# [OS X TeX] Problem accessing a .sty file command?

Christopher Allen christopher.t.allen.95 at Alum.Dartmouth.ORG
Thu Oct 15 10:21:04 EDT 2009

I have moved to a new computer. I loaded TeXShop along with the full MacTeX
distribution. It looks as if all the packages I regularly use (AMS stuff,
PSTricks stuff, beamer plus its required pieces, probsoln, calc, delarray,
ifthen) are included. Great, yay! I'm very thankful for this.

However, when I try to compile a previously working file using probsoln, it
appears to find probsoln.sty correctly, at
/usr/local/texlive/2008/texmf-dist/tex/latex/probsoln/probsoln.sty , but
using \selectrandomly{}{} returns "! Undefined control sequence." I don't
understand why, if it finds probsoln.sty, it would not pick up that command.
Normally I would think I made typo in the command name, but the file
compiled without problems many times on the previous computer.

Thanks for any help anyone can give. I'll attach the code and the log below.

Chris

___The Primary File___

\documentclass[letterpaper,10pt,fleqn]{article}

\usepackage[letterpaper,width=6.5in,height=9in,vmarginratio=1:1]{geometry}

\usepackage{pst-all}
\usepackage{amstext}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{delarray}

\allowdisplaybreaks

\usepackage{ifthen}
\usepackage{calc}

\newcounter{versionnumber}
\setcounter{versionnumber}{15}% student ##
\newcounter{problemset}
\setcounter{problemset}{02}% problem set ##
\newcounter{thisyearnumber}
\setcounter{thisyearnumber}{9}% year #
\newcounter{finalversion}
\setcounter{finalversion}{10000*\value{thisyearnumber}+100*\value{problemset
}+\value{versionnumber}}
\PSNrandseed{\value{finalversion}}

