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

`\epsilon=(p_(out))/(P_(in)) \\p_(in)=(p_(out))/(\epsilon) \\p_(in)=(32965.5)/(0.65)\\ p_(in)=50716.1538 lb.ft/s\\`

In hp:

`p_(in)=(50716.1538)/(500)\\ p_(in)=101.432 hp`

Part B:

`\epsilon=(p_(out))/(P_(in)) \\p_(in)=(p_(out))/(\epsilon) \\p_(in)=(50356.2)/(0.65)\\ p_(in)=77471.07692 lb.ft/s\\`

In hp:

`p_(in)=(77471.07692)/(500)\\ p_(in)=154.94215 hp`

Step-by-step 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=(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=(p_(out))/(P_(in)) \\p_(in)=(p_(out))/(\epsilon) \\p_(in)=(32965.5)/(0.65)\\ p_(in)=50716.1538 lb.ft/s\\`

In hp:

`p_(in)=(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=(p_(out))/(P_(in)) \\p_(in)=(p_(out))/(\epsilon) \\p_(in)=(50356.2)/(0.65)\\ p_(in)=77471.07692 lb.ft/s\\`

In hp:

`p_(in)=(77471.07692)/(500)\\ p_(in)=154.94215 hp`