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Homework Help: Frame of reference, Forces, and Angles

  1. Mar 19, 2005 #1
    This is more of a concept problem that I'm trying to grasp.

    Verbatim from the book:

    "A box is moving with a horizontal velocity, v, relative to an inertial frame of reference so that a medallion of mass, m, inside of the box hangs from the roof of the box with an angle, theta, relative to the vertical as shown in Figure F.


    What can be said about the velocity of the box? What can be said about its acceleration? Explain."

    I'm thinking, the velocity is constantly changing and the acceleration is constant because there would have to be an acceleration to keep that medallion at the position. What do you guys think?
  2. jcsd
  3. Mar 19, 2005 #2
    That's right!
  4. Mar 20, 2005 #3
    really? my friend from an ivy league school said the velocity was constant and the acceleration was zero... so .. can anyone else confirm which answer is correct?
  5. Mar 20, 2005 #4


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    Your ivy league friend is wrong.
    Now, ask yourself (or him), if the medallion hangs with a non-zero angle, gravity will produce a non-zero torque about the attachment point, right?

    The only way such a non-zero torque from external forces is compatible with the medallion's stationary position with respect to the box frame, is that the box frame is a non-inertial frame.
  6. Mar 22, 2005 #5
    is there anyway to prove this mathamatically? like the x component of the medallion?
  7. Mar 22, 2005 #6
    Your system is a medallion attached to a string.
    Net Force = 0.
    Fnet = Fg + Ft (Tension)
    Fnet = mgsin(theta)+(-mgsin(theta)). This would cancel only if the tension force is applied at the same angle as the gravity force, aka if the medallion is hanging straight down. For the medallion to be hanging at an angle there must be another force in play but keeping the system in equilibrium.
    Ft would expand to -mgsin(theta)+(-mgcos(theta)) where the sum of this quantity would equal mgsintheta.
  8. Mar 22, 2005 #7
    [tex] F_{net} = ma [/tex]
    [tex] ma = mgsin\theta [/tex]
    [tex] a = gsin\theta [/tex]


  9. Mar 23, 2005 #8
    how did you get mgsin"theta"?
  10. Mar 23, 2005 #9
    The magnitude of acceleration depends on the angle that gravity is pulling you at. Imagine a ball on a ramp. The steeper the ramp, the closer theta is to 90 degrees. From convention you know that a ball rolls down a steep ramp much faster than it would a shallow ramp (theta => 0 degrees).

    In the system, the sin(t) means the angle the medallion is hanging at in comparison to where 'straight down' is.
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