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Homework Help: Standing on a scale at the equator

  1. Sep 1, 2012 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    Fscale + Fg = ma

    That's my guess.

    3. The attempt at a solution

    Well, I'm reading it in the book. It's an example with the answer that the scale reading will be less than my weight. But I still don't understand how that works out to be.

    So, a scale is pushing back the same force that's going into it? And it notices how much weight it has to balance, thus say it's balancing 81.81 kg so, about 801N is being pushed into the scale and the scale is pushing back out about 801N? What?

    I really don't get this.

  2. jcsd
  3. Sep 1, 2012 #2


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    Remember, the Earth is spinning, so you have a centripetal force acting on you. This means that the resultant force on you is acting towards the center of the Earth.

    So with your scale, you have two forces acting, your weight mg and the normal reaction N (this is what the scale gives).

    The sum of these forces in the direction of the resultant force (i.e. centripetal force) would be :

    mg - N = Fcent

    If N> mg, then the vector would point away from the Earth and you'd fly into space. Since you don't fly into space it means that mg > N or the reading on the scale is less than your actual weight.
  4. Sep 1, 2012 #3


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    In the real world accurate scales would probably be calibrated to suit their latitude/locality so might over read at the poles rather than under read at the equator.
  5. Sep 1, 2012 #4
    I'm still not seeing it. This may be one of those, "Pound it into my head" situations. I've got a bachelors in neuroscience, so I'm impressed this is boggling me.

    Where do I start...
    So, sure, I'll have a weight despite where I stand on earth.
    But it seems like the problem has the condition, and this is an assumption, that I'm taking my scale from my bathroom in Illinois where I weigh 80 kg (for example), and moving it to the equator.

    So, the scale is going to exert back less contact force, just because I'm at the equator?
    Why? Without knowing that, it seems like I'm just memorizing these facts.

    So what is the equation for when the scale is at my place of residence?
    mg = N = F_cnet?

    Why should I read of a weight less than I am at home because I'm at the equator? Isn't centripetal force acting on me despite where I am in the world?
  6. Sep 1, 2012 #5


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    So you will have

    mg-N = mω2r

    The radius of the Earth is not constant, so at different places, the radius would be different (remember, the Earth is not a perfect sphere)
  7. Sep 1, 2012 #6
    Newton's 2nd law,
    When you are at the equator, you are rotating with the radius of the earth.
    Rotation with constant speed around a circle means you are accelerating towards the center.

    There are 2 forces acting on your body. One pulling you down by the earth and the other is the scale that stop you from sinking into the earth.

    The net force will produce -macentripetal.
  8. Sep 1, 2012 #7
    Let me see if I get this right...

    This all really has to do with the difference between linear acceleration and rotational acceleration. At the equator, a person is getting the full impact of the entire Earth's rotational being, thus the entirety of the centripetal force that can come upward from the Earth's core through the scale.

    But at a location like Illinois, this is reduced, because a person is only getting a portion of the full centripetal force, because the distance of the total possible circumference traveled is much less.

    And at a pole, such as the True North or South pole, there isn't any rotational force, because a person is away from a point of rotation, so the main force at play would be vertical/linear acceleration. But like... this totally gets into the whole argument of is a point in space a circle... so couldn't one argue that yes, centripetal force even occurs at the poles, yet it's so ridiculously small, that you might as well consider it an linear form of acceleration?

    Is that the right idea?

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  9. Sep 1, 2012 #8
  10. Sep 1, 2012 #9


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    The cause of the scale reading being different? Because the value of the Earth's radius differs for various latitudes and longitudes.
  11. Sep 2, 2012 #10


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    The sentence "is the reading greater or less than your weight" is confusing. Weight -by definition- is the force a body acts on a horizontal support or vertical suspension. When standing on a scale there is a normal force between the person and the scale and it is detected for example by a spring which is compressed with some amount proportional to that force. That normal force and the other forces (gravity, centrifugal force) balance each other as you are in rest.
    The reading of scales is mass (in kg-s for example). The compression is proportional to the normal force, and it is divided by g, the gravitational acceleration to get mass. So it would be better to ask "Is the reading greater or less than your mass? "
    g is the acceleration of a free-falling body near the Earth surface. There are two things that affect it at different places of Earth. One is the distance from the centre of Earth as gravity acts as if the mass of Earth would be concentrated there. A person at the Equator is farther from the centre of Earth than at the poles. The other thing is the centrifugal force, which acts in rotating frames of reference. It acts perpendicular to the axis of rotation, and away from the axis, and its magnitude is mrw^2. (r is the radius of the cross section of Earth at the given latitude and w is the angular speed of rotation, m is the mass.)
    This centrifugal force is greatest at the Equator and zero at the poles. So you can conclude that the bodies fall with greatest acceleration at the poles and smallest acceleration at the Equator.
    Read http://en.wikipedia.org/wiki/Gravity_of_Earth and http://en.wikipedia.org/wiki/Standard_gravity.

    I think the scales are calibrated using the standard gravity - g at 45° latitude: gs=9.80665 m/s^2. At the equator g is lower, ge= 9.78033 m/s2.

    So the normal force at the equator is N=mge, but the scale is calibrated with gs, its reading is m(read)=N/gs.

    I used the term centrifugal force as we are and observe things on the Earth and using the Earth as frame of reference, it is a rotating one.

    If you see things from outside the Earth, from the Space, you see that the person standing on the scale moves along a circle together with that point of Earth where they are. But circular motion is only possible if the resultant force is appropriate to the circular motion with the given angular speed and radius: in case of the person on the scales, gravity - normal force equals the centripetal force mw2r. This discussion will lead to the same result as the one with the centrifugal force, experienced on the surface of Earth.

    Last edited: Sep 2, 2012
  12. Sep 2, 2012 #11


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    Yes that's the idea.

    As you approach the pole the vertical component of centrifugal force reduces to zero.

    The vertical component depends on the Cosine of the latitude..

    As latitude -> 90 then COS(latitude) -> 0

    However if your scales were calibrated for Illinois then they would read your correct weight in Illinois. At the equator they would under read. At the pole they would over read.

    The bit I snipped below is wrong.

    At/near the pole there is still centripetal/fugal force but it all acts in a direction the scale doesn't respond to. There is no vertical component.
    Last edited: Sep 2, 2012
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