How to calculate counterweight at a different height?

AI Thread Summary
To calculate counterweights at different heights, the equilibrium condition m1y = m2x applies, where y and x represent the distances from the pivot to the line of action of the forces. The height differences h1 and h2 do not affect the equilibrium equation, as they are not included in the torque calculations. It is essential to clarify the full problem statement and ensure that the system is in equilibrium. The forces acting on the masses must be analyzed correctly to confirm that the counterweights balance each other. Ultimately, height differences do not influence the counterweight calculations in this scenario.
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Homework Statement


m2=3kg, m1=?
the diagram has been attached

Homework Equations


I know that we find counterweights by equating equation mass x distance frm pivot but how do we find if it is at different height

The Attempt at a Solution


Is it m1yh1=m2xh2
is counterweight at all dependent on height?[/B]
 

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Does the full problem statement mention that the two beams to which the masses are attached are horizontal ?

If yes, then your attempt looks good.
Mass x g is force
force x distance from pivot is torque
No net torque means no angular acceleration => balance. g divides out.

Advice: draw the forces in the diagram.
 
That is basically m1y=m2x (in this case)
But what about the effects of h1 & h2 ?
The two beams are at different heights so will the approach change?
 
Pity you don't show any equations. Your "Is it m1yh1=m2xh2" doesn't make sense to me. What does it represent ? A wild guess ? Is it a multiplication ? If so, I have no idea what y and x stand for. And the dimensions I don't recognize.

Time to read the guidelines and make better use of the template . . .

And completing the problem statement might be a good idea too: m1 can be anything if there are no restrictions. A restriction could be that the stuff is in equilibrium, but if you don't tell, nobody knows !

[edit] I am a bit too strict. Your "That is basically m1y=m2x (in this case)" is correct for equilibrium. It doesn't feature h1 or h2.
 
Last edited:
BvU said:
force x distance from pivot is torque
To be precise (and this seems to be the crucial point here), the distance in that formula is the distance from the pivot to the line of action of the force, not to the point of application of the force.
 
BvU said:
Pity you don't show any equations. Your "Is it m1yh1=m2xh2" doesn't make sense to me. What does it represent ? A wild guess ? Is it a multiplication ? If so, I have no idea what y and x stand for. And the dimensions I don't recognize.

Time to read the guidelines and make better use of the template . . .

And completing the problem statement might be a good idea too: m1 can be anything if there are no restrictions. A restriction could be that the stuff is in equilibrium, but if you don't tell, nobody knows !

[edit] I am a bit too strict. Your "That is basically m1y=m2x (in this case)" is correct for equilibrium. It doesn't feature h1 or h2.
Sorry for the inconvenience.
m1 & m2 are in equilibrium , y= 300mm & x=900 mm and hence m1y=m2x.
So, if m1 & m2 are in equilibrium, will different heights have any effect? h1=900mm & h2=913mm.
 
"That is basically m1y=m2x (in this case)" is correct for equilibrium.

h1 and h2 are not present in the expression, so they have no effect !
 
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