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Determine the balanced state

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?


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?

Frictionless stick in balance
Frictionless stick in balance


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

Gleichgewichtslage eines (ideal glatten) Stabs ermitteln - Technische Mechanik 1, Übung 23
Determination of the balanced state

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.

Reaction forces in the slot
Reaction forces in the slot
Balance of forces in x-direction:

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

Balance of forces in y-direction:

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

Balance of moments around A

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

Solve for l

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

Solving equations 1 and 2 for FA and FB

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

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

Substitution of FA

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

Substitution of FB in equation 4

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


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

Alternative notation of the result

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