\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}
\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}
\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}
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("$y$");
\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}
\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}
+% 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
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;
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}
projects / course.git / commitdiff