From 04444e9cf12902699bb620c70a09f8dde4df35c5 Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Thu, 13 Jan 2011 13:10:37 -0500 Subject: [PATCH] Add solutions to week 2 recitation problems. --- .../Serway_and_Jewett_8/problem03.24.tex | 98 +++++++++++++++++++ .../Serway_and_Jewett_8/problem03.28.tex | 40 ++++++++ .../Serway_and_Jewett_8/problem03.29.tex | 64 ++++++++++-- .../Serway_and_Jewett_8/problem03.36.tex | 60 ++++++++++-- .../Serway_and_Jewett_8/problem03.43.tex | 35 ++++++- .../Serway_and_Jewett_8/problem03.63.tex | 51 +++++++++- 6 files changed, 323 insertions(+), 25 deletions(-) diff --git a/latex/problems/Serway_and_Jewett_8/problem03.24.tex b/latex/problems/Serway_and_Jewett_8/problem03.24.tex index 8ba7025..ff6f894 100644 --- a/latex/problems/Serway_and_Jewett_8/problem03.24.tex +++ b/latex/problems/Serway_and_Jewett_8/problem03.24.tex @@ -9,4 +9,102 @@ with respect to the positive $x$ axis. \end{problem*} \begin{solution} +\Part{a} +\begin{center} +\begin{asy} +import Mechanics; + +real u = 0.25cm; + +pair a = (2u, 6u); +pair b = (3u, -2u); +Vector A = Vector((0,0), mag=length(a), dir=degrees(a), "$\vect{A}$"); +A.draw(rotateLabel=true); +Vector B = Vector(a, mag=length(b), dir=degrees(b), "$\vect{B}$"); +B.draw(rotateLabel=true); +Vector C = Vector((0,0), mag=length(a+b), dir=degrees(a+b), "$\vect{C}$"); +C.draw(rotateLabel=true, labelOffset=-(a+b)/3); +\end{asy} +\hspace{1cm} +\begin{asy} +import Mechanics; + +real u = 0.25cm; + +pair a = (2u, 6u); +pair b = (3u, -2u); +Vector A = Vector((0,0), mag=length(a), dir=degrees(a), "$\vect{A}$"); +A.draw(rotateLabel=true, labelOffset=-a/3); +Vector B = Vector(a-b, mag=length(b), dir=degrees(b), "$\vect{B}$"); +B.draw(rotateLabel=true); +Vector D = Vector((0,0), mag=length(a-b), dir=degrees(a-b), "$\vect{D}$"); +D.draw(rotateLabel=true); +\end{asy} +\end{center} + +\Part{b} +\begin{align} + \vect{C} &= \vect{A} + \vect{B} + = (2.00\ihat + 6.00\jhat) + (3.00\ihat-2.00\jhat) \\ + &= (2.00+3.00)\ihat + (6.00-2.00)\jhat + = \ans{5.00\ihat + 4.00\jhat} \\ + \vect{D} &= \vect{A} - \vect{B} + = (2.00\ihat + 6.00\jhat) - (3.00\ihat-2.00\jhat) \\ + &= (2.00-3.00)\ihat + (6.00+2.00)\jhat + = \ans{-1.00\ihat + 8.00\jhat} +\end{align} + +\Part{c} +\begin{align} + r_\vect{C} &= \sqrt{\vect{C}_x^2 + \vect{C}_y^2} + = \sqrt{5.00^2 + 4.00^2} + = \ans{6.40} \\ + \theta_\vect{C} &= \arctan\p({\frac{\vect{C}_y}{\vect{C}_x}}) + = \arctan\p({\frac{4.00}{5.00}}) + = \ans{38.7\dg} \\ + r_\vect{D} &= \sqrt{\vect{D}_x^2 + \vect{D}_y^2} + = \sqrt{(-1.00)^2 + 8.00^2} + = \ans{8.06} \\ + \theta_\vect{D} &= \arctan\p({\frac{\vect{D}_y}{\vect{D}_x}}) + = \arctan\p({\frac{8.00}{-1.00}}) + = \ans{97.1\dg} +\end{align} + +Be careful with any $\arctan()$ evaluations. The standard $\arctan()$ +has a range of $\pm90\dg$ (or $\pm\pi/2$ in radians), which means that +vectors with negative $x$ values will need some manual correction. +For example, $\arctan(8/(-1))=-82.9\dg$, but that points down and to +the right, not up and to the left like $\vect{D}$. You can correct +for such ``back side'' angles by adding $180\dg$, which gives +$-82.9\dg+180\dg=97.1\dg$, the correct angle for $\vect{D}$. +\begin{center} +\begin{asy} +import graph; +import Mechanics; + +real u = 0.25cm; +pair d = (1u, -8u); +pair D = -d; + +Angle a = Angle((1,0), (0,0), d, red, "$-82.9\dg$"); +a.draw(labelOffsetAdjustment=2mm); +Vector vd = Vector((0,0), mag=length(d), dir=degrees(d), "$(1,-8)$"); +vd.draw(); +Angle A = Angle((1,0), (0,0), D, blue, "$97.1\dg$"); +A.draw(labelOffsetAdjustment=2mm); +Vector vD = Vector((0,0), mag=length(D), dir=degrees(D), "$(-1,8)$"); +vD.draw(); +Angle r = Angle(d, (0, 0), D, radius=15mm, green, "$180.0\dg$"); +r.draw(labelOffsetAdjustment=4mm); + +xaxis("$x$"); +\end{asy} +\end{center} +The reason for the ambiguity is that $-82.9\dg$ corresponds to +$\arctan(-8/1)$ which \emph{does} point down and to the right. There +is no way for the $\arctan()$ function to give different value for +$8/(-1)$ than it does for $-8/1$, so you have to make the adjustment +by hand. If your calculator has an \verb+atan2()+ function, it's a +good idea to use that instead, as it handles any ``back side'' +corrections automatically. \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem03.28.tex b/latex/problems/Serway_and_Jewett_8/problem03.28.tex index 3fbb423..5d331f0 100644 --- a/latex/problems/Serway_and_Jewett_8/problem03.28.tex +++ b/latex/problems/Serway_and_Jewett_8/problem03.28.tex @@ -8,4 +8,44 @@ magnitude of the football's resultant displacement? \end{problem*} \begin{solution} +\begin{center} +\begin{asy} +import Mechanics; + +real u = 0.1cm; + +pair drop = (-10u, 0); +pair scramble = (0, 15.0u); +pair pass = (50u, 0); +pair result = drop+scramble+pass; + +dot((0,0)); // initial ball position +draw((0,-5u)--(0, 20u), dashed); // line of scrimmage + +Vector vDrop = Vector((0,0), mag=length(drop), dir=degrees(drop), "drop"); +vDrop.draw(labelOffset=-drop/2); +Vector vScramble = Vector( + drop, mag=length(scramble), dir=degrees(scramble), "scramble"); +vScramble.draw(rotateLabel=true, labelOffset=-scramble/2); +Vector vPass = Vector(drop+scramble, mag=length(pass), dir=degrees(pass), + "pass"); +vPass.draw(rotateLabel=true, labelOffset=-pass/2); +Vector vResult = Vector((0,0), mag=length(result), dir=degrees(result), + "result"); +vResult.draw(rotateLabel=true, labelOffset=-result/2); +\end{asy} +\end{center} + +The resultant displacement is +\begin{equation} + \vect{r} = \vect{d} + \vect{s} + \vect{p} + = (-10.0\ihat + 15.0\jhat + 50.0\ihat)\U{yards} + = (40.0\ihat + 15.0\jhat)\U{yards} +\end{equation} +Which has a magnitude +\begin{equation} + |\vect{r}| = \sqrt{\vect{r}_x^2 + \vect{r}_y^2} + = \sqrt{40.0^2 + 15.0^2}\U{yards} + = \ans{42.7\U{yards}} +\end{equation} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem03.29.tex b/latex/problems/Serway_and_Jewett_8/problem03.29.tex index 59bc6ad..d61ed3b 100644 --- a/latex/problems/Serway_and_Jewett_8/problem03.29.tex +++ b/latex/problems/Serway_and_Jewett_8/problem03.29.tex @@ -1,20 +1,66 @@ \begin{problem*}{3.29} The helicopter view in Fig.