Question

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.

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.

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.

Part D

Find T₁, the tension in the upper rope.

Answer 1

Part A Answer

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 Answer

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 Answer

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

F = Ma
F_T – F_g = Ma

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

F_T = Ma + F_g
F_T = Ma + Mg
T = Ma + Mg

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

T₂ = M₂a + M₂g

Part D Answer

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