## Mastering Physics Solutions: Two Hanging Masses

Two Hanging Masses

Part A = M2g
Part B = (M1 + M2)g
Part C = M2a + M2)g
Part D = (M1 + M2)a + (M1 + M2)g

Solutions Below:

Two blocks with masses M1 and M2 hang one under the other. For this problem, take the positive direction to be upward, and use g for the magnitude of the acceleration due to gravity.

Part A

Assume the blocks are at rest. Find T2, the tension in the lower rope.
Express your answer in terms of some or all of the variables M1, M2, and g.

Because the blocks and rope are stationary, we know that the only force applied to the blocks is by gravity. Tension is just the force applied to pull whatever object is attached to the string, therefore the tension in the rope is Mg, and since the question asks for the tension at T2, we know the mass of block 1 isn’t included (since it’s above that point). So:

T2 = M2g

Part B

Assume the blocks are still at rest. Find T1, the tension in the upper rope. Express your answer in terms of some or all of the variables M1, M2, and g.

The solution is the same as for Part A, except the mass of both blocks is included now. So:

T1 = (M1 + M2)g

Part C

The blocks are now accelerating upward (due to the tension in the strings) with acceleration of magnitude a. Find T2, the tension in the lower rope. Express your answer in terms of some or all of the variables M1, M2, a, and g.

Remember that F = ma. Since the upward acceleration is opposite to gravity, we can solve for the tension (FT) like this:

F = Ma
FT – Fg = Ma

In other words, the upwards force, less the effect of gravity, gives the net acceleration upwards. Simplifying for FT gives:

FT = Ma + Fg
FT = Ma + Mg
T = Ma + Mg

Since the question asks for T2 (below the point of block #1:

T2 = M2a + M2g

Part D

Find T1, the tension in the upper rope.

All we have to do here is start with the solution from Part C, but include the masses of both blocks:

T2 M2a + M2g

T1 = (M1 + M2)a + (M1 + M2)g

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