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G-force equation confusion

  1. Nov 4, 2012 #1
    Hi.

    Can you clarify for me the equation for g-force below:

    F= ma + mg

    I got this equation from my lecturer. He asked the class to determine how was the formula above formulated. But this is not a homework. Is the equation above true for g-force? If it's true, how does the formula come about?

    I thought g-force is given by (as stated by Newton's law of universal gravitation):

    F = G m1m2/r^2

    I really can't see how these two equations are equal. Could you help me out? Thanks!
     
  2. jcsd
  3. Nov 4, 2012 #2
    Now that I have had a good look at my question, I thought about the following:

    F = ma + mg
    F - mg = ma
    F - Fg = ma
    → ƩFnet = F - Fg = ma

    Actually I will be on a trip to ride a roller coaster. Our class was asked to determine first how was the 'formula for g-force' F = ma + mg formulated.

    If an object falls freely downward, it would be subjected to g (acceleration due to gravity). Time this with the object' mass we would obtain its weight/ gravitational force, Fg. However Newton's 3rd law states that for every action there is an equal but opposite reaction, hence this opposite force, denoted as F, which is directed upward (opposite Fg), but has an equal magnitude to that of Fg. The net force, ƩFnet, is then the difference between these two forces (F and Fg) and it must obey Newton's 2nd law, which is
    ƩFnet = ma​

    Therefore, the F in the equation F = ma + mg (given by my lecturer) is not really a g-force, is it? Because g-force is already given by Fg = mg. In other words, F = ma + mg is not an equation to find the gravitational force, right?

    Am I right? If not, please correct me.

    Thanks!
     
    Last edited: Nov 4, 2012
  4. Nov 4, 2012 #3
    Well, mg is the force due to gravity alone - which we call 'weight'.

    Is weight a g-force, or do we consider the other 'peculiar' forces (a) experienced on a ride the 'g-force' and leave weight out of it? Or is it all g-force?

    It's just semantics, but generally we'd consider weight part of the g-force so it's g+a.
    So F = m(g+a)
     
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