Elevator Problem

A 60-kg man takes the elevator to a higher floor. Calculate the “weight” of the man when the elevator is

  1. speeding up from rest to 2.0 m s-1 in 4.0 s,
  2. slowing down from 2.0 m s-1 to rest in 4.0 s.

 

Solution

The two forces acting on the man are (1) the downward gravitational pull mg and (2) the upward normal contact force N.

atRest

While the gravitational pull on the man (true weight) is always equal to mg = (60)(9.81) = 589 N, the normal contact force N between the man and the elevator floor can vary. In fact, N is equal to mg only when the man is in equilibrium, when the elevator is at rest, or travelling at a constant speed.

 

a)

Speeding up from rest to 2.0 m s-1 in 4.0 s corresponds to an acceleration of 2.0/4.0 = 0.50 m s-2. To have an upward acceleration, he must be experiencing a net upward force. This means that N must be stronger than mg.

accelerate

Forming the N2L equation for the man,

\begin{aligned}({{F}_{net}}&=ma)\\N-mg&=ma\\N=m(g+a)\\&=(60)(9.81+0.50)\\&=619\text{ N}\end{aligned}

N (apparent weight) is larger than mg (true weight). The man feels a heavier “weight” of 619 ÷ 9.81 = 63 kg.

 

b)

On the other hand, slowing down from 2.0 m s-1 to rest in 4.0 s corresponds to an acceleration of (0.0 – 2.0)/4.0 = -0.50 m s-2. To have a downward acceleration, he must be experiencing a net downward force. This means that mg must be stronger than N.

decelerate

\begin{aligned}({{F}_{net}}&=ma)\\mg-N&=ma\\N&=m(g-a)\\&=(60)(9.81-0.50)\\&=\text{ }559\text{ N}\end{aligned}

N (apparent weight) is smaller than mg (true weight). The man feels a lighter “weight” of 559 ÷ 9.81 = 57 kg.

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