(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I attached a picture of the given figure to this thread. The suspended 2.5kg mass is moving up, the 2.9kg mass on the ramp is sliding down, and the 9kg mass is moving down. It is given that there is friction between the block on the ramp and the ramp itself with a coefficient of friction = 0.11. The pulleys are massless and frictionless.

What is the tension in the cord connected to the 9kg block?

2. Relevant equations

F = ma

3. The attempt at a solution

So, I drew three free-body diagrams for the three separate masses. I'll call the furthermost left mass m1, the middle block on the ramp m2, and the rightmost mass m3.

The forces acting on m1 are the force of gravity and the tension in the far left cord (I'll call this tension T1, and m1's overall acceleration a1). Therefore, since m1 is falling:

Sum(forces) = m1g - T1 = m1a1

For m2, in the x direction the forces are the force of friction in opposite direction to the motion, the tension in cord T1, the tension in the other cord T2, and the x-component of gravity, m1gsin(33). In the y direction, the forces are the normal force and the y-component of gravity. Therefore:

Sum(x forces) = T2 + Ffriction - T1 - m2gsin(33) = m2(a2 - a1)

Sum(y forces) = n - mgcos(33) = 0

The free-body diagram for m3 is the same as m1, but the object is moving in opposite direction so:

Sum(forces) = T2 - m3g = m3a2

I've reached this point and I believe my equations are right (except possibly the sum of x forces for m2), but I can't seem to figure how to combine these equations to determine T1.

Any help is greatly appreciated. Thanks!

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# Calculating Tension in Cord in 2-Pulley System

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