\begin{document}

\center{\textbf{\LARGE{Problem Set \#2 - Version
Solutions}{}}}}\\

\begin{enumerate}

\selectrandomly{ps02ramp}{1}
%\selectallproblems{ps02ramp}

%\selectrandomly{ps02stackedblocks}{1}
%\selectallproblems{ps02stackedblocks}

%\selectrandomly{ps02upanddown}{1}
%\selectallproblems{ps02upanddown}

%\selectrandomly{ps02spinningball}{1}
%\selectallproblems{ps02spinningball}

\end{enumerate}

\end{document}

___ps02ramp.tex___

\newproblem{ramp1}{%
%
% Problem
%
Two blocks, of masses $m_1$ and $m_2$ lie on a ramp and are tied together by
a very light string. The blocks are released from rest. The coefficients of
static and kinetic friction between block 1 and the ramp are 0.4 and 0.3,
respectively. The coefficients of static and kinetic friction between block
2 and the ramp are 0.6 and 0.4, respectively. $m_1$ is just massive enough
compared to $m_2$ for the blocks to accelerate down the ramp.
\begin{center}
\begin{pspicture}(7.5,4.2)
\psline(0,4)(6.928,0)
\psline[linestyle=dotted](7.5,0)
\psarc(6.928,0){.6}{150}{180}
\rput[r](6.228,.2){$30^\circ$}
\psline[linewidth=2
pt](.866,3.5)(1.559,3.1)(1.959,3.793)(1.266,4.193)(.866,3.5)
\rput{-30}(1.412,3.646){$m_2$}
\psline(1.759,3.446)(3.491,2.446)
\psline[linewidth=2
pt](3.291,2.1)(3.984,1.7)(4.384,2.393)(3.691,2.793)(3.291,2.1)
\rput{-30}(3.837,2.246){$m_1$}
\end{pspicture}
\end{center}
\begin{enumerate}
\item In terms of $m_2$, what minimum mass can $m_1$ have if the blocks are
to accelerate? (8 pts)
\item Find the magnitude of the blocks' acceleration down the ramp in terms
of $g$. (8 pts)
\item Determine the magnitude of the tension in the string in terms of $m_2$
and $g$. (4 pts)
\end{enumerate}}
%
% Solution
%
{\begin{enumerate}
\item Use Newton's second law with axes parallel to and perpendicular to the
ramp's surface to write a system of equations.
\begin{equation*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-F_{s1}-F_{s2}=(m_1+m_2)a_x\\
-F_{g1y}+F_{n1}=m_1a_{1y}\\
-F_{g2y}+F_{n2}=m_2a_{2y}
\end{array}
\end{equation*}
Include the conditions on the acceleration.
\begin{equation*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-F_{s1}-F_{s2}>0\\
-F_{g1y}+F_{n1}=0\\
-F_{g2y}+F_{n2}=0
\end{array}
\end{equation*}
Include the equation for static friction and the components of gravity to
solve the system of equations.
\begin{gather*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-\mu_{s1}F_{n1}-\mu_{s2}F_{n2}>0\\
-F_{g1y}+F_{n1}=0\\
-F_{g2y}+F_{n2}=0
\end{array}\\
F_{g1x}+F_{g2x}-\mu_{s1}F_{g1y}-\mu_{s2}F_{g2y}>0\\
m_1g\sin30^\circ+m_2g\sin30^\circ-\mu_{s1}m_1g\cos30^\circ-\mu_{s2}m_2g\cos3
0^\circ>0\\
m_1g\sin30^\circ-\mu_{s1}m_1g\cos30^\circ>\mu_{s2}m_2g\cos30^\circ-m_2g\sin3
0^\circ\\
m_1-\sqrt{3}\mu_{s1}m_1>\sqrt{3}\mu_{s2}m_2-m_2\\
\begin{aligned}
m_1&>\frac{3\sqrt{3}-5}{5-2\sqrt{3}}m_2\\
&>\boxed{\frac{5\sqrt{3}-7}{13}m_2}
\end{aligned}
\end{gather*}
\item Use the same system of equations as in part (a), but kinetic friction
replaces the static friction and $a_x\neq0$.
\begin{gather*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-\mu_{k1}F_{n1}-\mu_{k2}F_{n2}=(m_1+m_2)a_x\\
