--- /dev/null
+\begin{problem*}{5.25}
+A bag of cement whose weight is $F_g$ hangs in equilibrium from three
+wires shown in Figure P5.24. Two of the wires make angles
+$\theta_1=60.0\dg$ and $\theta_2=40.0\dg$ with the horizontal.
+Assuming the system is in equilibrium, show that the tension in the
+left-hand wire is
+\begin{equation}
+ T_1 = \frac{F_g \cos\theta_2}{\sin(\theta_1+\theta_2)}
+\end{equation}
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+
+real theta1 = 60;
+real theta2 = 40;
+real d = 3u; // contact point seperation
+real a = u; // block side length
+
+pair c1 = (0,0); // left contact point
+pair c2 = c1 + (d,0); // right contact point
+pair c3 = extension(c1, c1+dir(-theta1),
+ c2, c2+dir(180+theta2));
+
+Surface s = Surface(c2+(0.2d,0), c1-(0.2d,0));
+Angle a1 = Angle(c2, c1, c3, "$\theta_1$");
+Angle a2 = Angle(c1, c2, c3, "$\theta_2$");
+Block cement = Block(c3-(0, a), width=a, height=a, "$F_g$");
+
+a1.draw();
+a2.draw();
+draw(c1 -- c3 -- c2);
+draw(c3 -- cement.center);
+s.draw();
+cement.draw();
+
+label("$T_1$", (c1+c3)/2, SW);
+label("$T_2$", (c2+c3)/2, SE);
+label("$T_3$", c3-(0,a/4), E);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+Balancing forces on the cement bag, $T_3=F_g$.
+
+Balancing forces on the wire joint is a bit mor complicated and
+deserves a free body diagram.
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+
+real theta1 = 60;
+real theta2 = 40;
+real T3 = 2u;
+real T1 = T3 * Cos(theta2)/Sin(theta1+theta2);
+real T2 = T3 * Cos(theta1)/Sin(theta1+theta2);
+
+Angle a1 = Angle(dir(180-theta1), (0,0), (-1,0), "$\theta_1$");
+Angle a2 = Angle(dir(theta2), (0,0), (1,0), "$\theta_2$");
+Angle A1 = Angle(dir(180-theta1), (0,0), dir(180+theta2), radius=10mm,
+ "$\theta_1+\theta_2$");
+Angle A3 = Angle((0,-1), (0,0), dir(-90+theta2), "$\theta_2$");
+
+Vector F1 = Force((0,0), mag=T1, dir=180-theta1, "$\vect{F}_1$");
+Vector F2 = Force((0,0), mag=T2, dir=theta2, "$\vect{F}_2$");
+Vector F3 = Force((0,0), mag=T3, dir=-90, "$\vect{F}_3$");
+
+a1.draw();
+a2.draw();
+A1.draw(labelOffsetAdjustment=12pt);
+A3.draw();
+draw((-7mm,0)--(7mm,0));
+draw((12mm*dir(-90+theta2))--(0,0)--(12mm*dir(180+theta2)));
+F1.draw();
+F2.draw();
+F3.draw();
+dot((0,0));
+
+draw_ijhat((1.5u,-.5u), theta2);
+\end{asy}
+\end{center}
+Where we've chosen a coordinate system such that $\vect{F}_2$ has no
+$y$ component.
+
+Balancing forces in the $y$ direction,
+\begin{align}
+ 0 &= T_1 \sin(\theta_1+\theta_2) - T_3\cos(\theta_2) \\
+ T_1 &= T_3\frac{\cos(\theta_2)}{\sin(\theta_1+\theta_2)}
+ = F_g\frac{\cos(\theta_2)}{\sin(\theta_1+\theta_2)} \;,
+\end{align}
+which is what we set out to show.
+\end{solution}
--- /dev/null
+\begin{problem*}{5.28}
+An object of mass $m_1=5.00\U{kg}$ placed on a frictionless,
+horizontal table is connected to a string that passes over a pulley
+and then is fastened to a hanging object of mass $m_2=9.00\U{kg}$ as
+shown in Figure P5.28. \Part{a} Draw free-body diagrams of both
+objects. Find \Part{b} the magnitude of that acceleration of the
+objects and \Part{c} the tension in the string.
