A person stands on a bathroom scale in a motionless elevator. when the elevator begins to move

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 .

The weight of the woman when the elevator begins to move is .

The free body diagram of the woman is as follows:

The relation to calculate the vertical forces is given by:

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

A person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.76 of the person's regular weight. Calculate the magnitude of acceleration of the elevator.

Homework Equations

F=ma N=ma-mg

The Attempt at a Solution

Ok, so the scale in the elevator shows weight (mg), the elevator accelerates downwards, and the scale shows .76 of the persons weight. [tex]\Sigma[/tex]F=ma mg-N=ma a=(mg-N)/m a=(m(9.8)-.76)/m So now I have 2 unknowns in one equation. I need to find A, but I do not know the mass of the person. I figured that if the weight shown by the person is .76 of his normal, I should be able to figure this out if i just assign a random value to "m"... correct? so plugging a number in for m i get... a=((10)(9.8)-.76(10))/(10) a=9.04 This does not possibly seem correct. if it were true, the elevator would nearly be in freefall, and the scale would be reading something around .2 of the persons weight, right?

What am i doing wrong?