-F_{g1y}+F_{n1}=0\\
-F_{g2y}+F_{n2}=0
\end{array}\\
(m_1+m_2)a_x=F_{g1x}+F_{g2x}-\mu_{k1}F_{g1y}-\mu_{k2}F_{g2y}\\
(m_1+m_2)a_x=m_1g\sin30^\circ+m_2g\sin30^\circ-\mu_{k1}m_1g\cos30^\circ-\mu_
{k2}m_2g\cos30^\circ\\
\biggl(\frac{5\sqrt{3}-7}{13}+1\biggr)m_2a_x=\frac{1}{2}\biggl(\frac{5\sqrt{
3}-7}{13}\biggr)m_2g+\frac{1}{2}m_2g-\frac{3\sqrt{3}}{20}\biggl(\frac{5\sqrt
{3}-7}{13}\biggr)m_2g-\frac{4\sqrt{3}}{20}m_2g\\
\frac{5\sqrt{3}+6}{13}a_x=\frac{10(5\sqrt{3}-7)+130-3\sqrt{3}(5\sqrt{3}-7)-5
2\sqrt{3}}{260}g\\
\begin{aligned}
a_x&=\frac{50\sqrt{3}-70+130-45+21\sqrt{3}-52\sqrt{3}}{20(5\sqrt{3}+6)}g\\
&=\boxed{\frac{5-\sqrt{3}}{20}g}
\end{aligned}
\end{gather*}
\item Determine the net force on either block and use that with Newton's
second law. In this case block 2 is used.
\begin{gather*}
F_t+F_{g2x}-F_{k2}=m_2a_x\\
\begin{aligned}
F_t&=m_2a_x+\mu_{k2}F_{g2y}-F_{g2x}\\
&=m_2\biggl(\frac{5-\sqrt{3}}{20}g\biggr)+\frac{\sqrt{3}}{2}\cdot\frac{4}{10
}m_2g-\frac{1}{2}m_2g\\
&=\boxed{\frac{3\sqrt{3}-5}{20}m_2g}
\end{aligned}
\end{gather*}
\end{enumerate}}
%
%
%
\newproblem{ramp2}{%
%
% Problem
%
Two blocks, of masses $m_1$ and $m_2$ lie on a ramp and are tied together by
a very light string. The blocks are released from rest. The coefficients of
static and kinetic friction between block 1 and the ramp are 0.4 and 0.3,
respectively. The coefficients of static and kinetic friction between block
2 and the ramp are 0.7 and 0.5, respectively. $m_1$ is just massive enough
compared to $m_2$ for the blocks to accelerate down the ramp.
\begin{center}
\begin{pspicture}(7.5,4.2)
\psline(0,4)(6.928,0)
\psline[linestyle=dotted](7.5,0)
\psarc(6.928,0){.6}{150}{180}
\rput[r](6.228,.2){$30^\circ$}
\psline[linewidth=2
pt](.866,3.5)(1.559,3.1)(1.959,3.793)(1.266,4.193)(.866,3.5)
\rput{-30}(1.412,3.646){$m_2$}
\psline(1.759,3.446)(3.491,2.446)
\psline[linewidth=2
pt](3.291,2.1)(3.984,1.7)(4.384,2.393)(3.691,2.793)(3.291,2.1)
\rput{-30}(3.837,2.246){$m_1$}
\end{pspicture}
\end{center}
\begin{enumerate}
\item In terms of $m_2$, what minimum mass can $m_1$ have if the blocks are
to accelerate? (8 pts)
\item Find the magnitude of the blocks' acceleration down the ramp in terms
of $g$. (8 pts)
\item Determine the magnitude of the tension in the string in terms of $m_2$
and $g$. (4 pts)
\end{enumerate}}
%
% Solution
%
{\begin{enumerate}
\item Use Newton's second law with axes parallel to and perpendicular to the
ramp's surface to write a system of equations.
\begin{equation*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-F_{s1}-F_{s2}=(m_1+m_2)a_x\\
-F_{g1y}+F_{n1}=m_1a_{1y}\\
-F_{g2y}+F_{n2}=m_2a_{2y}
\end{array}
\end{equation*}
Include the conditions on the acceleration.
\begin{equation*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-F_{s1}-F_{s2}>0\\
-F_{g1y}+F_{n1}=0\\
-F_{g2y}+F_{n2}=0
\end{array}
\end{equation*}
Include the equation for static friction and the components of gravity to
solve the system of equations.
\begin{gather*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-\mu_{s1}F_{n1}-\mu_{s2}F_{n2}>0\\
-F_{g1y}+F_{n1}=0\\
-F_{g2y}+F_{n2}=0
