Calculation of Counterbalance Mass

AI Thread Summary
The discussion centers on calculating the mass m2 in a mechanical system with two links and a variable angle Theta. The original calculation suggests m2 is independent of Theta, yielding m2 = (m1)*(d1/d2)*(1/2). However, it is clarified that both instances of Theta must be the same, affecting the balance of forces. Adjustments in angle lead to changes in m2 and m1, indicating that the relationship is more complex than initially thought. The final consensus confirms the need to consider the angle's impact on the mass calculations.
bsheikho
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Hi,

I've been grappling with a calculation that appears to give me a good result, however my gut feeling says other wise. Could someone please check the layout and possibly confirm my final answer. Attached is the layout of the system.

The system: -There is a long arm, d1+d2 is one link. then end of d2 to m2 is the second link.
-m2 is constrained in all directions except vertical up/down movement. (by putting the mass on a linear shaft with linear bearings)
The goal is to calculate m2, everything else is given, and Theta is a variable.

my final answer i reach to is: m2= (m1)*(d1/d2)*(1/2)

The point is that it seems that the mass of object two is independent of the angle. and can be fixed, however I have a feeling that tells me that its not that simple.

I can also upload how I got the final answer, but I'd prefer to see how others would tackle the situation.

Any help is appreciated.
Many Thanks!.
 

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I agree with your answer.
The mechanics require both instances of ##\theta## to be the same, increasing the angle lowers m2 by ##2d_2 \cos(\theta) d\theta## and raises m1 by ##d_1 \cos(\theta) d\theta##. That leads to balance with the condition you posted.
 
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mfb said:
I agree with your answer.
The mechanics require both instances of ##\theta## to be the same, increasing the angle lowers m2 by ##2d_2 \cos(\theta) d\theta## and raises m1 by ##d_1 \cos(\theta) d\theta##. That leads to balance with the condition you posted.

Excellent! Thank you.
 
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