What is the force exerted by the pivot on the stick?

Thread starter hidemi
Start date Mar 11, 2021
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Force Pivot

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SUMMARY

The discussion centers on the forces acting on a stick pivoted at a point, specifically examining the relationship between forces F1 and F2. When a force is applied at 60 cm from the pivot, F2 must act in the opposite direction to F1 to maintain equilibrium. If both forces act in the same direction, the stick cannot remain at rest, as the moments about the pivot would not balance. This analysis is crucial for understanding static equilibrium in physics.

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Understanding of static equilibrium principles
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Basic grasp of Newton's laws of motion

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Mar 11, 2021

hidemi

Homework Statement: A massless meter stick on a horizontal frictionless table top is pivoted at the 80-cm mark. A force F1 is applied perpendicularly to the end of the stick at 0 cm. A second force F2 (not shown) is applied perpendicularly at the 60-cm mark. The forces are in the plane of the table top. If the stick does not move, the force exerted by the pivot on the stick:

A) must be zero
B) must be in the same direction as F1 and have magnitude |F2| - |F1|
C) must be directed opposite to F1 and have magnitude| F2| -|F1|
D) must be in the same direction as F1 and have magnitude |F2| + |F1|
E) must be directed opposite to F1 and have magnitude |F2| + |F1|

The answer is B.

Relevant Equations: τ ＝rF

If exerted a force at 60 cm, should the force F2 be the opposite direction of F1? Thanks!

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Mar 11, 2021

Doc Al

Mentor

hidemi said:

If exerted a force at 60 cm, should the force F2 be the opposite direction of F1?

If both F1 and F2 were in the same direction, could the stick remain at rest? (Consider moments about the pivot.)

Mar 11, 2021

hidemi

Thanks for your hint. I got it.

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