\end{array}\\
F_{g1x}+F_{g2x}-\mu_{s1}F_{g1y}-\mu_{s2}F_{g2y}>0\\
m_1g\sin30^\circ+m_2g\sin30^\circ-\mu_{s1}m_1g\cos30^\circ-\mu_{s2}m_2g\cos3
0^\circ>0\\
m_1g\sin30^\circ-\mu_{s1}m_1g\cos30^\circ>\mu_{s2}m_2g\cos30^\circ-m_2g\sin3
0^\circ\\
m_1-\sqrt{3}\mu_{s1}m_1>\sqrt{3}\mu_{s2}m_2-m_2\\
\begin{aligned}
m_1&>\frac{7\sqrt{3}-10}{10-4\sqrt{3}}m_2\\
&>\boxed{\frac{15\sqrt{3}-8}{26}m_2}
\end{aligned}
\end{gather*}
\item Use the same system of equations as in part (a), but kinetic friction
replaces the static friction and $a_x\neq0$.
\begin{gather*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-\mu_{k1}F_{n1}-\mu_{k2}F_{n2}=(m_1+m_2)a_x\\
-F_{g1y}+F_{n1}=0\\
-F_{g2y}+F_{n2}=0
\end{array}\\
(m_1+m_2)a_x=F_{g1x}+F_{g2x}-\mu_{k1}F_{g1y}-\mu_{k2}F_{g2y}\\
(m_1+m_2)a_x=m_1g\sin30^\circ+m_2g\sin30^\circ-\mu_{k1}m_1g\cos30^\circ-\mu_
{k2}m_2g\cos30^\circ\\
\biggl(\frac{15\sqrt{3}-8}{26}+1\biggr)m_2a_x=\frac{1}{2}\biggl(\frac{15\sqr
t{3}-8}{26}\biggr)m_2g+\frac{1}{2}m_2g-\frac{3\sqrt{3}}{20}\biggl(\frac{15\s
qrt{3}-8}{26}\biggr)m_2g-\frac{5\sqrt{3}}{20}m_2g\\
\frac{15\sqrt{3}+18}{26}a_x=\frac{10(15\sqrt{3}-8)+260-3\sqrt{3}(15\sqrt{3}-
8)-130\sqrt{3}}{520}g\\
\begin{aligned}
a_x&=\frac{150\sqrt{3}-80+260-135+24\sqrt{3}-130\sqrt{3}}{20(15\sqrt{3}+18)}
g\\
&=\frac{10-\sqrt{3}}{60}g\\
&=\boxed{\frac{10-\sqrt{3}}{60}g}
\end{aligned}
\end{gather*}
\item Determine the net force on either block and use that with Newton's
second law. In this case block 2 is used.
\begin{gather*}
F_t+F_{g2x}-F_{k2}=m_2a_x\\
\begin{aligned}
F_t&=m_2a_x+\mu_{k2}F_{g2y}-F_{g2x}\\
&=m_2\biggl(\frac{10-\sqrt{3}}{60}g\biggr)+\frac{\sqrt{3}}{2}\cdot\frac{5}{1
0}m_2g-\frac{1}{2}m_2g\\
&=\boxed{\frac{7\sqrt{3}-10}{30}m_2g}
\end{aligned}
\end{gather*}
\end{enumerate}}
\newproblem{ramp3}{%
%
% Problem
%
Two blocks, of masses $m_1$ and $m_2$ lie on a ramp and are tied together by
a very light string. The blocks are released from rest. The coefficients of
static and kinetic friction between block 1 and the ramp are 0.3 and 0.2,
respectively. The coefficients of static and kinetic friction between block
2 and the ramp are 0.6 and 0.4, respectively. $m_1$ is just massive enough
compared to $m_2$ for the blocks to accelerate down the ramp.
\begin{center}
\begin{pspicture}(7.5,4.2)
\psline(0,4)(6.928,0)
\psline[linestyle=dotted](7.5,0)
\psarc(6.928,0){.6}{150}{180}
\rput[r](6.228,.2){$30^\circ$}
\psline[linewidth=2
pt](.866,3.5)(1.559,3.1)(1.959,3.793)(1.266,4.193)(.866,3.5)
\rput{-30}(1.412,3.646){$m_2$}
\psline(1.759,3.446)(3.491,2.446)
\psline[linewidth=2
pt](3.291,2.1)(3.984,1.7)(4.384,2.393)(3.691,2.793)(3.291,2.1)
\rput{-30}(3.837,2.246){$m_1$}
\end{pspicture}
\end{center}
\begin{enumerate}
\item In terms of $m_2$, what minimum mass can $m_1$ have if the blocks are
to accelerate? (8 pts)
\item Find the magnitude of the blocks' acceleration down the ramp in terms
of $g$. (8 pts)
\item Determine the magnitude of the tension in the string in terms of $m_2$
and $g$. (4 pts)
\end{enumerate}}
%
% Solution
%
{\begin{enumerate}
\item Use Newton's second law with axes parallel to and perpendicular to the
