Question

The 1000-lb elevator is hoisted by the pulley system and motor M. The motor exerts a constant force of 500 lb on the cable. The motor has an efficiency of ε = 0.65. Determine the power that must be supplied to the motor at the instant the load has been hoisted s = 27 ft starting from rest.

Answer 1

The power that must be supplied to the motor is 136 hp

Explanation:

Given-

weight of the elevator, m = 1000 lb

Force on the table, F = 500 lb

Distance, s = 27 ft

Efficiency, ε = 0.65

Power  = ?

According to the equation of motion:

F = ma

$3(500) - 1000 = \frac{1000}{32.2} * a$

a = 16.1 ft/s²

We know,

$v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s$

To calculate the output power:

Pout = F. v

Pout = 3 (500) * 29.48

Pout = 44220 lb.ft/s

As efficiency is given and output power is known, we can calculate the input power.

ε = Pout / Pin

0.65 = 44220 / Pin

Pin = 68030.8 lb.ft/s

Pin = 68030.8 / 500 hp

     = 136 hp

Therefore, the power that must be supplied to the motor is 136 hp