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+
+real a = u; // block side length
+real pr = a/4; // pulley radius
+real psw = 0.2u; // pulley support width
+real d = 2u; // rope length
+
+Surface s = Surface((-.7a, 0), (d-2.5pr, 0));
+Block m1 = Block((0, a/2), width=a, height=a, "$m_1$");
+Block m2 = Block(m1.center + (d,-d), width=a, height=a, "$m_2$");
+
+pair ropecross = extension(m1.center, m1.center+E,
+ m2.center, m2.center+N);
+pair pulley = ropecross + (-pr, -pr/Tan(90/2));
+
+// pulley support
+draw(pulley -- (pulley + 1.3(s.pTo-pulley)), psw+currentpen);
+
+draw(m1.center -- (pulley.x, m1.center.y));
+draw(m2.center -- (m2.center.x, pulley.y));
+
+filldraw(shift(pulley)*scale(pr)*unitcircle, fillpen=white);
+
+s.draw();
+m1.draw();
+m2.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+real m1 = 5;
+real m2 = 9;
+real g = 0.2u;
+real t = (m1*m2*g)/(m1 + m2);
+
+Vector T = Force((0,0), mag=t, dir=0, "$T$");
+T.draw();
+Vector G = Force((0,0), mag=m1*g, dir=-90, "$F_g$");
+G.draw();
+Vector norm = Force((0,0), mag=m1*g, dir=90, "$N$");
+norm.draw();
+dot("$m_1$", (0,0), W);
+\end{asy}
+\hspace{1cm}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+real m1 = 5;
+real m2 = 9;
+real g = 0.2u;
+real t = (m1*m2*g)/(m1 + m2);
+
+Vector T = Force((0,0), mag=t, dir=90, "$T$");
+T.draw();
+Vector G = Force((0,0), mag=m2*g, dir=-90, "$F_g$");
+G.draw();
+dot("$m_2$", (0,0), E);
+\end{asy}
+\end{center}
+
+\Part{b}
+Because the rope does not stretch, both objects have the same
+magnitude of acceleration. Using $F=ma$ on both objects, we can solve
+for $a$.
+\begin{align}
+ T &= m_1 a \\
+ m_2 g - T &= m_2 a \\
+ m_2 g - m_1 a &= m_2 a \\
+ m_2 g &= (m_1 + m_2) a \\
+ a &= \frac{m_2 g}{m_1 + m_2}
+ = \frac{9.00\U{kg}\cdot9.80\U{m/s$^2$}}{5.00\U{kg} + 9.00\U{kg}}
+ = \ans{6.30\U{m/s$^2$}}
+\end{align}
+
+\Part{c}
+Plugging the solution for $a$ back into either of the $F=ma$
+equations,
+\begin{equation}
+ T = m_1 a
+ = \frac{m_1 m_2 g}{m_1 + m_2}
+ = \frac{5.00\U{kg}\cdot9.00\U{kg}\cdot9.80\U{m/s$^2$}}
+ {5.00\U{kg} + 9.00\U{kg}}
+ = \ans{32.1\U{N}}
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{5.30}
+Two objects are connected by a light string that passes over a
+frictionless pulley as shown in Figure P5.30. Assume the incline is
+frictionless and take $m_1=2.00\U{kg}$, $m_2=6.00\U{kg}$, and
+$\theta=55.0\dg$. \Part{a} Draw free-body diagrams of both objects.
+Find \Part{b} the magnitude of the acceleration of the
+objects, \Part{c} the tension in the string, and \Part{d} the speed of
+each object $2.00\U{s}$ after it is released from rest.