~P3.29 shows two people pulling on a stubborn mule. The person on the right pulls with a force -$\vect{F}_1$ of magnitude $120\U{N}$ and direction -$\theta_1=60.0\deg$. The person on the left pulls with a force -$\vect{F}_2$ of magnitude $80.0\U{N}$ and direction of -$\theta_2=75.0\deg$. Find \Part{a} the single force that is -equivalent to the two forces shown and \Part{b} the force that a third -person would have to exert on the mule to make the resultant force -equal to zero. The forces are measured in units of newtons -(symbolized $\bareU{N}$). +$\vect{F}_1$ of magnitude $120\U{N}$ and direction $\theta_1=60.0\dg$. +The person on the left pulls with a force $\vect{F}_2$ of magnitude +$80.0\U{N}$ and direction of $\theta_2=75.0\dg$. Find \Part{a} the +single force that is equivalent to the two forces shown and \Part{b} +the force that a third person would have to exert on the mule to make +the resultant force equal to zero. The forces are measured in units +of newtons (symbolized $\bareU{N}$). \begin{center} \begin{asy} -draw((0,0)--(1,1)); +import graph; +import Mechanics; + +real u = 0.02cm; + +Angle a = Angle((1,0), (0,0), dir(60), "$\theta_1$"); +a.draw(); +Vector A = Vector((0,0), mag=120u, dir=60, "$\vect{F}_1$"); +A.draw(); +Angle b = Angle((-1,0), (0,0), dir(180-75), "$\theta_2$"); +b.draw(); +Vector B = Vector((0,0), mag=80u, dir=180-75, "$\vect{F}_2$"); +B.draw(); + +xaxis("$x$"); +yaxis("$y$"); \end{asy} \end{center} \end{problem*} \begin{solution} +\Part{a} +\begin{align} + \vect{F}_\text{tot} &= \vect{F}_1 + \vect{F}_2 + = |\vect{F}_1|(\cos(\theta_1)\ihat + \sin(\theta_1)\jhat) + + |\vect{F}_2|(-\cos(\theta_2)\ihat + \sin(\theta_2)\jhat) \\ + &= [|\vect{F}_1|\cos(\theta_1) - |\vect{F}_2|\cos(\theta_2)]\ihat + + [|\vect{F}_1|\sin(\theta_1) + |\vect{F}_2|\sin(\theta_2)]\jhat \\ + &= \{ [120\cos(60.0\dg) - 80.0\cos(75.0\dg)]\ihat + + [120\sin(60.0\dg) + 80.0\sin(75.0\dg)]\jhat\}\U{N} \\ + &= \ans{(39.3\ihat + 181\jhat)\U{N}} +\end{align} +Or, in polar coordinates +\begin{align} + |\vect{F}_\text{tot}| + &= \sqrt{\vect{F}_{\text{tot},x}^2 + \vect{F}_{\text{tot},y}^2} + = \sqrt{39.3^2 + 181^2}\U{N} + = \ans{185\U{N}} \\ + \theta_\text{tot} + &= \arctan\p({\frac{\vect{F}_{\text{tot},y}}{\vect{F}_{\text{tot},x}}}) + = \arctan\p({\frac{181}{39.3}}) + = \ans{77.8\dg} +\end{align} +where $\theta_\text{tot}$ is measured from the positive $x$ axis. + +\Part{b} +The third person must balance the force from the first two people, so +\begin{align} + \vect{F}_3 &= -(\vect{F}_1 + \vect{F}_2) + = \ans{(-39.3\ihat + -181\jhat)\U{N}} \\ + |\vect{F}_3| &= |\vect{F}_1 + \vect{F}_2| = \ans{185\U{N}} \\ + \theta_3 &= 77.8\dg \pm 180\dg = \ans{-102.2\dg \text{ or } 257.8\dg} +\end{align} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem03.36.tex b/latex/problems/Serway_and_Jewett_8/problem03.36.tex index 658bc93..e5ef67d 100644 --- a/latex/problems/Serway_and_Jewett_8/problem03.36.tex +++ b/latex/problems/Serway_and_Jewett_8/problem03.36.tex @@ -9,14 +9,18 @@ direction