# Elevator Problem

• Jul 26th 2007, 03:58 PM
topsquark
Elevator Problem
Quote:

Question: A 75.0 kg person is standing on a scale in an elevator. What is the reading of the scale in N (Newtons) if the elevator is (a) at rest, (b) moving up at a constant of 2.00 m/s, and (c) accelerating up at 2.00 m/s²?
We need a Free-Body Diagram (on the person) for this.

The FBD has two forces in it: a weight (w) of the person acting straight downward, and a normal force (N) acting straight upward. I am taking +y to be upward.

The scale is reading the value of the Normal force, in this case usually referred to as the "apparent weight."

a) and b)
The elevator is not accelerating, so Newton's 2nd Law says that
$\displaystyle \sum F_y = N - w = ma = 0$

So $\displaystyle N = w = mg$

c)
The elevator is accelerating upward, so the person is also (since the person is at rest with respect to the elevator.)
$\displaystyle \sum F_y = N - w = ma$

So $\displaystyle N = w + ma = mg + ma$

-Dan
• Jul 26th 2007, 08:29 PM
CaptainBlack
Quote:

Originally Posted by topsquark

We need a Free-Body Diagram (on the person) for this.

The FBD has two forces in it: a weight (w) of the person acting straight downward, and a normal force (N) acting straight upward. I am taking +y to be upward.

The scale is reading the value of the Normal force, in this case usually referred to as the "apparent weight."

a) and b)
The elevator is not accelerating, so Newton's 2nd Law says that
$\displaystyle \sum F_y = N - w = ma = 0$

So $\displaystyle N = w = mg$

c)
The elevator is accelerating upward, so the person is also (since the person is at rest with respect to the elevator.)
$\displaystyle \sum F_y = N - w = ma$

So $\displaystyle N = w + ma = mg + ma$

-Dan

Principle of equivalence gives this straight off without a second thought.

RonL