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Dynamics Question -- packages discharged from a conveyor belt

  1. Feb 5, 2017 #1
    1. The problem statement, all variables and given/known data
    A series of small packages, each with a mass of 4 kg, are discharged from a conveyor belt as shown. Assume that the coefficient of static friction between each package and the conveyor belt is 0.4.

    Determine the force exerted by the belt on a package just after it has passed Point A

    I attached the picture below

    2. Relevant equations
    F = ma

    3. The attempt at a solution
    So I began by drawing a FBD:
    -W: Weight pointed downwards
    -N: Normal pointed upwards
    -f: Friction pointed left
    -B: Force from the belt pointed right.
    From what I understand, we are trying to solve for B.
    Also, it is about to enter a circular region, which makes me think that the net acceleration is -mv^2/r in the y direction and dv/dt in the x direction.

    So, I began by solving the forces in the y direction:
    -mv^2/r = -W + N
    -(4)(1^2)/(0.250) = -(4)(9.81) + N
    --> N = 23.24 N

    From there, I setup the equation in the x direction:
    dv/dt = B - (u)N
    dv/dt = B - (0.4)(23.24)
    Now the thing is, I assumed that the speed is consistent which would imply B = (0.4)(23.24) = 9.296 N.

    However, this answer was wrong, so I am not sure how to approach it from here.
     

    Attached Files:

    Last edited by a moderator: Feb 5, 2017
  2. jcsd
  3. Feb 5, 2017 #2

    PeroK

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    When the package passes point A, what value for ##\theta## are you assuming?
     
  4. Feb 5, 2017 #3
    0 degrees as shown in the picture. But 90 degrees, counterclockwise from the (+) x axis (assuming the center of the circle is the origin).
     
  5. Feb 5, 2017 #4

    PeroK

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    So, what force is causing the acceleration in the x-direction?
     
  6. Feb 5, 2017 #5
    I assumed the conveyor belt was pushing horizontally on it, to offset frictional impacts.
     
  7. Feb 5, 2017 #6

    PeroK

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    Friction is a resisting force. Unless an external force is trying to accelerate an object, the frictional force will be zero. In the case of a block at rest or a block moving with constant velocity horizontally, there can be no frictional force; if there were the block would accelerate.
     
  8. Feb 5, 2017 #7
    Okay, so then we're left with its weight and the normal force. So would the normal force be defined as the "the force exerted by the belt on a package." In which case the answer would simply be 23.24 N?
     
  9. Feb 5, 2017 #8

    PeroK

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    I think so!
     
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