x\\\\ y\\end{bmatrix} = \n", "\\begin{bmatrix} E\\\\ F\\end{bmatrix} $$\n", "\n", "To get our problem in this form, write\n", "\n", "$$\\begin{bmatrix}5 & 4 \\\\ -7 & -2 \\end{bmatrix}\n", "\\begin{bmatrix} x\\\\ y\\end{bmatrix} = \n", "\\begin{bmatrix} -1\\\\ -13\\end{bmatrix} $$\n", "\n", "This is solved this with the `linalg` (i.e., [linear algebra](http://docs.scipy.org/doc/numpy/reference/routines.linalg.html)) routines as follows." ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[ 3. -4.]\n" ] } ], "source": [ "CC = array((( 5., 4. ),\n", " ( -7., -2. ) ))\n", "SS = array(( -1., -13. ))\n", "solution = linalg.solve(CC,SS)\n", "print(solution)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "That is, $x=3$ and $y=-4$ solve our system of equations." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## problem 1 (simultaneous equations)\n", "\n", "Solve the following equations for $x$, $y$ and $z$:\n", "\n", "$2x+-y+3z=37$\n", "\n", "$x+y-3z=-16$\n", "\n", "$-3x+5z=24$" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] }, { "cell_type": "markdown", "metadata": {}, "source": [ "(solution: $x=7$, $y=4$, $z=9$)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## example 2 (vector addition)\n", "\n", "Add the two vectors shown.\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "a. vector sum [205.96448785 149.9048241 ] mm\n" ] } ], "source": [ "dtr = pi/180. # a factor to convert degrees to radians\n", "# part a\n", "R1 = 183.*array(( cos(55*dtr), sin(55*dtr) )) # mm\n", "R2 = array(( 101., 0. )) # mm\n", "R = R1+R2\n", "print ('a. vector sum', R, 'mm')" ] }, { "cell_type": "code", "execution_count": 4, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "b. vector sum [-0.86974134 -0.68025866] kip\n" ] } ], "source": [ "# part b\n", "R1 = 1.23*array(( -cos(45*dtr), sin(45*dtr) )) # kip\n", "R2 = array(( 0., -1.55 )) # kip\n", "R = R1+R2\n", "print ('b. vector sum', R, 'kip')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## problem 2 (vector addition)\n", "\n", "\n", "Add the two vectors shown.\n", "\n", "" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## example 3 (vector algebra)\n", "\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 5, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "a. R=A+B= [350. 280.]\n", " magnitude of R= 448.2186966202994 lb\n" ] } ], "source": [ "A = array(( 150., -200. )) # lb\n", "B = array(( 200., 480. )) # lb\n", "\n", "# part a\n", "R = A+B\n", "R_mag = sqrt( sum(R**2) )\n", "print ('a. R=A+B=',R)\n", "print (' magnitude of R=',R_mag,'lb')" ] }, { "cell_type": "code", "execution_count": 6, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "b. R=2A-(1/2)B= [ 200. -640.]\n", " magnitude of R= 670.5221845696084 lb\n" ] } ], "source": [ "# part b\n", "R = 2*A - (1/2)*B\n", "R_mag = sqrt( sum(R**2) )\n", "print ('b. R=2A-(1/2)B=',R)\n", "print (' magnitude of R=',R_mag,'lb')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "To do part (c), solve for $s$ when the $y$ component of $\\vec{R}$ is 0.\n", "\n", "$R_y = s A_y + B_y = 0$\n", "\n", "so \n", "\n", "$s = -\\frac{B_y}{A_y}$" ] }, { "cell_type": "code", "execution_count": 7, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "c. s= 2.4\n" ] } ], "source": [ "# part c\n", "s = - B[1]/A[1]\n", "print ('c. s=',s)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "For part (d), first compute $\\vec{C} = \\vec{B}-\\vec{A}$, then get its magnitude $\\left | \\vec{C} \\right |$.\n", "\n", "Finally, find the unit vector with $\\hat{u} = \\frac{\\vec{C}}{\\left | \\vec{C} \\right |}$." ] }, { "cell_type": "code", "execution_count": 8, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "d. unit vector= [0.07333144 0.99730763]\n" ] } ], "source": [ "# part d\n", "C = B-A\n", "C_mag = sqrt(sum(C**2))\n", "u = C / C_mag\n", "print ('d. unit vector=',u)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## problem 3 (vector algebra)\n", "\n", "\n", "" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## example 4 (unit vectors)\n", "\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 9, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "a. rAB= [3.46410162 2. ] ft\n", "b. rBA= [-3.46410162 -2. ] ft\n", "c. uAB= [0.8660254 0.5 ]\n", "d. uBA= [-0.8660254 -0.5 ]\n", "e. FAB= [6.92820323 4. ] lb\n", "f. FBA= [-6.92820323 -4. ] lb\n" ] } ], "source": [ "rAB = 4.*array(( cos(30*dtr), sin(30*dtr) )) # ft\n", "rBA = -rAB\n", "uAB = rAB / sqrt(sum(rAB**2))\n", "uBA = -uAB\n", "FAB = 8.*uAB # lb\n", "FBA = 8.