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 Determine the power that must be supplied to the motor at the instant the load has been hoisted s = 35 ft starting from rest.

Answer 1

Answer:

$\epsilon=\frac{p_{out}}{P_{in}} \\p_{in}=\frac{p_{out}}{\epsilon} \\p_{in}=\frac{32965.5}{0.65}\\ p_{in}=50716.1538 lb.ft/s$

In hp:

$p_{in}=\frac{50716.1538}{500}\\ p_{in}=101.432 hp$

Part B:

$\epsilon=\frac{p_{out}}{P_{in}} \\p_{in}=\frac{p_{out}}{\epsilon} \\p_{in}=\frac{50356.2}{0.65}\\ p_{in}=77471.07692 lb.ft/s$

In hp:

$p_{in}=\frac{77471.07692}{500}\\ p_{in}=154.94215 hp$

Explanation:

Weight of elevator=1000-lb

Force=500 lb

s=15 ft

Force on pulley=F=3*500=1500 lb

g=$32.2 ft/s^2$

According to Newton's Second law:

$\sum F_y=ma_y$

According to attachment:

$F-W=ma_y$

$1500-1000=\frac{1000}{32.2}a_y$

$a_y=16.1 ft/s^2$

According to third equation of motion:

$v^2=v_o^2+2a_y(S-So)$

Where:

Vo is initial velocity

V is final velocity

S is final distance

So is starting distance

$v^2=(0)^2+2*16.1*(15)\\v^2=483\\v=21.977 ft/s$

Output power:

$P_{out}=F.v\\P_{out}=1500*21.977\\P_{out}=32965.5 lb.ft/s$

$\epsilon=\frac{p_{out}}{P_{in}} \\p_{in}=\frac{p_{out}}{\epsilon} \\p_{in}=\frac{32965.5}{0.65}\\ p_{in}=50716.1538 lb.ft/s$

In hp:

$p_{in}=\frac{50716.1538}{500}\\ p_{in}=101.432 hp$

Part B:

When S=35 ft

$v^2=(0)^2+2*16.1*(35)\\v^2=1127\\v=33.5708 ft/s$

Output power:

$P_{out}=F.v\\ P_{out}=1500*33.5708 \\ P_{out}=50356.2 lb.ft/s$

$\epsilon=\frac{p_{out}}{P_{in}} \\p_{in}=\frac{p_{out}}{\epsilon} \\p_{in}=\frac{50356.2}{0.65}\\ p_{in}=77471.07692 lb.ft/s$

In hp:

$p_{in}=\frac{77471.07692}{500}\\ p_{in}=154.94215 hp$