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Determining Effective Force of Gravity

  1. Nov 20, 2008 #1
    How do you find the effective force of Gravity when at a particular latitude?

    I keep trying to figure it out, but I can't seem to wrap my head around it. I understand that some small portion of the force of gravity goes towards your centripetal acceleration at certain latitudes, but I can't seem get the idea, since you seem to have three unknowns, but only two equations.

    For example (45 degree latitude):

    You are not moving in the y-direction, so the y-component of your weight must be equal to the y-component of your normal force; [​IMG]
    and then the net force in the centripetal acceleration is:

    But you only have these two equations, but you have three variables, the normal force, the acceleration of gravity (and as a result the force of gravity) and the angle that the normal force is at (it is offset from the force of gravity).

    Does anyone have any idea as to what I am missing, that is making me incapable of understanding this? Thanks.
  2. jcsd
  3. Nov 22, 2008 #2

    Doc Al

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    Staff: Mentor

    There are only two forces acting on the object: The weight, W = mg, which acts toward the center of the Earth, and the normal force, N. You also know the acceleration, a = ω²r = ω²Rcosθ, which acts toward the axis of rotation. (R is the radius of the Earth.)

    You have all you need to solve for the normal force (which gives you the effective weight of the object). Think vectorially:
    [tex]\vec{W} + \vec{N} = m\vec{a}[/tex]

    Looking at just the radial components, you have:
    [tex]-mg + N_r = -m\omega^2R\cos^2\theta[/tex]

    [tex]N_r = mg -m\omega^2R\cos^2\theta[/tex]
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