Question

The figure shows an arrangement in which four disks are suspended by cords. The longer, top cord loops over a frictionless pulley and pulls with a force of magnitude 84 N on the wall to which it is attached. The tensions in the shorter cords are T1 = 69.3 N, T2 = 47 N, and T3 = 6.72 N. What are the masses of (a) disk A, (b) disk B, (c) disk C, and (d) disk D?

Answer 1

Ma g = 84 - 69.3 = 14.7 N

Ma = 14.7 / 9.80 = 1.5 kg

Mb g = 69.3 - 47 = 22.3 N

Mb = 22.3 / 9.80 = 2.28 kg

Mc g = 47 - 6.72 = 40.3 N

Mc = 40.3 / 9.80 = 4.11kg

Md = 6.72 / 9.80 = .69 kg

Check: (1.5 + 2.28 + 4.11 + .69) * 9.80 = 84.1 N

Total mass suspended * 9.80 m/s^2 = force of wall