Accelerating Frames: Apparent Weight of Crate at Equator

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    Accelerating Frames
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Discussion Overview

The discussion centers on the apparent weight of a crate at the equator due to Earth's rotation, specifically examining the effects of centripetal and gravitational forces in an accelerating frame. Participants explore the relationship between these forces and how they affect the reading on a scale.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants assert that the apparent weight is calculated as m(g - (v^2)/r), where g is gravitational acceleration and v is the tangential velocity at the equator.
  • Others argue that centripetal force is not an additional force acting downward but is provided by gravity itself, which also contributes to the normal force pressing the object against the scale.
  • A participant compares the situation to feeling pushed against a door in a turning car, suggesting that the centrifugal force experienced is similar to the effects of Earth's rotation.
  • There is a distinction made between centripetal and centrifugal forces, with some clarifying that centrifugal force is considered a fictitious force and does not directly affect the weight measured on a scale.
  • Some participants express confusion about the role of centripetal force, with one noting that it is gravity that provides the necessary centripetal acceleration for an object on Earth's surface.
  • There is a discussion about the nature of centrifugal force, with some describing it as a "phony" force that is a reaction to the centripetal force applied to keep an object moving in a circle.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the interpretation of forces involved in the apparent weight measurement. While some clarify the roles of centripetal and gravitational forces, others maintain differing views on the nature and impact of centrifugal force.

Contextual Notes

The discussion reflects varying interpretations of force dynamics in rotating frames and the implications for apparent weight, with no consensus reached on the conceptual understanding of centrifugal force.

anikmartin
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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 acceleration 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. :smile:
 
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anikmartin said:
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.
 
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.
 
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).
 
anikmartin said:
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. :smile:

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.
 
kawikdx225 said:
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.
 
HallsofIvy said:
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?
 
No! It will depend solely on your tolerance.
 
kawikdx225 said:
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.
 
  • #10
HallsofIvy said:
"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.
 

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