Ratio of Lengths b/a for Torques Equilibrium

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The discussion revolves around 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 on a massless beam. Participants emphasize the importance of setting the sum of torques to zero, as the system is in equilibrium. The torques are calculated using the formula Torque = Force x length, with specific attention to the direction of each torque. There is a suggestion to clarify the number of torques involved and the convention for positive and negative values based on their direction. The conversation highlights the need for a visual representation to aid in understanding the problem better.
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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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