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+
+real theta = 55.0; // ramp angle in degrees
+real w = 2u; // width of ramp base
+
+// ramp corners
+pair botL = (0, 0);
+pair botR = (w, 0);
+pair top = (0, w*Tan(theta));
+
+real a = u; // diameter of m_1, side length of m_2
+real dx = 0.1u; // narrow sliver separating m_1 from wedge wall
+real pr = a/4; // pulley radius
+real psw = 0.2u; // pulley support width
+
+Surface s = Surface((botL - (0.2w + a + dx, 0)), botR + (0.2w, 0));
+
+Angle A = Angle(botL, botR, top, "$\theta$");
+
+Mass m1 = Mass(top - (a/2 + dx, 1.5a), radius=a/2, "$m_1$");
+pair m2bot = (top+botR)/2;
+pair m2perp = rotate(-90)*dir(top-botR);
+Block m2 = Block(
+ m2bot + a/2*m2perp, width=a, height=a, direction=-theta, "$m_2$");
+
+pair ropecross = extension(m1.center, m1.center+N,
+ m2.center, m2.center+(top-botR));
+pair pulley = ropecross + (pr, -pr/Tan(theta/2));
+
+s.draw();
+
+// pulley support
+draw(pulley -- (pulley + 1.3(top-pulley)), psw+currentpen);
+
+filldraw(botL -- botR -- top -- cycle, fillpen=rgb(0.8,0.8,0.3));
+A.draw();
+
+draw(m1.center -- (m1.center.x, pulley.y));
+draw(m2.center -- (pulley + pr*m2perp));
+
+filldraw(shift(pulley)*scale(pr)*unitcircle, fillpen=white);
+
+m1.draw();
+m2.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+real g = 0.2u;
+real theta = 55;
+real m1 = 2;
+real m2 = 6;
+real t = g*m1*m2/(m1+m2)*(1+Sin(theta));
+
+Vector T = Force((0,0), mag=t, dir=90, "$T$");
+T.draw();
+Vector G = Force((0,0), mag=m1*g, dir=-90, "$m_1 g$");
+G.draw();
+dot("$m_1$", (0,0), W);
+\end{asy}
+\hspace{1cm}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+real g = 0.2u;
+real theta = 55;
+real m1 = 2;
+real m2 = 6;
+real t = g*m1*m2/(m1+m2)*(1+Sin(theta));
+
+Vector T = Force((0,0), mag=t, dir=180-theta, "$T$");
+T.draw();
+Vector N = Force((0,0), mag=m2*g*Cos(theta), dir=90-theta, "$N$");
+N.draw();
+Vector G = Force((0,0), mag=m2*g, dir=-90, "$m_2 g$");
+G.draw();
+dot("$m_2$", (0,0), SE);
+\end{asy}
+\end{center}
+
+\Part{b}
+Because the rope does not stretch, both objects have the same
+magnitude of acceleration. Using $F=ma$ on both objects, we can solve
+for $a$. $m_2$ is the heavier object, so we'll pick the positive
+direction to be dropping $m_2$ and raising $m_1$.
+\begin{align}
+ T - m_1 g &= m_1 a \\
+ m_2 g \sin(\theta) - T &= m_2 a \\
+ T &= m_1 (g+a) \\
+ m_2 g \sin(\theta) - m_1 (g+a) &= m_2 a \\
+ g (m_2\sin(\theta) - m_1) &= a (m_1 + m_2) \\
+ a &= g \frac{m_2\sin(\theta)-m_1}{m_1 + m_2} \\
+ &= 9.80\U{m/s$^2$}\cdot\frac{6.00\U{kg}\cdot\sin(55.0\dg)-2.00\U{kg}}
+ {2.00\U{kg} + 6.00\U{kg}} \\
+ &= \ans{3.57\U{m/s$^2$}}
+\end{align}
+
+\Part{c}
+Plugging the solution for $a$ back into either of the $F=ma$
+equations,
+\begin{align}
+ T &= m_1 (g+a)
+ = m_1 g \p({1 +\frac{m_2\sin(\theta)-m_1}{m_1 + m_2}})
+ = m_1 g \frac{m_1 + m_2 + m_2\sin(\theta)-m_1}{m_1 + m_2} \\
+ &= m_1 g \frac{m_2(1+\sin(\theta))}{m_1 + m_2}
+ = g \frac{m_1 m_2}{m_1 + m_2}(1+\sin(\theta)) \\
+ &= 9.80\U{m/s$^2$}\cdot\frac{2.00\U{kg} \cdot 6.00\U{kg}}
+ {2.00\U{kg} + 6.00\U{kg}}
+ \cdot(1+\sin(55.0\dg))
+ = \ans{26.7\U{N}}
+\end{align}
+
+\Part{d}
+Because the string does not stretch, the speed of both objects are the
+same. Because the acceleration is constant,
+\begin{equation}
+ v = a\cdot t + v_0 = a\cdot t = 3.57\U{m/s$^2$} \cdot 2.00\U{s}
+ = \ans{7.14\U{m/s}}
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{5.38}
+A car is traveling at $50.0\U{mi/h}$ on a horizontal
+highway. \Part{a} If the coefficient of static friction between road
+and tires on a rainy day is $0.100$, what is the minimum distance in
+which the car will stop? \Part{b} What is the stopping distance when
+the surface is dry and $\mu_s=0.600$?
