A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.75 of her regular weight. Calculate the acceleration of the elevator, and find the direction of acceleration.

In this case, only two forces will be acting on the woman; weight (W) and the normal force pushing up on her (N). Consider the upward direction to be positive and apply Newton’s second law to calculate the acceleration.

Given data:

The regular weight of the woman is $W$.

The weight of the woman when the elevator begins to move is $N=0.75W$.

The free body diagram of the woman is as follows:

The relation to calculate the vertical forces is given by:

$\begin{array}{c}\Sigma F_v=ma\\ N-W=ma\end{array}$

Here, m is the mass, a is the acceleration, and N is the normal force.

On plugging the values in the above relation, you get:

$\begin{array}{c}0.75W-W=ma\\ 0.75mg-mg=ma\\ a=-0.25\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\\ a=-2.45\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, $a=-2.45\mathrm{m}/{\mathrm{s}}^2$is the acceleration of the elevator. The negative sign indicates that the direction of acceleration is downward.