Question

The 1000-lb elevator is hoisted by the pulley system and motor M. If the motor exerts a constant force of 500 lb on the cable, determine the power that must be supplied to the motor at the instant the load has been hoisted s = 15 ft starting from rest. The motor has an efficiency of e = 0.65.

Answer 1

Answer:

the power that must be supplied to the motor is  101.45 hp

Explanation:

Given that,

The weight of elevator = 1000lb

The motor exerts a constant force of 500 lb on the cable

The load has been hoisted s = 15 ft starting from rest

The motor has an efficiency of e = 0.65

According to the equation of motion:

F = ma

$3(500) - 1000 = \frac{1000}{32.2} * aa = 16.1 ft/s^2$

$v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (15-0)\\\\v = 21.98m/s$

To calculate the output power:

P(out) = F. v

P(out) = 3 (500) * 21.98

P(out) = 32970 lb.ft/s

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

ε = P(out) / P(in)

0.65 = 32970 / P(in )

P(in) = 50,723 lb.ft/s

P(in) =  50,723 / 500 hp

    = 101.45 hp

the power that must be supplied to the motor is  101.45 hp