How to calculate counterweight at a different height?

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Homework Help Overview

The discussion revolves around calculating the counterweight in a system involving two masses, m1 and m2, with a focus on how the height difference between the two beams affects the calculation. The original poster presents a formula and questions its validity in the context of different heights.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between mass, distance from the pivot, and torque, questioning how height differences impact the equilibrium condition. There is a discussion about the correct formulation of the problem and the relevance of the heights h1 and h2.

Discussion Status

Some participants have provided guidance on the principles of torque and equilibrium, while others have raised questions about the clarity and completeness of the problem statement. There is an ongoing exploration of whether the height differences affect the calculations, with some asserting that they do not appear in the equilibrium expression.

Contextual Notes

There is a mention of the need for a complete problem statement and clarity regarding the assumptions of equilibrium. The specific heights of the beams are noted, but their implications remain under discussion.

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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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