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+A normal force with magnitude $mg$ is required to keep the car from
+accelerating in the vertical direction (and either sinking into the
+pavement or levitating above it). The frictional resistance has
+magnitude
+\begin{equation}
+ F_f = \mu_s N = \mu_s m g \;,
+\end{equation}
+which gives a deceleration of
+\begin{align}
+ -F_f = m a &= -\mu_s m g \\
+ a &= -\mu_s g \;.
+\end{align}
+Because the car is moving, you might expect the coefficient of kinetic
+friction would be more appropriate. However, if the wheels are not
+skidding (e.g. with anti-lock brakes), the tire does not slide over
+the road, so you use the coeffient of static friction. You would use
+a coefficient of kinetic friction if you were analyzing the
+disk/brake-pad interaction.
+
+For constant acceleration problems,
+\begin{equation}
+ v^2 = v_0^2 + 2a(x-x_0)
+\end{equation}
+(see my solution to Prob.~2.33 for a derivation).
+We can use this forumla to solve for the stopping distance
+\begin{equation}
+ \Delta_x = x-x_0 = \frac{v^2-v_0^2}{2a} = \frac{-v_0^2}{2a}
+ = \frac{v_0^2}{2\mu_s g} \;.
+\end{equation}
+
+Converting the initial speed to $\bareU{m/s}$,
+\begin{equation}
+ v_0 = 50.0\U{mi/h} \cdot \frac{1.61\U{km}}{1.00\U{mi}}
+ \cdot \frac{1.00\U{h}}{3600\U{s}}
+ = 22.4\U{m/s} \;.
+\end{equation}
+Plugging into our formula for stopping distance
+\begin{equation}
+ \Delta_x = \frac{v_0^2}{2\mu_s g}
+ = \frac{(22.4\U{m/s})^2}{2\cdot0.100\cdot9.80\U{m/s$^2$}}
+ = \ans{255\U{m}} \;.
+\end{equation}
+
+\Part{b}
+Plugging the new $\mu_s$ into our formula for stopping distance
+\begin{equation}
+ \Delta_x = \frac{v_0^2}{2\mu_s g}
+ = \frac{(22.4\U{m/s})^2}{2\cdot0.600\cdot9.80\U{m/s$^2$}}
+ = \ans{42.5\U{m}} \;,
+\end{equation}
+which is much shorter.
+\end{solution}
--- /dev/null
+\begin{problem*}{5.47}
+Two blocks connected by a rope of negligable mass are being dragged by
+a horizontal force (Fig.~P5.47). Suppose $F=68.0\U{N}$,
+$m_1=12.0\U{kg}$, $m_2=18.0\U{kg}$, and the coefficient of kinetic
+friction between each block and the surface is $0.100$. \Part{a} Draw
+a free-body diagram for each block. Determine \Part{b} the
+acceleration of the system and \Part{c} the tension $T$ in the rope.