ramp's surface to write a system of equations.
\begin{equation*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-F_{s1}-F_{s2}=(m_1+m_2)a_x\\
-F_{g1y}+F_{n1}=m_1a_{1y}\\
-F_{g2y}+F_{n2}=m_2a_{2y}
\end{array}
\end{equation*}
Include the conditions on the acceleration.
\begin{equation*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-F_{s1}-F_{s2}>0\\
-F_{g1y}+F_{n1}=0\\
-F_{g2y}+F_{n2}=0
\end{array}
\end{equation*}
Include the equation for static friction and the components of gravity to
solve the system of equations.
\begin{gather*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-\mu_{s1}F_{n1}-\mu_{s2}F_{n2}>0\\
-F_{g1y}+F_{n1}=0\\
-F_{g2y}+F_{n2}=0
\end{array}\\
F_{g1x}+F_{g2x}-\mu_{s1}F_{g1y}-\mu_{s2}F_{g2y}>0\\
m_1g\sin30^\circ+m_2g\sin30^\circ-\mu_{s1}m_1g\cos30^\circ-\mu_{s2}m_2g\cos3
0^\circ>0\\
m_1g\sin30^\circ-\mu_{s1}m_1g\cos30^\circ>\mu_{s2}m_2g\cos30^\circ-m_2g\sin3
0^\circ\\
m_1-\sqrt{3}\mu_{s1}m_1>\sqrt{3}\mu_{s2}m_2-m_2\\
\begin{aligned}
m_1&>\frac{6\sqrt{3}-10}{10-3\sqrt{3}}m_2\\
&>\boxed{\frac{30\sqrt{3}-46}{73}m_2}
\end{aligned}
\end{gather*}
\item Use the same system of equations as in part (a), but kinetic friction
replaces the static friction and $a_x\neq0$.
\begin{gather*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-\mu_{k1}F_{n1}-\mu_{k2}F_{n2}=(m_1+m_2)a_x\\
-F_{g1y}+F_{n1}=0\\
-F_{g2y}+F_{n2}=0
\end{array}\\
(m_1+m_2)a_x=F_{g1x}+F_{g2x}-\mu_{k1}F_{g1y}-\mu_{k2}F_{g2y}\\
(m_1+m_2)a_x=m_1g\sin30^\circ+m_2g\sin30^\circ-\mu_{k1}m_1g\cos30^\circ-\mu_
{k2}m_2g\cos30^\circ\\
\biggl(\frac{30\sqrt{3}-46}{73}+1\biggr)m_2a_x=\frac{1}{2}\biggl(\frac{30\sq
rt{3}-46}{73}\biggr)m_2g+\frac{1}{2}m_2g-\frac{2\sqrt{3}}{20}\biggl(\frac{30
\sqrt{3}-46}{73}\biggr)m_2g-\frac{4\sqrt{3}}{20}m_2g\\
\frac{30\sqrt{3}+27}{73}a_x=\frac{5(30\sqrt{3}-46)+365-\sqrt{3}(30\sqrt{3}-4
6)-146\sqrt{3}}{730}g\\
\begin{aligned}
a_x&=\frac{150\sqrt{3}-230+365-90+46\sqrt{3}-146\sqrt{3}}{10(30\sqrt{3}+27)}
g\\
&=\boxed{\frac{1}{6}g}
\end{aligned}
\end{gather*}
\item Determine the net force on either block and use that with Newton's
second law. In this case block 2 is used.
\begin{gather*}
F_t+F_{g2x}-F_{k2}=m_2a_x\\
\begin{aligned}
F_t&=m_2a_x+\mu_{k2}F_{g2y}-F_{g2x}\\
&=m_2\biggl(\frac{1}{6}g\biggr)+\frac{\sqrt{3}}{2}\cdot\frac{4}{10}m_2g-\fra
c{1}{2}m_2g\\
&=\boxed{\frac{3\sqrt{3}-5}{15}m_2g}
\end{aligned}
\end{gather*}
\end{enumerate}}
%
%
%
\newproblem{ramp4}{%
%
% Problem
%
Two blocks, of masses $m_1$ and $m_2$ lie on a ramp and are tied together by
a very light string. The blocks are released from rest. The coefficients of
static and kinetic friction between block 1 and the ramp are 0.3 and 0.2,
respectively. The coefficients of static and kinetic friction between block
2 and the ramp are 0.7 and 0.5, respectively. $m_1$ is just massive enough
compared to $m_2$ for the blocks to accelerate down the ramp.
\begin{center}
\begin{pspicture}(7.5,4.2)
\psline(0,4)(6.928,0)
\psline[linestyle=dotted](7.5,0)
\psarc(6.928,0){.6}{150}{180}
\rput[r](6.228,.2){$30^\circ$}
\psline[linewidth=2
pt](.866,3.5)(1.559,3.1)(1.959,3.793)(1.266,4.193)(.866,3.5)
