Question

mass is 5 lb attached to a rope wound around a pulley. The radius of the pulley is 3 in. If the mass falls at a constant velocity of 5 ft/s, determine the power transmitted to the pulley, in hp.

Answer 1

Answer:

The power transmitted to the pulley is 0.0455 hp.

Explanation:

Given;

mass attached to the rope, m = 5 lb

radius of the pulley, r = 3 in

constant rate of fall of the mass, v = 5 ft/s

acceleration due to gravity, g = 32.2 ft/s²

1 lbf = 32.2 lb.ft/s²

The power transmitted to the pulley is calculated as;

P = Fv

P = (mg)v

$P = 5 \ lb \ \times \ 32.2 \ \frac{ft}{s^2} \ \times \ 5 \ \frac{ft}{s} \ \times \ \frac{1 \ lbf}{32.2 \ lb.ft/s^2} \ \ = 25 \ \frac{ft.lbf}{s}$

in horse power, the power transmitted is calculated as;

$P = \frac{25 \ ft.lbf}{s} \ \times \ \frac{1 \ hp}{550 \ ft.lbf/s} \ \ = 0.0455 \ hp$

Therefore, the power transmitted to the pulley is 0.0455 hp.