Question

If the applied force is 60 N and the lower mass is 1.0 kg, what is the tension in the string?(frictionless surface)

Answer 1

The net force on the larger mass in the horizontal direction (where we take the direction of F to be positive) is

F - T = 2Ma

where T is the magnitude of tension and a is the mass's acceleration.

The net force on the smaller mass in the vertical direction (upward is positive) is

T - Mg = Ma

If M = 1.0 kg and F = 60 N, then we end up with two equations with two unknowns,

60 N - T = (2.0 kg) a

T - (1.0 kg) (9.8 m/s²) = (1.0 kg) a

Eliminate a and solve for T :

(60 N - T) - 2 (T - (1.0 kg) (9.8 m/s²)) = (2.0 kg) a - 2 (1.0 kg) a

60 N - T - 2T + 19.6 N = (2.0 kg) a - (2.0 kg) a

40.4 N - 3T = 0

3T = 40.4 N

T = (40.4 N) / 3

T ≈ 13 N