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Relative motion on a conveyor belt

  1. Sep 14, 2010 #1


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    1. The problem statement, all variables and given/known data
    Belt A conveys sand with a velocity of 6.0 ft/s (constant). The velocity of belt B is 8 ft/s (constant). Determine the velocity of the sand relative to B as it lands on belt B. Answer: VS/B=20.1 ft/s at 85.1 degrees


    2. Relevant equations

    VS/B = VS-VB
    (Velocity of sand relative to belt B = velocity of sand as it hits belt B - velocity of belt B)

    3. The attempt at a solution

    VB=8cos(15°)+8sin(15°) = -7.727[tex]\hat{i}[/tex]+2.071[tex]\hat{j}[/tex](ft/s)

    My trouble is with finding VS. I know that on the horizontal section of belt A, the sand's velocity (VS) is -6[tex]\hat{i}[/tex]+0[tex]\hat{j}[/tex](ft/s).

    When the sand hits the bend where belt A is not horizontal anymore, but starts to bend into belt B, the velocity of the sand changes.

    I think I'm stuck on the trig of what the angle is and how the 5ft contributes to finding the correct angle that the sand runs into belt B.
  2. jcsd
  3. Oct 24, 2010 #2
    when the sand leaves the belt A, its in free fall. use projectile motion analysis to find
    the velocity 5 ft below, when it hits belt B.
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