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What is the force of the boom on the car at the top of circle?

  1. Mar 21, 2015 #1
    1. The problem statement, all variables and given/known data
    An amusement ride consists of a car moving in a vertical circle on the end of a rigid boom. The radius of the circle is 10 m. The combined weight of the car and riders is 5.0 kN. At the top of the circle the car has a speed of 5.0 m/s which is not changing at that instant. What is the force of the boom on the car at the top of the circle?

    a.3.7 kN (Down)

    b.1.3 kN (Down)

    c.6.3 kN (Up)

    d.3.7 kN (Up)

    e.5.2 kN (Down)

    2. Relevant equations
    Fc = mv^2/r
    Fg = mg = 5000N
    m = weight / g = 510kg

    3. The attempt at a solution
    Since the car is at the top of the circle, I thought it would be
    Fc = Fb + Fg
    Fc = 510 * (5)^2 / 10 = 1275 N
    Fb = Fg - 1275 N = 3725 N
    So I thought the answer was 3725 N to the down way, but the answer says it is d which is 3.7kN to the up way.
    Can you tell me what I did wrong?
     
  2. jcsd
  3. Mar 21, 2015 #2

    haruspex

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    Your Fc=Fb+Fg is correct if you are taking down as positive throughout. But how did that equation become Fb = Fg - 1275N?
     
  4. Mar 21, 2015 #3
    Oh I'm so dumb haha so since Fb = 1275N - Fg = -3725 N and I picked down as positive, the force is up 3725 N right??
     
  5. Mar 21, 2015 #4

    haruspex

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    Yes.
     
  6. Mar 21, 2015 #5
    Thank you so much!!
     
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