of the resultant displacement. import graph; import Mechanics; -real u = 0.1cm; +real u = 0.06cm; -Vector a = Vector((0,0), mag=20u, dir=90, "$\vect{A}$"); -a.draw(); -Vector b = Vector((0,0), mag=40u, dir=45, "$\vect{B}$"); -b.draw(); -Vector c = Vector((0,0), mag=30u, dir=-45, "$\vect{C}$"); -c.draw(); +pair a = 20u * dir(90); +pair b = 40u * dir(45); +pair c = 30u * dir(-45); + +Vector A = Vector((0,0), mag=length(a), dir=degrees(a), "$\vect{A}$"); +A.draw(); +Vector B = Vector((0,0), mag=length(b), dir=degrees(b), "$\vect{B}$"); +B.draw(); +Vector C = Vector((0,0), mag=length(c), dir=degrees(c), "$\vect{C}$"); +C.draw(); xaxis("$x$"); yaxis("$y$"); @@ -25,4 +29,46 @@ yaxis("$y$"); \end{problem*} \begin{solution} +\Part{a} +\begin{center} +\begin{asy} +import graph; +import Mechanics; + +real u = 0.06cm; + +pair a = 20u * dir(90); +pair b = 40u * dir(45); +pair c = 30u * dir(-45); +pair r = a + b + c; + +Vector A = Vector((0,0), mag=length(a), dir=degrees(a), "$\vect{A}$"); +A.draw(); +Vector B = Vector(a, mag=length(b), dir=degrees(b), "$\vect{B}$"); +B.draw(); +Vector C = Vector(a+b, mag=length(c), dir=degrees(c), "$\vect{C}$"); +C.draw(); +Vector R = Vector((0,0), mag=length(r), dir=degrees(r), "$\vect{r}$"); +R.draw(labelOffset=-r/2); + +xaxis("$x$"); +yaxis("$y$"); +\end{asy} +\end{center} +The resultant displacement is +\begin{align} + \vect{r} &= \vect{A} + \vect{B} + \vect{C} \\ + &= \{ 20.0\jhat + + 40.0[\cos(45\dg)\ihat + \sin(45\dg)\jhat] + + 30.0[\cos(-45\dg)\ihat + \sin(-45\dg)\jhat]\}\U{units} \\ + &= (49.5\ihat + 27.1\jhat)\U{units} +\end{align} + +\Part{b} +\begin{align} + |\vect{r}| &= \sqrt{\vect{r}_x^2 + \vect{r}_y^2} + = \sqrt{49.5^2 + 27.1^2}\U{units} + = \ans{56.4\U{units}} \\ + \theta_\vect{r} &= \arctan\p({\frac{27.1}{49.5}}) = \ans{28.7\dg} +\end{align} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem03.43.tex b/latex/problems/Serway_and_Jewett_8/problem03.43.tex index 15efaf1..e293531 100644 --- a/latex/problems/Serway_and_Jewett_8/problem03.43.tex +++ b/latex/problems/Serway_and_Jewett_8/problem03.43.tex @@ -12,21 +12,46 @@ airplane's position vector at $t=45.0\U{s}$. \begin{asy} import Mechanics; -real u = 1cm; +real u = 1.3cm; real h = 1u; real d = 8.04/7.6*h; real dx = 0.2u; -draw((-dx,0)--(d+dx,0)); -draw((-dx,h)--(d+dx,h)); +draw((-dx,h)--(d+dx,h), dashed); +Surface s = Surface((-dx,0), (d+dx,0)); +s.draw(); -Vector A = Vector();draw((0,0)--(0,h)); +Vector A = Vector((0,0), mag=h, dir=90, "$\vect{R}_0$"); A.draw(); -Vector B = Vector();((0,0)--(d,h)); +Vector B = Vector((0,0), mag=length((d,h)), dir=degrees((d,h)), "$\vect{R}_{30}$"); B.draw(); \end{asy} \end{center} \end{problem*} \begin{solution} +The airplane is flying at a constant velocity, so we'll compute that +velocity first, +\begin{align} + \vect{v} &= \frac{\vect{P}_{30} - \vect{P}_0}{t_{30} - t_{0}} + = \frac{(8.04\E{3}\ihat+7.60\E{3}\jhat)\U{m} - 7.60\E{3}\jhat\U{m}} + {30.0\U{s} - 0\U{s}} \\ + &= \frac{8.04\E{3}\ihat}{30.0}\U{m/s} + = 268\ihat\U{m/s} +\end{align} +Which is, as claimed in