Question

Consider a vertical elevator whose cabin has a total mass of 800 kg when fully loaded and 150 kg when empty. The weight of the elevator cabin is partially balanced by a 400-kg counterweight that is connected to the top of the cabin by cables that pass through a pulley located on top of the elevator well. Neglecting the weight of the cables and assuming the guide rails and the pulleys to be frictionless, determine (a) the power required while the fully loaded cabin is rising at a constant speed of 1.2 m/s and (b) the power required while the empty cabin is descending at a constant speed of 1.2 m/s. What would your answer be to (a) if no counterweight were used? What would your answer be to (b) if a friction force of 800 N has developed between the cabin and the guide rails?

Answer 1

Answer:

Part a)

$P = 4.71 \times 10^3 Watt$

Part b)

$P = 2.94 \times 10^3 W$

Part c)

$P = 9.4 \times 10^3 W$

Part d)

$P = 3.9 \times 10^3 W$

Explanation:

Part a)

When cabin is fully loaded and it is carried upwards at constant speed

then we will have

net tension force in the rope = mg

$T = (800)(9.81)$

$T = 7848 N$

now it is partially counterbalanced by 400 kg weight

so net extra force required

$F = 7848 - (400 \times 9.81)$

$F = 3924 N$

now power required is given as

$P = Fv$

$P = 3924 (1.2)$

$P = 4.71 \times 10^3 Watt$

Part b)

When empty cabin is descending down with constant speed

so in that case the force balance is given as

$F + (150 \times 9.8) = (400 \times 9.8)$

$F = 2450 N$

now power required is

$P = F.v$

$P = (2450)(1.2)$

$P = 2.94 \times 10^3 W$

Part c)

If no counter weight is used here then for part a)

$F = 7848 N$

now power required is

$P = F.v$

$P = 7848 (1.2)$

$P = 9.4 \times 10^3 W$

Part d)

Now in part b) if friction force of 800 N act in opposite direction

then we have

$F + (150 \times 9.8) = 800 +(400 \times 9.8)$

$F = 3250 N$

now power is

$P = (3250)(1.2)$

$P = 3.9 \times 10^3 W$