Break the system into four free body diagrams as shown. Carefully label the forces to ensure that the tension in any single cord is uniform. The equations of equilibrium for the four objects, starting with the leftmost pulley and moving clockwise, are:
S - 3T = 0, R -3S = 0, F -3R = 0,
and 2T + 2S + 2R - mAg = 0.
We want to eliminate S, R, and F from our result and find T in terms of mA and g. From the first two equations, we get S = 3T, and R = 3S = 9T. Substituting these into the last equilibrium equation results in 2T + 2(3T) + 2(9T) = mAg.
Solving, we get T = mAg/26 .
Note: We did not have to solve for F to find the appropriate value of T. The final equation would give us the value of F in terms of mA and g. We would get F = 27mAg/26. If we then drew a free body diagram of the entire assembly, the equation of equilibrium would be F - T - mAg = 0. Substituting in the known values for T and F, we see that this equation is also satisfied. Checking the equilibrium solution by using the “extra” free body diagram is often a good procedure.