Solving a Quick Force Problem: Finding Acceleration in an Elevator

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Homework Help Overview

The problem involves determining the acceleration of a person standing on a scale in an elevator, given their normal weight and the scale reading. The subject area pertains to forces and Newton's laws of motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss applying the formula Fnet=ma and the significance of free body diagrams. There is an exploration of the forces acting on the person, including the normal force and gravitational force. Questions arise regarding the relationship between the forces and the acceleration of the elevator.

Discussion Status

Some participants have offered guidance on visualizing the problem through free body diagrams and have prompted further exploration of the forces involved. There is an ongoing examination of the net forces acting on the person, but no consensus has been reached on the correctness of the calculations presented.

Contextual Notes

Participants are considering the implications of the scale reading and the forces acting on the person, while also referencing Newton's third law. There may be assumptions regarding the direction of forces and the relationship between the elevator's acceleration and the person's acceleration.

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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?
 
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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.
 
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!
 
joe215 said:
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!
 

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