Weight of a block in equilibrium in a system.

[IAmPat]

Homework Statement

The picture below shows a mobile
in equilibrium. Each of the rods is 0.16m long and each hangs from a supporting string that is attached one fourth of the way across it. The mass of each rod is 0.10kg. The mass of the strings connecting the blocks to the rods is negligible. What is the mass of block A.

Homework Equations

Net Torque = 0
Rod length (left side) = 0.12m
Rod length (right side) = 0.04m

Weight of Rod (left side) = 0.075kg
Weight of Rod (right side) = 0.025kg

Weight of block A = 9.8A

The Attempt at a Solution

I've tried two methods, neither worked. I don't know how to get the torque of the rod itself.

(0.30 + 0.075) * 9.8 = 3.675N
3.675 * 0.12 = 0.441 >> Torque of left side

.441 = [[A + 0.025] * 9.8] * 0.04
11.025 = [A + 0.025] * 9.8
1.125 = A + 0.025
1.1 = A >> Mass of A

But this was wrong. I also tried to do the problem by ignoring the mass of the rod, but that was also wrong.

 

[PhanthomJay]

You have incorrectly assumed that the mass of each part of the lower rod acts at the point where the block masses are attached. The resultant mass of the rod acts at its center of mass.

 

[AlexChandler]

You have incorrectly assumed that the mass of each part of the lower rod acts at the point where the block masses are attached. The resultant mass of the rod acts at its center of mass.

It should work out correctly using tension forces on the rod at the points where the block masses are attached. Certainly you would have to agree that this is where the forces are actually exerted.. It is not incorrect to handle the problem this way.

What I would try is to choose two points, call them O and P. Say P is on the lower rod at the point where mass A is hanging from the lower rod. The sum of all torques on the rod calculated with P as the origin must be zero. So this equation allows you to find the tension in the string that the lower rod is hanging from. Now say point O is on the lower rod at the point where the known mass hangs from. Calculated the Sum of all the torques about this point and it must also be equal to zero. You know all of the distances and you know the tension in the string. This should lead you to the correct answer.

 

[PhanthomJay]

It should work out correctly using tension forces on the rod at the points where the block masses are attached. Certainly you would have to agree that this is where the forces are actually exerted.. It is not incorrect to handle the problem this way.

There are forces acting on the lower rod from from three masses, the 0.30 kg mass of the left block, the unknown mass of block A, and the 0.1 kg mass of the rod itself. The weight force of the hanging masses act on the lower rod at the points where their respective masses are attached. The weight of the rod itself does not act at those points. It is simplest to find out where the resultant weight of the rod acts, apply that force at that point, and them sum torques = 0 about the point where the string supporting the lower rod is attached to it. This is what the OP was attempting to do, but incorrectly determined the moment arm of the rod's weight.