\rput{-30}(1.412,3.646){$m_2$}
\psline(1.759,3.446)(3.491,2.446)
\psline[linewidth=2
pt](3.291,2.1)(3.984,1.7)(4.384,2.393)(3.691,2.793)(3.291,2.1)
\rput{-30}(3.837,2.246){$m_1$}
\end{pspicture}
\end{center}
\begin{enumerate}
\item In terms of $m_2$, what minimum mass can $m_1$ have if the blocks are
to accelerate? (8 pts)
\item Find the magnitude of the blocks' acceleration down the ramp in terms
of $g$. (8 pts)
\item Determine the magnitude of the tension in the string in terms of $m_2$
and $g$. (4 pts)
\end{enumerate}}
%
% Solution
%
{\begin{enumerate}
\item Use Newton's second law with axes parallel to and perpendicular to the
ramp's surface to write a system of equations.
\begin{equation*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-F_{s1}-F_{s2}=(m_1+m_2)a_x\\
-F_{g1y}+F_{n1}=m_1a_{1y}\\
-F_{g2y}+F_{n2}=m_2a_{2y}
\end{array}
\end{equation*}
Include the conditions on the acceleration.
\begin{equation*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-F_{s1}-F_{s2}>0\\
-F_{g1y}+F_{n1}=0\\
-F_{g2y}+F_{n2}=0
\end{array}
\end{equation*}
Include the equation for static friction and the components of gravity to
solve the system of equations.
\begin{gather*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-\mu_{s1}F_{n1}-\mu_{s2}F_{n2}>0\\
-F_{g1y}+F_{n1}=0\\
-F_{g2y}+F_{n2}=0
\end{array}\\
F_{g1x}+F_{g2x}-\mu_{s1}F_{g1y}-\mu_{s2}F_{g2y}>0\\
m_1g\sin30^\circ+m_2g\sin30^\circ-\mu_{s1}m_1g\cos30^\circ-\mu_{s2}m_2g\cos3
0^\circ>0\\
m_1g\sin30^\circ-\mu_{s1}m_1g\cos30^\circ>\mu_{s2}m_2g\cos30^\circ-m_2g\sin3
0^\circ\\
m_1-\sqrt{3}\mu_{s1}m_1>\sqrt{3}\mu_{s2}m_2-m_2\\
\begin{aligned}
m_1&>\frac{7\sqrt{3}-10}{10-3\sqrt{3}}m_2\\
&>\boxed{\frac{40\sqrt{3}-37}{73}m_2}
\end{aligned}
\end{gather*}
\item Use the same system of equations as in part (a), but kinetic friction
replaces the static friction and $a_x\neq0$.
\begin{gather*}
\begin{array}\{{l}.
F_{g1x}+F_{g2x}-\mu_{k1}F_{n1}-\mu_{k2}F_{n2}=(m_1+m_2)a_x\\
-F_{g1y}+F_{n1}=0\\
-F_{g2y}+F_{n2}=0
\end{array}\\
(m_1+m_2)a_x=F_{g1x}+F_{g2x}-\mu_{k1}F_{g1y}-\mu_{k2}F_{g2y}\\
(m_1+m_2)a_x=m_1g\sin30^\circ+m_2g\sin30^\circ-\mu_{k1}m_1g\cos30^\circ-\mu_
{k2}m_2g\cos30^\circ\\
\biggl(\frac{40\sqrt{3}-37}{73}+1\biggr)m_2a_x=\frac{1}{2}\biggl(\frac{40\sq
rt{3}-37}{73}\biggr)m_2g+\frac{1}{2}m_2g-\frac{2\sqrt{3}}{20}\biggl(\frac{40
\sqrt{3}-37}{73}\biggr)m_2g-\frac{5\sqrt{3}}{20}m_2g\\
\frac{40\sqrt{3}+36}{73}a_x=\frac{10(40\sqrt{3}-37)+730-2\sqrt{3}(40\sqrt{3}
-37)-365\sqrt{3}}{1460}g\\
\begin{aligned}
a_x&=\frac{400\sqrt{3}-370+730-240+74\sqrt{3}-365\sqrt{3}}{20(40\sqrt{3}+36)
}g\\
&=\boxed{\frac{10+\sqrt{3}}{80}g}
\end{aligned}
\end{gather*}
\item Determine the net force on either block and use that with Newton's
second law. In this case block 2 is used.
\begin{gather*}
F_t+F_{g2x}-F_{k2}=m_2a_x\\
\begin{aligned}
F_t&=m_2a_x+\mu_{k2}F_{g2y}-F_{g2x}\\
&=m_2\biggl(\frac{10+\sqrt{3}}{80}g\biggr)+\frac{\sqrt{3}}{2}\cdot\frac{5}{1
0}m_2g-\frac{1}{2}m_2g\\
&=\boxed{\frac{21\sqrt{3}-30}{80}m_2g}
\end{aligned}
\end{gather*}
\end{enumerate}}

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