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\begin{document}

\frame{\frametitle{\b Static Equilibrium - 1} \bigskip

{\bf Push-up (1):} 
$$ d_1\,m\,g = (d_1+d_2)\,F_2\,,~~~d_1\,F_1 = d_2\,F_2\,.$$
\smallskip

{\bf Static Equilibrium between Two Blocks (2):} $T_2$ is
determined by $M_2$ and gravity. Then,
$$ T_3\,cos(\theta)=T_2,~~~T_1=T_3\,\sin(\theta)\,,~~~
   T_1 = M\,g\,\mu_s\,. $$
\smallskip % \bigskip

{\bf Two Scales and Three Blocks (3).} Note that the weight of
block $X$ contributes with $d=d_1+d_2+d_3$ to scales $A$ and $B$ 
as follows:
$$ A = \frac{d_2+d_3}{d}\,X + \dots~~{\rm and}~~
   B = \frac{d_1}{d}\,X + \dots~\,. $$
Treat the weights of the other blocks correspondingly.
\smallskip % \bigskip

\vskip 20pt }


\frame{\frametitle{\b Static Equilibrium - 2} \bigskip

{\bf Horizontal Bar (4).} With $T_{y\rm max}=T_{\max}\,\sin(\theta)$
solve the torque equation \vskip -20pt
$$ L\,T_{y\rm max} = x\,W_M + \frac{L}{2}\,W_L $$
for $x$. Then, $T_{x\rm max}=?$ and $F_y=?$ at $A$. 
\bigskip

{\bf Painter on a Ladder (5).} Replace ladder and painter by a 
weightless ladder with one weight added at its top and a second 
at its bottom. Write the upper weight vector as a sum of the tension 
down the ladder and a vector perpendicular towards the wall. Then deal 
with the lower end of the ladder.  Draw all vectors involved at the 
upper as well as at the lower end.
\bigskip

{\bf Pole on an Incline (6):} 
$$ \mu_s\,F_{\rm per} = F_{\rm par}\,,$$
where $F_{\rm per}$ is the force perpendicular and $F_{\rm par}$ the 
force parallel to the incline (acting where the pole touches the
incline).

\vskip 20pt }


\frame{\frametitle{\b Static Equilibrium - 3} \bigskip

{\bf Wheel on an Incline (7).} Consider an infinitesimal movement
by $ds$ of the wheel down the incline. Calculate the corresponding
potential energy differences for (a)~the cm of the the wheel, and
(b)~the mass $m$. Calculate the mass $m$ for which the summed up 
potential energy difference is zero. \vfill}

\end{document}

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