Solving a Quick Force Problem: Finding Acceleration in an Elevator

[joe215]

Homework Statement

A person has a normal weight of 500N. The same person is standing on a scale in an elevator and the scale reads 700. what is the acceleration of the elevator?

The Attempt at a Solution

How do I apply the formula Fnet=ma to this problem?

 

[Mattowander]

I'd suggest drawing a free body diagram to start with. Make some effort to do the problem and I'll help you out if you need it.

 

[joe215]

There is an upwards normal force on the person and a downwards force of mg on the person.

Also, I think the acceleration of the elevator=acceleration of the person. Also, upwards is the positive direction.

Since the scale reads 700N, then the net forces on the person equal 700N.

Fnet=ma
700N=(500/9.8)a

a=700/(700/9.8)

Is this right?

Thanks!

 

[Galileo's Ghost]

There is an upwards normal force on the person and a downwards force of mg on the person.

Also, I think the acceleration of the elevator=acceleration of the person. Also, upwards is the positive direction.

Since the scale reads 700N, then the net forces on the person equal 700N.

Fnet=ma
700N=(500/9.8)a

a=700/(700/9.8)

Is this right?

Thanks!

The scale will read the force pushing down on it. From Newton's 3rd Law, we know that if the scale is pushed down with a force of 700 N then the person will experience a force from the scale equal to ___________? And the direction of this force will be _________? Now, what is the other force acting on the person? To determine the net force, make sure you total up all the forces acting on the person!