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Static Equilibrium - board problem

  • Thread starter Seraph404
  • Start date
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1. Homework Statement

A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.0 g stacked over the 12.0 cm mark, the stick is found to balance at the 45.5 cm mark. What is the mass of the meter stick?


2. Homework Equations

netF = 0
netTorque = 0
The answer is 74 g

3. The Attempt at a Solution

I know there is an upward force exerted by the knife, a downward force exerted by the weight of the coins, and another downward force exerted by gravity acting on the meter stick itself, which I originally thought was at the .5 mark.

Does there need to be a second upward force at the .445 m mark, or is that where I put my pivot point?

This shouldn't be a hard problem, but I don't think I really understand how the force diagram should look.
 
Last edited:

Answers and Replies

tiny-tim
Science Advisor
Homework Helper
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A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.0 g stacked over the 12.0 cm mark, the stick is found to balance at the 45.5 cm mark. What is the mass of the meter stick?



I know there is an upward force exerted by the knife, a downward force exerted by the weight of the coins, and another downward force exerted by gravity acting on the meter stick itself, which I originally thought was at the .5 mark.

Does there need to be a second upward force at the .445 m mark, or is that where I put my pivot point
Hi Seraph! :smile:

I think the question is slightly misleading.

It intends you to assume that there is only one knife-edge, and it is moved to the .445 m mark.

So the only force you need worry about at the .5 m mark is the weight of the stick. :smile:
 

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