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Weight at North Pole and on the Equator

  1. Sep 27, 2013 #1
    1. The problem statement, all variables and given/known data

    You stand on a spring scale at the norh pole and again at the equator. Which scale reading will be lower, and by what percent will it be lower than the higher reading? Assume g has the same value at pole and equator.

    I have no need for the answer to this question but am looking for an explanation, rather.
    In the solution, the centripetal force due to the Earth's rotation is pointed outwards (opposite to gravity) and I was wondering why this is? Why not add the centripetal force to the force of gravity instead of subtracting it, and find the weight the scale reads this way?

    Here is a diagram:

    http://imgur.com/wz5kuTf

    So any explanation on why the force of the Earth's rotation is outwards in this case as opposed to towards the centre of the Earth would be great.

    Thanks!
     
  2. jcsd
  3. Sep 27, 2013 #2

    tiny-tim

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    Hi MathewsMD! :smile:
    That isn't centripetal force, it's centrifugal force.

    In a rotating frame of reference, there is a centrifugal force (which of course has to be added to all the other forces).
     
  4. Sep 27, 2013 #3
    Do a free body diagram on the body at the equator. Let F be the outward force of the scale, and mg the inward force of the scale. What is the net outward force? If the body is rotating around the earth's axis, is the acceleration inward or outward. Write a Newton's 2nd law force balance for the body.

    Chet
     
  5. Sep 27, 2013 #4
    Here's the actual solution if you're interested.

    http://imgur.com/dBBZ6qt

    I understand it's the centrifugal force since it is directed radially outward, but they used centripetal in the solution which added to my confusion with this question. I just don't understand why the scale reading (FN) doesn't equal FN= mv2cosθ/r + mg.
     
    Last edited: Sep 27, 2013
  6. Sep 27, 2013 #5
    The force FN is outward (in the + r direction).
    The force mg in inward (in the negative r direction)
    The net outward force is FN-mg
    The acceleration v2/r is in the negative radial direction, as is the mass times the acceleration. So, from Newton's 2nd law,

    [tex]F_N-mg=-m\frac{v^2}{r}[/tex]

    Forget about using the specific words centrifugal and centripetal; they are just a source of confusion. Just focus on the acceleration and its direction.
     
  7. Sep 27, 2013 #6

    vela

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    You should try to get away from the notion of a centripetal force and think instead of centripetal acceleration. A centripetal force isn't another force that acts on a body. It's what we may call the resultant force if it happens to point toward the center of the circular path an object follows.

    An object moving in a circle of radius r at constant speed v has a centripetal acceleration of magnitude ##a_c = v^2/R##. That's the acceleration that goes into ##\sum F = ma##. Follow Chet's suggestion and write down what the net force on a person at the equator would be. Keep in mind the only two forces acting on the person are gravity and the scale pushing up on the person's feet.

    EDIT: Forgot to mention earlier that the solutions are just blatantly wrong when it says that the centripetal acceleration "points directly opposite to gravity." They point in the same direction.
     
    Last edited: Sep 27, 2013
  8. Sep 27, 2013 #7

    D H

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    The authors used centripetal force because they looked at things from the perspective of an inertial frame. There is no centrifugal force in an inertial frame.

    A frame with origin at the center of the Earth and in which the Earth is rotating is approximately an inertial frame. It isn't truly inertial because the Earth is accelerating toward the Sun and the Moon, and everything else out there. Ignoring those subtleties, this Earth centered, non-rotating frame is an inertial frame. Treating this frame as inertial, the only forces acting on the person standing on the scale are the downward force of gravity exerted by the Earth and the upward normal force exerted by the scale. It is this upward normal force that the scale measures.

    A person standing on one of the poles is stationary in this frame. Newton's first law dictates that these two forces must sum to zero. The person's weight is simply mg at the poles. What about a person standing on the equator? This person is not stationary. He is instead accelerating, undergoing uniform circular motion about the Earth's rotation axis. The net force cannot be zero. It must point to the center of the Earth to yield that circular motion. This means the upward force exerted by the scale has to be less than the downward gravitational force.
     
  9. Sep 27, 2013 #8

    D H

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    I agree with that advice.
     
  10. Sep 27, 2013 #9
    That bit really helped me clear things up.

    [tex]F_N+m\frac{v^2}{r}=mg[/tex]

    Thanks for the help everyone!
     
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