0
$\begingroup$

Is weight ≠ gravitational force, for an object on the Earth’s surface? Why is this? Is it because part of the gravitational force acts as a centripetal force? Does the normal contact force not nullify the centripetal force?

Also, If my tangential velocity (on the surface of the Earth) increases, does my effective weight decrease?

$\endgroup$
4
  • $\begingroup$ Yes. You are right. The weight/ normal force acting on you is less than the gravitational force. Because some of the gravitational force is used to move you around earth's centre. $\endgroup$
    – Alapan Das
    May 12, 2020 at 9:58
  • $\begingroup$ The terminology is not right. The weight is still the gravitational force. Of course, you can still speak of the effective weight, which in this case happens to be the normal force. It is more easily seen by transforming into a rotating frame of reference in which the vector sum of the centrifugal force and weight is the effective weight. $\endgroup$
    – 13509
    May 12, 2020 at 10:14
  • $\begingroup$ If my tangential velocity increases (on the surface of the Earth) increases, does my effective weight decrease? $\endgroup$ May 15, 2020 at 4:56
  • $\begingroup$ Why would that happen ? You are still considering the centripetal force as an independent force. Go through the answer first. $\endgroup$
    – user262060
    May 15, 2020 at 5:07

3 Answers 3

4
$\begingroup$

There is no fixed definition of weight. Some say it's the force of gravity on an object, others say it's the normal force that opposes gravity, i.e., what a scale measures.

If you take the view that weight is the force of gravity on an object, then we are finished. That's clear enough.

If we take the view that weight is what we measure when we put an object on a scale we have to be careful. The scale is in the rotating frame, so we must use the dynamics of a rotating frame of reference. Weight = Gravitational Force - Centrifugal Force.

Most people would say that weight is what a scale measures (although physics often re-purposes common words). In that case weight has no meaning in the stationary frame. We say that the weight of an astronaut is zero, yet there is a gravitational force on her.

As others have mentioned, centripetal force is a description of the net force on a rotating object. It is not a force itself; it is not the result of an interaction between two objects. A person on a Ferris wheel experiences the force of gravity and the normal force of the seat against her body. That's it. There's no additional centripetal force. We give the name centripetal force to the net of those two forces.

This point about the nature of centripetal force is not emphasized nearly enough in introductory presentations. The two-word phrase does make it sound like it itself is a force.

$\endgroup$
0
$\begingroup$

Look there is nothing like centripetal force. If a body is moving in a closed track and forces like gravitation , normal , tension is acting on it then the net result of these forces towards the centre of the closed track is called centripetal force. It doesn't have its own identity like other forces have.

If you are standing on the earth then the normal force balances the gravitational force and when standing on scale then scale measures the ratio of normal to g and shows your effective mass or for weight , it is the normal force which is equal to the gravitational force again.

$\endgroup$
0
$\begingroup$

Centripetal force is not an external force, it's always provided by a body, in this case it's provided by the gravitational force. The gravitational force allows the body to move around in orbits because of its velocity. This is one of the biggest misconceptions that centripetal force is an external force. Weight is always equal to gravitational force.

$\endgroup$
1
  • $\begingroup$ My over-sensitive track pad gave this answer a down vote. I can't undo it. I'm sorry; the answer does not deserve a down vote. $\endgroup$
    – garyp
    May 12, 2020 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.