If the force T exerted on the cable by the motor M is indica

If the force T exerted on the cable by the motor M is indicated by the graph, determine the speed of the 180-lb crate when t = 5 s. The coefficients of static and kinetic friction are ?s = 0.32 and ?k = 0.24, respectively.

Solution

Friction in Static condition : 0.32 x 180 = 57.6 lb.

Total pull on the crate by the motor-pully arrangement = 3T.

So, in equilibrium 3T = 57.6 i.e. T = 19.2 lb.

This pull by the motor achieves in , 3 x (19.2 - 12)/(30-12) = 1.2 s. [Linear interpolation between 0-3 s]

At this time velocity is 0 m/s. After this the crate will start to move.

Hence kinetic friction will be applicable.

Kinetic friction force = 0.24 x 180 = 43.2 lb.

At t = 5 s, assume, veloicty of the crate is V.

From, Newton\'s second law of motion, P = mf = m x (V - 0)/(5 - 1.2) [ P = (3T - kinetic friction force)]

Or, (3 x 30 - 43.2) x 32.2 = 180 x V/3.8 [ gravitaional accleration= 32.2 ft/s²]

Or, V = 31.8 ft/s.