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

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The discussion centers on the forces exerted on a stick pivoted at one end. It questions whether the force F2 should act in the opposite direction to F1 when a force is applied at 60 cm from the pivot. The participants explore the implications of both forces acting in the same direction and whether the stick could remain at rest under those conditions. The concept of moments about the pivot is highlighted as crucial to understanding the balance of forces. The inquiry concludes with a resolution on the relationship between the forces and their directions.
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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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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.)
 
Thanks for your hint. I got it.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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