This exercise is about the following questions:

- Which forces occur at frictionless contact points?
- How do you establish equilibrium conditions?
- How do you calculate lever arms for forces on inclined objects?

## Task

An ideally smooth rod with its own weight G should be inserted into a slot with the width B, so that the contact shown in A and B results and the state of equilibrium is established. What length l must the rod have so that it is in equilibrium at the angle Alpha?

## Solution

The following video is in German language, but English subtitles are available.

The way the problem is solved in the video is a little bit easier than the steps shown below.

The property "frictionless" results in the following reaction forces in the points A and B.

\( \DeclareMathOperator{\abs}{abs} \newcommand{\ensuremath}[1]{\mbox{$#1$}} \)

\[\tag{1} 0={F_A} \sin{\left( \alpha \right) }-{F_B}\]

\[\tag{2} 0={F_A} \cos{\left( \alpha \right) }-G\]

\[\tag{3} 0=G\, \left( \frac{l \cos{\left( \alpha \right) }}{2}-b\right) -\frac{{F_B} b \sin{\left( \alpha \right) }}{\cos{\left( \alpha \right) }}\]

\[\tag{4} l=\frac{2 {F_B} b \sin{\left( \alpha \right) }+2 G b \cos{\left( \alpha \right) }}{G\, {{\cos{\left( \alpha \right) }}^{2}}}\]

_{A}and F

_{B}

\[\tag{5} {F_B}={F_A} \sin{\left( \alpha \right) }\]

\[\tag{6} {F_A}=\frac{G}{\cos{\left( \alpha \right) }}\]

_{A}

\[\tag{7} {F_B}=\frac{G \sin{\left( \alpha \right) }}{\cos{\left( \alpha \right) }}\]

_{B}in equation 4

\[\tag{8} l=\frac{\frac{2 G b\, {{\sin{\left( \alpha \right) }}^{2}}}{\cos{\left( \alpha \right) }}+2 G b \cos{\left( \alpha \right) }}{G\, {{\cos{\left( \alpha \right) }}^{2}}}\]

\[\tag{9} l=\frac{2 b\, {{\sin{\left( \alpha \right) }}^{2}}+2 b\, {{\cos{\left( \alpha \right) }}^{2}}}{{{\cos{\left( \alpha \right) }}^{3}}}\]

\[\tag{10} l=\frac{2 b\, \left( {{\tan{\left( \alpha \right) }}^{2}}+1\right) }{\cos{\left( \alpha \right) }}\]