# Fictitious forces

## Homework Statement

A slope is inside an elevator. A body with the mass of m is on the slope. θ=30°.
What would be the acceleration of the body in relative to the elevator if the elevator rises with an acceleration of 0.5g?
What would be the acceleration of the body in relative to the earth if the elevator rises with an acceleration of 0.5g?

ma=F

## The Attempt at a Solution

This question got me really confused. When am I supposed to add the fictitious force? When looking at the body in relative to the elevator or in relative to the earth?
The calculation itself is simple, but I can't figure in what case do I need to add the fictitious force. Thanks a lot.

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Doc Al
Mentor
When am I supposed to add the fictitious force?
Fictitious forces are used when viewing things from an accelerated frame. So use it in the frame of the elevator.

Fictitious forces are used when viewing things from an accelerated frame. So use it in the frame of the elevator.
And when looking at the elevator in relative to the EARTH. I shouldn't add this?

haruspex
Homework Helper
Gold Member
And when looking at the elevator in relative to the EARTH. I shouldn't add this?
Correct. For the purposes of the question, the Earth frame is near enough inertial.

Doc Al
Mentor
And when looking at the elevator in relative to the EARTH. I shouldn't add this?
Right. In the inertial frame of the earth (close enough, as haruspex says), only "real" forces appear: the normal force and gravity.

You could solve the problem in both frames, but you may find it easier to solve it in one frame then simply transform your answer to the other frame.

Correct. For the purposes of the question, the Earth frame is near enough inertial.
Right. In the inertial frame of the earth (close enough, as haruspex says), only "real" forces appear: the normal force and gravity.

You could solve the problem in both frames, but you may find it easier to solve it in one frame then simply transform your answer to the other frame.
I if have a body on an accelerating wagon (the body doesn't move in relative to the wagon), with acceleration equals to a (horizontal).
If I look at it in relative to the earth without adding such force, how does it make sense. I get just the normal and gravity, but the body does move with that acceleration, without having any forces acting that way.

haruspex
Homework Helper
Gold Member
I get just the normal and gravity,
If it is staying with the accelerating wagon then the wagon is exerting a horizontal force on it.

Doc Al
Mentor
I get just the normal and gravity, but the body does move with that acceleration, without having any forces acting that way.
For the body to accelerate there must be a force. For example, friction. (If the wagon surface were frictionless, the body could not accelerate.)

scottdave
Homework Helper
For the body to accelerate there must be a force. For example, friction. (If the wagon surface were frictionless, the body could not accelerate.)
It was never stated (that I see) if the situation in the elevator had any friction associated with the ramp. Is the wagon just another idea to try to figure it out?

haruspex
Homework Helper
Gold Member
It was never stated (that I see) if the situation in the elevator had any friction associated with the ramp. Is the wagon just another idea to try to figure it out?
It was another scenario introduced by Eitan in post #6.

Doc Al
Mentor
It was never stated (that I see) if the situation in the elevator had any friction associated with the ramp.
The ramp, presumably, is frictionless.

Is the wagon just another idea to try to figure it out?
A different scenario.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

A slope is inside an elevator. A body with the mass of m is on the slope. θ=30°.
What would be the acceleration of the body in relative to the elevator if the elevator rises with an acceleration of 0.5g?
What would be the acceleration of the body in relative to the earth if the elevator rises with an acceleration of 0.5g?

ma=F

## The Attempt at a Solution

This question got me really confused. When am I supposed to add the fictitious force? When looking at the body in relative to the elevator or in relative to the earth?
The calculation itself is simple, but I can't figure in what case do I need to add the fictitious force. Thanks a lot.
Inside the elevator, the "effective" gravitational force that is felt by the passengers (and any masses involved in an experiment) is just $g_{\text{eff.}} = g+ 0.5 \times g = 1.5\, g.$ So you weigh more inside the elevator than you do standing on the ground, and dropped objects fall to the elevator floor faster as well.

Of course, if you filmed that from outside, through the transparent glass walls of an elevator, you would see a dropped object just traveling as usual using the standard earth's acceleration of gravity, $g$, but you would also see the elevator floor rising up to meet the object. From inside the elevator it looks like the object is falling faster.

I realize that your problem does not involve a dropped object, but the same downward forces are acting on the object as would be the case if the object were dropped. The ramp is opposing those forces and re-directing the direction of acceleration.

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Inside the elevator, the "effective" gravitational force that is felt by the passengers (and any masses involved in an experiment) is just $g_{\text{eff.}} = g+ 0.5 \times g = 1.5\, g.$ So you weigh more inside the elevator than you do standing on the ground, and dropped objects fall to the elevator floor faster as well.

Of course, if you filmed that from outside, through the transparent glass walls of an elevator, you would see a dropped object just traveling as usual using the standard earth's acceleration of gravity, $g$, but you would also see the elevator floor rising up to meet the object. From inside the elevator it looks like the object is falling faster.

I realize that your problem does not involve a dropped object, but the same downward forces are acting on the object as would be the case if the object were dropped. The ramp is opposing those forces and re-directing the direction of acceleration.