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Application of Newtons' Law

  1. Oct 15, 2007 #1
    A bicycle of bicyclist of combined masss 62 kg go over a hump of radius 6m at a speed of 5m/s. what is their combined appararent weight at the top of the hump?

    is it just...

    Fc = MAc

    Fc =M(V^2/r)

    Fc = 62(25/6)

    Fc = 290N
  2. jcsd
  3. Oct 15, 2007 #2
    You only took into account the centripetal force, there is another force acting on the system.

    BTW "apparent weight" usually means that they are asking for the normal force (the reaction force that the hump applies).
    Last edited: Oct 15, 2007
  4. Oct 15, 2007 #3
    grrr, I have no idea then.
  5. Oct 15, 2007 #4
    Well think about it. If i stand on a hump and don't move then my weight is gonna be w = mg but if i travel over the hump at some speed then the centripetal force will act in the opposite direction of gravity and make me lighter.
  6. Oct 15, 2007 #5
    oh, then normal force is what i solved for. didn't I?
  7. Oct 15, 2007 #6
    hmmm, ya, that's correct. Hold on let me check the problem again.
  8. Oct 15, 2007 #7
    man, I have no idea how to solve this one..
  9. Oct 15, 2007 #8
    you solved for the centripetal force

    Sum of force = MA (a = 0 because no acceleration in y direction, or you can plug in a = v^2/r and keep left side as n - mg)

    N - mg + Centripetal = 0

    you already solved the centripetal force. Now plug in mg and solve for N to get "apparent weight" or normal force.
  10. Oct 15, 2007 #9
    hmmm...thanks man. I appreciate it.
  11. Oct 15, 2007 #10

    Fn - mg + Fc = 0

    Fn = mg - Fc

    Fn = 607.6-258.33

    Fn is roughly 350
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