Quick force problem

  • Thread starter joe215
  • Start date
  • #1
26
0

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?
 

Answers and Replies

  • #2
159
0
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.
 
  • #3
26
0
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!
 
  • #4
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!
 

Related Threads on Quick force problem

  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
3
Views
18K
Replies
3
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
677
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
6
Views
903
  • Last Post
Replies
2
Views
905
  • Last Post
Replies
5
Views
2K
Top