*uBA # lb\n", "print ('a. rAB=',rAB,'ft')\n", "print ('b. rBA=',rBA,'ft')\n", "print ('c. uAB=',uAB)\n", "print ('d. uBA=',uBA)\n", "print ('e. FAB=',FAB,'lb')\n", "print ('f. FBA=',FBA,'lb')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## problem 4 (unit vectors)\n", "\n", "" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## example 5 (vector rotation)\n", "\n", "\n", "\n", "\n", "Vector rotations are effected by multiplying the vector by the matrix of trig functions described above. For coordinate rotation through angle $\\phi$:\n", "\n", "$$\n", "\\begin{bmatrix}v_t\\\\ v_n\\end{bmatrix} = \n", "\\begin{bmatrix}\n", "\\cos \\phi & \\sin \\phi \\\\ \n", "-\\sin \\phi & \\cos \\phi \n", "\\end{bmatrix} \n", "\\begin{bmatrix}v_x\\\\ v_y\\end{bmatrix}\n", "$$\n" ] }, { "cell_type": "code", "execution_count": 10, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "original vector R= [-809.86806494 373.86858304] lb\n", "rotation matrix=\n", " [[ 0.76604444 0.64278761]\n", " [-0.64278761 0.76604444]]\n" ] } ], "source": [ "R = 892.0 * array(( -cos(24.78*dtr), sin(24.78*dtr) ))\n", "phi = 40. * dtr # coordinate rotation angle, radians\n", "rot = matrix(((cos(phi),sin(phi)),(-sin(phi),cos(phi))))\n", "print ('original vector R=',R,'lb')\n", "print ('rotation matrix=\\n',rot)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Multiplication of a matrix and a vector is done with the `dot` function. The solution agrees with the result of example 2.7 in the text." ] }, { "cell_type": "code", "execution_count": 11, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "rotated vector= [[-380.07683798 806.97310812]] lb\n" ] } ], "source": [ "print ('rotated vector=', dot(rot,R), 'lb')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## problem 5 (vector rotation)\n", "\n", "\n", "" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## example 6 (unit vectors)\n", "\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 12, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "force the cable exerts on B is [-42.85714286 85.71428571 -28.57142857] lb\n", "force the cable exerts on E is [ 42.85714286 -85.71428571 28.57142857] lb\n" ] } ], "source": [ "x = 6. # in\n", "y = 12. # in\n", "rOB = array(( x, 0, 4. ))\n", "rOE = array(( 0, y, 0. ))\n", "rBE = rOE - rOB\n", "uBE = rBE / sqrt(sum(rBE**2)) # unit vector pointing from B to E\n", "FBE = 100.*uBE\n", "FEB = -FBE\n", "print ('force the cable exerts on B is',FBE,'lb')\n", "print ('force the cable exerts on E is',FEB,'lb')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## problem 6 (unit vectors)\n", "\n", "\n", "See problem 9, above. Use the parameters for 2.88." ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## example 7 (dot product)\n", "\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 13, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "a. angle between vectors= 45.90050773774112 degrees\n", "b. component of A parallel to B= 6.2631578947368425 N\n", " component of A perpendicular to B= 6.4631921823194745 N\n", "c. vector component of A parallel to B= [ 0.32963989 5.93351801 -1.97783934] N\n", " vector component of A perpendicular to B= [-1.32963989 2.06648199 5.97783934] N\n" ] } ], "source": [ "A = array(( -1., 8., 4. )) # N\n", "B = array(( 1., 18.,-6. )) # mm\n", "AdotB = dot(A,B)\n", "A_mag = sqrt(sum(A**2))\n", "B_mag = sqrt(sum(B**2))\n", "uA = A / A_mag # unit vector\n", "uB = B / B_mag # unit vector\n", "cos_AB = arccos( dot(uA,uB) ) / dtr # angle in degrees\n", "A_para_mag = dot(A,uB)\n", "A_para = A_para_mag*uB\n", "A_perp = A - A_para\n", "A_perp_mag = sqrt(sum(A_perp**2))\n", "print ('a. angle between vectors=',cos_AB,'degrees')\n", "print ('b. component of A parallel to B=',A_para_mag,'N')\n", "print (' component of A perpendicular to B=',A_perp_mag,'N')\n", "print ('c. vector component of A parallel to B=',A_para,'N')\n", "print (' vector component of A perpendicular to B=',A_perp,'N')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## problem 7 (dot product)\n", "\n", "\n", "See example 7 above. Use figure P2.104." ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## example 8 (cross product)\n", "\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 14, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "a. AxB= [ -4. -72. -40.] in^2\n", "b. BxA= [ 4. 72. 40.] in^2\n", "c. the cross product appears to be anti-commutative...!\n", "d. dot products: 0.0 and 0.0\n" ] } ], "source": [ "A = array(( 6., -2., 3. )) # in\n", "B = array((-14., -2., 5. )) # in\n", "C = cross(A,B)\n", "print ('a. AxB=',C,'in^2')\n", "print ('b. BxA=',cross(B,A),'in^2')\n", "print ('c. the cross product appears to be anti-commutative...!')\n", "print ('d. dot products:',dot(A,C),'and',dot(B,C))" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## problem 8 (cross product)\n", "\n", "(book problem 2.146)\n", "\n", "See example 8 above. Use figure P2.146." ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.8.2" } }, "nbformat": 4, "nbformat_minor": 1 }