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Motion in a Plane

  1. Feb 20, 2005 #1

    ktd

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    Here's the question:

    Sand moves without slipping at 6.0 m/s down a conveyer that is tilted at 15 degrees. The sand enters a pipe 3.0 m below the end of the conveyer belt. What is the horizontal distance between the conveyer belt and the pipe?

    For some reason, I'm having the hardest time with this question. On first glance, I automatically think of the kinematic equations, but for some reason I'm not sure how to use them here. What are the triangles here?

    Any help is appreciated!
     
  2. jcsd
  3. Feb 20, 2005 #2
    Start out by finding the X and Y compoents of the sands velocity. Then look at the kinematic equations again.
     
  4. Feb 21, 2005 #3

    ktd

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    ok so I've found the components, but now what? This problem is totally annoying me!
     
  5. Feb 21, 2005 #4

    xanthym

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    Science Advisor

    {Horizontal Distance Moved From Conveyer After Time T} =
    = T*(6 m/s)*Cos(15 deg)
    {Vertical Distance Moved From Conveyer After Time T} =
    = -T*(6 m/s)*Sin(15 deg) - (1/2)*g*(T^2) = (-3 m)

    Solve 2nd eq for T and sub into 1st eq.
    (Note all terms of 2nd eq have (-) since motion is downward.)
     
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