Calculating Force Exerted by a Pivot on a Meter Rule

[Taylor_1989]

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.

 

[Doc Al]

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

 

[Taylor_1989]

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?

 

[Doc Al]

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.

 

[Taylor_1989]

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.