What Does the Bathroom Scale Read in an Elevator?

If the scale is accurate, its reading will equal $F_{\text{p}}$, the magnitude of the force the person exerts downward on it. Figure 4.24(a) shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and a free-body diagram is drawn as in Figure 4.24(b). Analysis of the free-body diagram using Newton’s laws can produce answers to both parts (a) and (b) of this example, as well as some other questions that might arise. The only forces acting on the person are his weight $\mathbf{\text{w}}$ and the upward force of the scale ${\mathbf{\text{F}}}_{\text{s}}$. According to Newton’s third law ${\mathbf{\text{F}}}_{\text{p}}$ and ${\mathbf{\text{F}}}_{\text{s}}$ are equal in magnitude and opposite in direction, so that we need to find $F_{\text{s}}$ in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,

From the free-body diagram we see that $F_{\text{net}}=F_{\text{s}}-w$, so that

Solving for $F_{\text{s}}$ gives an equation with only one unknown:

or, because $w=\mathrm{mg}$, simply

No assumptions were made about the acceleration, and so this solution should be valid for a variety of accelerations in addition to the ones in this exercise.

In this part of the problem, $a=1.20{\text{m/s}}^2$, so that

yielding

This is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:

So, the scale reading in the elevator is greater than his 735-N (165 lb) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators.

Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because $a=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$, and $\mathrm{\Delta }v=0$.

Thus,

Now

which gives

The scale reading is 735 N, which equals the person’s weight. This will be the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.