In a pulley problem, if you have a chain of boxes all connected by ropes, first find the tension in the rope that goes over the pulley (treat all of the boxes on the table as a single box and all of the boxes hanging down as a single box). Then work your way to each end of the chain one box at a time to find the tension in each rope along the chain.

Example: Find the tensions in each cable and the acceleration of the blocks in the diagram below. Assume that the pulley is massless and the table and pulley are frictionless.

Solution: First, treat the two blocks on the table as a single unit of 5 kg, and treat the two hanging blocks as a single unit of 5 kg:

The net force equation for the “block” on the table is:

T_{2} = ma = 5a (1)

The net force equation for the hanging “block” is:

mg – T_{2} = ma → 5g – T_{2} = 5a (2)

Plugging equation (1) into equation (2) gives

a = g/2 = 4.9 m/s^{2} and T_{2} = 5g/2 = 24.5 N.

Now you can find the other tensions. Find T_{3} by doing an FBD on the 3 kg block:

3g – T_{3} = 3a = 3g/2

Solve to get T_{3} = 3g/2 = 14.7 N. Find T_{1} by doing an FBD on the 4 kg block:

T_{1} = 4a = 2g, so T_{1} = 19.6 N.