the problem text, parallel to the $x$ axis. + +The position at $t=45.0\U{s}$ is therefore +\begin{align} + \vect{P}_{45} &= \vect{P}_0 + \vect{v}(t_{45}-t_0) + = 7.60\E{3}\jhat\U{m} + 268\ihat\U{m/s} \cdot (45.0\U{s} - 0\U{s}) \\ + &= (12.1\E{3}\ihat + 7.60\E{3}\jhat)\U{m} + = (12.1\ihat + 7.60\jhat)\U{km} \\ + |\vect{P}_{45}| &= \sqrt{\vect{P}_{45,x}^2 + \vect{P}_{45,y}^2} + = \sqrt{12.1^2 + 7.60^2}\U{km} + = \ans{14.3\U{km}} \\ + \theta_{45} &= \arctan\p({\frac{\vect{P}_{45,y}}{\vect{P}_{45,x}}}) + = \arctan\p({\frac{7.60}{12.1}}) + = \ans{32.2\dg} +\end{align} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem03.63.tex b/latex/problems/Serway_and_Jewett_8/problem03.63.tex index bdf50f0..bdfc893 100644 --- a/latex/problems/Serway_and_Jewett_8/problem03.63.tex +++ b/latex/problems/Serway_and_Jewett_8/problem03.63.tex @@ -1,3 +1,11 @@ +% Requires the following code at some point after `\begin{document}`: +% \begin{asydef} +% usepackage("wtk_cmmds"); +% \end{asydef} +% This allows macros like `\vect{}` to be used in 3D graphics. See +% https://sourceforge.net/tracker/?func=detail&atid=685683&aid=3155798&group_id=120000 +% for details. + \begin{problem*}{3.63} A rectangular parallelpiped has dimensions $a$, $b$, and $c$ as shown in Figure 3.63. \Part{a} Obtain a vector expression for the face @@ -6,11 +14,13 @@ vector? \Part{c} Notice that $\vect{R}_1$, $c\khat$, and $\vect{R}_2$ make a right triangle. Obtain a vector expression for the body diagonal vector $\vect{R}_2$. \begin{center} -\begin{asy} +\begin{asy}[inline=true,attach=false] import graph3; import Mechanics; -real u = 1cm; +currentprojection = perspective(10cm,4cm,5cm); + +real u = 2.5cm; real a = 1u; real b = 1u; @@ -20,11 +30,44 @@ xaxis3(Label("$x$"), 0, 1.5a); yaxis3(Label("$y$"), 0, 1.5b); zaxis3(Label("$z$"), 0, 1.5c); -draw(xscale3(a)*yscale3(b)*zscale3(c)*unitcube); -//Vector A = Vector((0,0,0), + +draw(xscale3(a)*yscale3(b)*zscale3(c)*unitcube, surfacepen=opacity(0.2), meshpen=currentpen); + +// Distance3? +draw((0, 0, 1.3c) -- (a, 0, 1.3c), Arrows3); +label(Label("$a$", (0, 0, 1)), (a/2, 0, 1.3c)); +draw((0, 0, 1.3c) -- (0, b, 1.3c), Arrows3); +label(Label("$b$", (0, 0, 1)), (0, b/2, 1.3c)); +draw((0, 1.3b, 0) -- (0, 1.3b, c), Arrows3); +label(Label("$c$", (0, 1, 0)), (0, 1.3b, c/2)); + + +draw(surface((0,0,0)--(a,b,0)--(a,b,c)--(0,0,c)--cycle), yellow+opacity(0.3), nolight, meshpen=dashed); +// Vector3? +//Vector A = Vector((0,0,0), (a,b,0)); +draw((0,0,0) -- (a,b,0), Arrow3); +label(Label("$\vect{R}_1$", (1,1,0)), position=(a,b,0)); +draw((0,0,0) -- (a,b,c), Arrow3); +label(Label("$\vect{R}_2$", (0,1,-1)), (a,b,c)); \end{asy} \end{center} \end{problem*} \begin{solution} +\Part{a} +\begin{equation} + \vect{R}_1 = \ans{a\ihat + b\jhat} +\end{equation} + +\Part{b} +\begin{equation} + |\vect{R_1}| = \sqrt{\vect{R}_{1,x}^2 + \vect{R}_{1,y}^2} + = \ans{\sqrt{a^2 + b^2}} +\end{equation} + +\Part{c} +\begin{equation} + \vect{R}_2 = \vect{R}_1 + c\khat + = \ans{a\ihat + b\jhat + c\khat} +\end{equation} \end{solution} -- 2.26.2