Force and Newton's Laws

[Soaring Crane]

A person stands on a bathroom (?) scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.75 of the person's regular weight. Calculate the elevator's acceleration, and find the direction of acceleration.

The only thing that I really got is that the scale reading is (3/4)*mg (person's weight). Is there anyone who can tell me how to start?

Thanks. :smile:

[PICsmith]

It's intuitive that the elevator is going down...If it were going up he'd feel a force pulling him down, and the scale would read higher than normal weight. You know that g is the acceleration due to gravity (=9.8m/s^2). If the elevator were to accelerate at a rate that was half of g, then the scale would read half his weight. If the elevator were to accelerate at the same rate as g, the scale would go to 0....see the relationship? Hope this helps.

[Doc Al]

Another way to think of it is to realize that the scale doesn't read the person's weight (which is always w=mg down), instead it reads the normal force that the scale pushes up on the person with. When the scale reads a force equal to the person's weight, that means that the upward force exactly equals the downward force: so the net force is zero, no acceleration. So... if the scale reads a force equal to 0.75 of the person's weight, there is now a net force acting down. Find that net force and figure out the acceleration. (Draw a picture showing all forces acting on the person.)