Two Hanging Masses

Part A = M_{2}g

Part B = (M_{1} + M_{2})g

Part C = M_{2}a + M_{2})g

Part D = (M_{1} + M_{2})a + (M_{1} + M_{2})g

**Solutions Below:**

_{1}and M

_{2}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

_{2}, the tension in the lower rope.

Express your answer in terms of some or all of the variables M

_{1}, M

_{2}, 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 T_{2}, we know the mass of block 1 isn’t included (since it’s above that point). So:

T_{2} = M_{2}g

Part B

_{1}, the tension in the upper rope. Express your answer in terms of some or all of the variables M

_{1}, M

_{2}, and g.

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

T_{1} = (M_{1} + M_{2})g

Part C

_{2}, the tension in the lower rope. Express your answer in terms of some or all of the variables M

_{1}, M

_{2}, a, and g.

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 T_{2} (below the point of block #1:

T_{2} = M_{2}a + M_{2}g

Part D

_{1}, 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:

T_{2} M_{2}a + M_{2}g

T_{1} = (M_{1} + M_{2})a + (M_{1} + M_{2})g