Ratio of Lengths b/a for Torques Equilibrium

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SUMMARY

The discussion centers on calculating the ratio of lengths b/a for a system in torque equilibrium involving a 3.0-kg ball and a 1.0-kg ball placed on a massless beam. The equilibrium condition is established by setting the sum of torques (Σtorques) to zero, using the formula Torque = Force x lever arm. The participants emphasize the importance of considering the direction of torques, with anticlockwise torques taken as positive and clockwise torques as negative, to ensure the total torque equals zero.

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Homework Statement


A 3.0-kg ball and a 1.0-kg ball are placed at opposite ends of a massless beam so that the system is in equilibrium as shown. Note: The drawing is not drawn to scale. What is the ratio of the lengths, b/a?

(a) 2.0 (c) 3.0 (e) 5.0

(b) 2.5 (d) 4.0

Homework Equations

The Attempt at a Solution


so I did the Σtorques = 0 since the whole system is in equilibrium

Torque = Fl

1st Torque is
(mga)(level arm of a)

2nd Torque is

mgb(level arm of b)

3rd Torque is
I don't know
 
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No figure was attached.

Why do you think that three torques are involved in this problem?
 
When calculating torques you need to keep track of which sense applies to each. Usual is to pick one sense, anticlockwise say, as positive, then any clockwise torque will be negative. (There must be some of each for the total to be zero!)
 

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