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+real h = u; // height of blocks
+real a = u; // width of blocks
+real d = 2a; // distance between block centers
+real f = a; // magnitude of force
+
+Surface s = Surface((-0.7a,0), (d+0.7a+f,0));
+s.draw();
+Block b1 = Block((0, h/2), width=a, height=h, "$m_1$");
+Block b2 = Block((d, h/2), width=a, height=h, "$m_2$");
+draw(b1.center -- b2.center); // rope
+label("$T$", (b1.center+b2.center)/2, N);
+b1.draw();
+b2.draw();
+Vector F = Force(b2.center+(a/2,0), mag=f, dir=0, "$\vect{F}$");
+F.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 0.05cm;
+real vscale = 0.1; // rescale vertical forces
+real f = 68.0u;
+real g = 9.8u;
+real m1 = 12;
+real m2 = 18;
+real mu = 0.1;
+real t = f*m1/(m1+m2);
+
+Vector G = Force((0,0), mag=m1*g*vscale, dir=-90, "$m_1g$");
+Vector N = Force((0,0), mag=m1*g*vscale, dir=90, "$\vect{N}_1$");
+Vector T = Force((0,0), mag=t, dir=0, "$T$");
+Vector F = Force((0,0), mag=mu*m1*g, dir=180, "$\vect{F}_{f1}$");
+
+G.draw();
+N.draw();
+T.draw();
+F.draw();
+dot("$m_1$", (0,0), NE);
+\end{asy}
+\hspace{1cm}
+\begin{asy}
+import Mechanics;
+
+real u = 0.05cm;
+real vscale = 0.1; // rescale vertical forces
+real f = 68.0u;
+real g = 9.8u;
+real m1 = 12;
+real m2 = 18;
+real mu = 0.1;
+real t = f*m1/(m1+m2);
+real dy=1mm;
+
+Vector G = Force((0,0), mag=m2*g*vscale, dir=-90, "$m_2g$");
+Vector N = Force((0,0), mag=m2*g*vscale, dir=90, "$\vect{N}_2$");
+Vector E = Force((0,0), mag=f, dir=0, "$\vect{F}$");
+Vector T = Force((0,dy), mag=t, dir=180, "$T$");
+Vector F = Force((0,-dy), mag=mu*m2*g, dir=180, "$\vect{F}_{f2}$");
+
+G.draw();
+N.draw();
+E.draw();
+T.draw();
+F.draw();
+dot("$m_2$", (0,0), NE);
+\end{asy}
+\end{center}
+
+\Part{b}
+Because the string does not stretch, the blocks will have the same
+acceleration, and can be treated as a single block.
+\begin{align}
+ F - \mu (m_1 + m_2) g &= (m_1 + m_2)a \\
+ a + \mu g &= \frac{F}{m_1 + m_2} \\
+ a &= \frac{F}{m_1 + m_2} - \mu g \\
+ &= \frac{68.0\U{N}}{12.0\U{kg}+18.0\U{kg}}
+ - 0.100\cdot 9.80\U{m/s$^2$}
+ = \ans{1.29\U{m/s$^2$}}
+\end{align}
+
+\Part{c}
+We can use the horizontal force on $m_1$ to calculate the tension
+\begin{align}
+ T - \mu m_1 g &= m_1 a \\
+ T &= m_1 (\mu g + a)
+ = m_1 \frac{F}{m_1 + m_2}
+ = F \frac{m_1}{m_1 + m_2} \\
+ &= 68.0\U{N} \frac{12.0\U{kg}}{12.0\U{kg} + 18.0\U{kg}}
+ = \ans{27.2\U{N}}
+\end{align}
+\end{solution}
--- /dev/null
+\begin{problem*}{5.63}
+A crate of wieght $F_g$ is pushed by a force $\vect{P}$ on a
+horizontal floor as shown in Figure P5.63. The coefficient of static
+friction is $\mu_s$, and $\vect{P}$ is directed at an angle $\theta$
+below the horizontal. \Part{a} Show that the minimum value of $P$
+that will move the crate is given by
+\begin{equation}
+ P = \frac{\mu_s F_g \sec\theta}{1 - \mu_s \tan\theta}
+\end{equation}
+\Part{b} Find the condition on $\theta$ in terms of $\mu_s$, for
+which motion of the crate is impossible for any value of $P$.
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+pair p = (0.3u, -0.3u);
+
+Surface s = Surface((-p.x,0), (1u+p.x,0));
+s.draw();
+Block b = Block((0.5u, 0.35u), width=1u, height=0.7u, "crate");
+b.draw();
+Vector P = Force((0,0.7u)-p, mag=length(p), dir=degrees(p), "$\vect{P}$");
+P.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+The normal force must resist both the force of gravity and the
+vertical component of $\vect{P}$, so
+\begin{equation}
+ N = F_g + P\sin(\theta) \;.
