Tensions with 3 blocks on a single pulley

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AI Thread Summary
The discussion focuses on a physics problem involving two blocks of mass m and one block of mass M connected by cords over a frictionless pulley. The system is released from rest, with block M descending under constant acceleration due to the condition 2m < M. Participants confirm the correct free body diagram (FBD) and emphasize the importance of consistent sign conventions in the equations. The derived expressions for the tensions in cords A and B are Tension A = m(a + g) and Tension B = M(g - a). The conversation highlights the need for careful analysis in deriving these tensions during the motion of the blocks.
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Homework Statement



The diagram shows two blocks, each of mass m, connected by cord A, and the top block of mass m connected to the block of mass M by cord B. Both cords are of negligible mass. Cord B passes over a frictionless peg that is fixed in place. The system of blocks and cords is released from rest from the position shown, and because 2m<M, block M descends under constant acceleration the distance h in a time T. Derive, in terms of m, M, h, T, and the acceleration due of gravity g, expressions for the tensions in the cords A and B during that time T
pulley.png
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upload_2015-9-22_21-30-25.png

Homework Equations

The Attempt at a Solution


The picture shows my work. And I was wondering if I drew the FBD correctly and my equations were correct[/B]
 
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(1) In your FBD for mass 2, you left out the tension from the cord below it.
(2) When writing your equations, take care to use a consistent sign convention for the acceleration.
 
So tension A= m(a+g)
And
Tension B= M(g-a)
 
PAstudent said:
So tension A= m(a+g)
And
Tension B= M(g-a)
Those are correct.
 
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