From: W. Trevor King Date: Fri, 21 Jan 2011 18:52:10 +0000 (-0500) Subject: Add week four recitation problems. X-Git-Url: http://git.tremily.us/?a=commitdiff_plain;h=c187a2ca29f299234c4e630193dce5e342df8e06;p=course.git Add week four recitation problems. --- diff --git a/latex/problems/Serway_and_Jewett_8/problem05.25.tex b/latex/problems/Serway_and_Jewett_8/problem05.25.tex new file mode 100644 index 0000000..391cbb7 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem05.25.tex @@ -0,0 +1,96 @@ +\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} diff --git a/latex/problems/Serway_and_Jewett_8/problem05.28.tex b/latex/problems/Serway_and_Jewett_8/problem05.28.tex new file mode 100644 index 0000000..e2c2867 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem05.28.tex @@ -0,0 +1,104 @@ +\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} diff --git a/latex/problems/Serway_and_Jewett_8/problem05.30.tex b/latex/problems/Serway_and_Jewett_8/problem05.30.tex new file mode 100644 index 0000000..b6516d1 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem05.30.tex @@ -0,0 +1,140 @@ +\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} diff --git a/latex/problems/Serway_and_Jewett_8/problem05.38.tex b/latex/problems/Serway_and_Jewett_8/problem05.38.tex new file mode 100644 index 0000000..304a2d8 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem05.38.tex @@ -0,0 +1,62 @@ +\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} diff --git a/latex/problems/Serway_and_Jewett_8/problem05.47.tex b/latex/problems/Serway_and_Jewett_8/problem05.47.tex new file mode 100644 index 0000000..d6c2147 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem05.47.tex @@ -0,0 +1,109 @@ +\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} diff --git a/latex/problems/Serway_and_Jewett_8/problem05.63.tex b/latex/problems/Serway_and_Jewett_8/problem05.63.tex new file mode 100644 index 0000000..bccafb2 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem05.63.tex @@ -0,0 +1,195 @@ +\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}