Accelerating Frames: Apparent Weight of Crate at Equator

[anikmartin]

Hello all!
I have a question on acclerating frames.
The apparent weight of an object in an elevator that is accelerating upward is m(g+a), where g is gravity and a is the acceleration of the elevator.
In my textbook they give an example of a crate placed on a scale at the equator. They want to find the apparent weight of the crate as the Earth rotates about its axis. They found the apparent weight to be m(g-(v^2)/r), the difference between the accleration due to gravity and the centripetal acceleration of Earths rotation. At first thought I assumed that the apparent weight would be more because both gravity and the centripetal force and pulling down on the crate. Why am I wrong to think this? Please help, and thank you everyone for all you help.

 

[Fredrik]

At first thought I assumed that the apparent weight would be more because both gravity and the centripetal force and pulling down on the crate. Why am I wrong to think this?

Because gravity (minus the normal force) is the centripetal force in this case. They are not two different forces acting together.

 

[arildno]

Note that, as Fredrik pointed out, that:
1)"Centripetal force" means nothing else than:
Whatever force is present to provide centripetal acceleration.
2)"Tangential force" means nothing else than:
Whatever force is present to provide tangential acceleration.

That is, this distinction between forces is subsidiary; the primary distinction is between centripetal/tangential accelerations, and through Newton's 2.law, we later on attach these labels onto whatever forces happened to be the agents of these accelerations.

 

[Hyperreality]

I assumed that the apparent weight would be more because both gravity and the centripetal force and pulling down on the crate. Why am I wrong to think this? Please help, and thank you everyone for all you help.

Ever felt being pushed to door when a car is turning? Remember Earth rotates about the equator, so it's same system as you're in a car making a turn.

The sensation of being pushed against the door is called a centrifugal force (a modification of Newton's 2nd Law). Remember, this only applies when you're inside the car.

So, if you are at equator, you would feel a centrifugal force (or fictitious ) of mv^2/r.

Therefore the weight is m(g-mv^2/r).

 

[kawikdx225]

At first thought I assumed that the apparent weight would be more because both gravity and the centripetal force and pulling down on the crate. Why am I wrong to think this? Please help, and thank you everyone for all you help.

Why would you think centripetal force pulls down?
Think about spinning a wet basketball on your finger. The water drops will fly away from the ball.

 

[HallsofIvy]

Why would you think centripetal force pulls down?
Think about spinning a wet basketball on your finger. The water drops will fly away from the ball.

No, you are thinking of centriFUGAL force. Centripetal force is the force that keeps an object moving in circle. In the case of an object rotating on the surface of the earth, the centripetal force IS gravity. The force of gravity goes partially into keeping the object moving in a circle and partially into pressing the object against the earth.

If an object has mass m kg, then the gravitational force is mg. Part of that, mv/r², goes into keeping the object moving in a circle and the rest, mg- mv/r², presses the object against the Earth and is what would be read on a scale.

 

[kawikdx225]

No, you are thinking of centriFUGAL force. Centripetal force is the force that keeps an object moving in circle. In the case of an object rotating on the surface of the earth, the centripetal force IS gravity. The force of gravity goes partially into keeping the object moving in a circle and partially into pressing the object against the earth.

If an object has mass m kg, then the gravitational force is mg. Part of that, mv/r², goes into keeping the object moving in a circle and the rest, mg- mv/r², presses the object against the Earth and is what would be read on a scale.

OK, thanks
So does centrifugal force play any role in the weight measured on the scale?

 

[Taoist]

No! It will depend solely on your tolerance.

 

[HallsofIvy]

OK, thanks
So does centrifugal force play any role in the weight measured on the scale?

"centrifugal force" is usually referred to as a "phony" force. It isn't, strictly speaking, a force in itself. If you are swinging a weight on a rope, you have to APPLY force to keep the weight going in a circle, instead of off on a straight line- that's the "centripetal" force. But to your hand and arm, it feels like the weight is pulling on you- that's the "centrifugal" force. It's really the "reaction" to your pull.

 

[kawikdx225]

"centrifugal force" is usually referred to as a "phony" force. It isn't, strictly speaking, a force in itself. If you are swinging a weight on a rope, you have to APPLY force to keep the weight going in a circle, instead of off on a straight line- that's the "centripetal" force. But to your hand and arm, it feels like the weight is pulling on you- that's the "centrifugal" force. It's really the "reaction" to your pull.

Ahhh, "I see" said the blindman
Thanks for clearing that up.