+\end{equation}
+This moves the crate when the horizontal component of $\vect{P}$
+balances the force of friction.
+\begin{align}
+ P\cos(\theta) &= \mu_s N = \mu_s (F_g + P\sin(\theta)) \\
+ P(\cos(\theta) - \mu_s \sin(\theta)) &= \mu_s F_g \\
+ P(1 - \mu_s \tan(\theta)) &= \mu_s F_g \sec(\theta) \\
+ P &= \ans{\frac{\mu_s F_g \sec(\theta)}{1 - \mu_s \tan(\theta)}} \;,
+\end{align}
+which is what we set out to show.
+
+Note that this formula is only valid when there is an actual normal
+force to provide friction. Therefore $P\cos(\theta) > 0$. We can
+posit, without loss of generality, that $P>0$, in which case the
+restriction is $-90\dg < \theta < 90\dg$. By symmetry, the situation
+for the backside $180\dg$ is just a mirror image of the frontside.
+
+\Part{b}
+As $P$ becomes larger, the $F_g$ component of our horizontal force
+balance becomes negligable, so we cannot move the block when
+\begin{align}
+ P\cos(\theta) &\le \mu_s P \sin(\theta) \\
+ \frac{1}{\mu_s} &\le \tan(\theta) \\
+ \theta &\ge \ans{\arctan\p({\frac{1}{\mu_s}})} \equiv \theta_c \;,
+\end{align}
+where the last step uses the fact that $\tan(\theta)$ is strictly
+increasing on the range $\theta\in(-90\dg,90\dg)$.
+
+What does this mean about our answer to \Part{a}? Let's rework the
+condition to look more like the denominator in the \Part{a} answer.
+\begin{align}
+ \tan(\theta) &\ge \frac{1}{\mu_s} \\
+ 0 &\ge \frac{1}{\mu_s} - \tan(\theta) \\
+ 0 &\ge 1 - \mu_s \tan(\theta) \;,
+\end{align}
+so the denominator is negative or zero for $\theta \ge \theta_c $.
+For $\theta$ just below the cutoff, the denominator is small but
+positive, and you get a really large value for $P$. For
+$\theta=\theta_c$, the denominator is zero, and you get an infinite
+value for $P$. For $\theta$ above the cutoff, the denominator is
+negative, so $P$ is also negative, which, as I pointed out
+in \Part{a}, is not allowed.
+
+The whole thing is a bit easier to understand if we rephrase the
+answer to \Part{a} as
+\begin{equation}
+ P = \frac{\mu_s F_g}{\cos(\theta) - \mu_s \sin(\theta)}
+ = \frac{C}{\cos(\theta) - \mu_s \sin(\theta)}
+ = (A\cos(\theta) - B\sin(\theta))^{-1} \;,
+\end{equation}
+where $C=\equiv \mu_s F_g$, $A\equiv 1/C$, and $B\equiv \mu_s/C=1/F_g$.
+We can consolidate to a single trig term using
+\begin{align}
+ \sin(a \pm b) &= \sin(a)\cos(b) \pm \cos(a)\sin(b) \\
+ (D\sin(a - b))^{-1} &= (D \sin(a)\cos(b) - D\cos(a)\sin(b))^{-1} \;.
+\end{align}
+Matching with our formula,
+\begin{align}
+ \theta &= b \\
+ A &= D\sin(a) \\
+ B &= D\cos(a) \\
+ \tan(a) &= \frac{A}{B} = \frac{1/C}{\mu_s/C} = \frac{1}{\mu_s} \\
+ a &= \arctan\p({\frac{1}{\mu_s}}) \\
+ D &= \frac{B}{\cos(a)}
+ = \frac{B}{\cos\p({\arctan\p({\frac{1}{\mu_s}})})}
+ = B\sqrt{1+\frac{1}{\mu_s^2}}
+ = \frac{\sqrt{1+\frac{1}{\mu_s^2}}}{F_g} \\
+ P &= D^{-1} \p({ \sin(a - b) })^{-1}
+ = \frac{F_g}{\sqrt{1 + \frac{1}{\mu_s^2}}}
+ \csc\p({\arctan\p({\frac{1}{\mu_s}}) - \theta}) \;.
