Calculating Force Exerted by a Pivot on a Meter Rule

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Homework Help Overview

The problem involves calculating the force exerted by a pivot on a uniform meter rule, which has a weight of 1.5N. The original poster expresses confusion regarding the calculations and the provided answer in the textbook.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use various equations related to forces and moments but struggles to reconcile the provided answer. Some participants suggest simply adding up the forces without considering moments, while others question the interpretation of the forces involved.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on how to approach the calculation. There is a recognition of the need to consider the direction of forces, and the original poster acknowledges this clarification.

Contextual Notes

The original poster mentions previous questions that they believe are not relevant to this problem. There is also a reference to a spring balance that may affect the forces being considered.

Taylor_1989
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Homework Statement


The weight of a uniform meter rule is 1.5N. Calculate the force exerted by the pivot on the meter rule.


The answer in the back of the book say it is magnitude of 0.5N downwards, I have tried every combo of equation from m=f*d to W=m*g

There are 3 question before, which I don't think are relevant to this question. Which I answers and got right. I have drawn a diagram to show what the meter rule looks like.
 

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Just add up the forces. (No need for moments to answer this question.)
 
Doc Al said:
Just add up the forces. (No need for moments to answer this question.)

I don't understand, by add up the force. If I add 6N plus 1.5N, it dose not equal 0.5N. Could you please expand on your answer?
 
Taylor_1989 said:
I don't understand, by add up the force. If I add 6N plus 1.5N, it dose not equal 0.5N. Could you please expand on your answer?
Don't forget the force from the spring balance, presumably acting upward.

So, including the force of the pivot itself, I count four forces acting.
 
Doc Al said:
Don't forget the force from the spring balance, presumably acting upward.

So, including the force of the pivot itself, I count four forces acting.

I see where you are coming from, I was not taking direction of the force being applied.

Thanks for clearing that up, it has been bugging me all day.
 

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