+\end{align}
+This doesn't look as clean as the phrasing in \Part{a}, but it makes
+the dependence of $P$ on $\theta$ much clearer. For example, $P$ is
+obviously negative for $\theta > \theta_c \equiv \arctan(1/\mu_s)$.
+The dependency on $\theta$ over the rest of the range is
+\begin{equation}
+ P \propto \csc(\theta_c - \theta) = \frac{1}{\sin(\theta_c - \theta)}
+\end{equation}
+Because $\mu_s$ is a positive number, $1/\mu_s$ will also be positive,
+and $\theta_c$ will be between $0$ and $90\dg$. The status on all
+possible angles looks something like
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real tc = 67;
+real u = 1cm;
+real r = u;
+real R = 1.5u;
+real L = 3u;
+
+Angle Atc = Angle((1,0), (0,0), dir(tc), "$\theta_c$");
+Atc.draw();
+
+draw(scale(r)*unitcircle);
+draw((-R,0)--(R,0));
+draw((0,-R)--(0,R), red);
+draw((-dir(tc)*R)--(dir(tc)*R), blue);
+
+label("$\cos(\theta)>0$", L*dir(0), red);
+label("$P > 0$", L*dir(tc-90), blue);
+\end{asy}
+\end{center}
+Taking the $\cos(\theta)>0$ portion of our $P$ dependence (where the
+equation we started with in \Part{a} applies, and combining it with
+the reflection (which applies when $\cos(\theta)<0$, we get
+\begin{equation}
+ P = \begin{cases}
+ \infty & \text{if $\theta_c \ge \theta < 180\dg-\theta_c$} \\
+ \frac{F_g}{\sqrt{1 + \frac{1}{\mu_s^2}}}
+ \csc\p({\arctan\p({\frac{1}{\mu_s}}) - \theta})
+ & \text{if $-90\dg \le \theta < \theta_c$} \\
+ \frac{F_g}{\sqrt{1 + \frac{1}{\mu_s^2}}}
+ \csc\p({\arctan\p({\frac{1}{\mu_s}}) - 180\dg + \theta})
+ & \text{if $180\dg - \theta_c < \theta \le 180\dg$
+ or $-180\dg \le \theta \le -90\dg$} \\
+ \end{cases}
+\end{equation}
+which looks like
+\begin{center}
+\begin{asy}
+import graph;
+import Mechanics;
+
+real u = 2cm;
+
+//size(0, 2u);
+scale(false);
+
+real tc = 67;
+real tc_r = tc*pi/180; // theta_c in radians
+real tc_buf = (tc-25)*pi/180; // weakened version of tc_r
+
+real r_scale(real r) { return u/2 * r; }
+real R(real theta) { return r_scale( 1/sin(tc_r-theta) ); }
+real L(real theta) { return R(pi-theta); }
+real onef (real theta) { return r_scale(1); }
+real twof (real theta) { return r_scale(2); }
+
+draw((0,0)--(u,0));
+draw((0,0)--(u*dir(tc)));
+draw((0,0)--(u*dir(180-tc)));
+
+draw(polargraph(onef, -pi-tc_r, tc_r));
+draw(polargraph(twof, -pi-tc_r, tc_r));
+label("$1$", r_scale(1)*dir(tc-90), SE);
+label("$2$", r_scale(2)*dir(tc-90), SE);
+
+draw(polargraph(R, -pi/2, tc_buf), blue);
+draw(polargraph(L, pi-tc_buf, 3pi/2), blue);
+
+Angle Ainf = Angle(dir(180-tc), (0,0), dir(tc), radius=0.7u, red, "$\infty$");
+Ainf.draw();
+Angle Atc = Angle(dir(0), (0,0), dir(tc), "$\theta_c$");
+Atc.draw();
+\end{asy}
+\end{center}
+where I've just plotted the $\theta$ dependence of $P$, setting the
+constant $F_g/\sqrt{ }$ term equal to $1$. Note that the $\csc$ makes
+nice, straight lines in this polar plot.
+\end{solution}
projects / course.git / commitdiff