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Friction/Kinematics Question. Conveyor Belt.

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data

    A baggage handler drops your 8kg suitcase onto a conveyor belt running at 1.8 m/s. The materials are such that the static coefficient of friction is .4 and the kinetic coefficient of friction is .6. How far is your suitcase dragged before it is riding smoothly on the belt?

    2. Relevant equations

    F=ma
    v_f= v_i+a(delta t)
    f_s: force of static friction
    f_k: force of kinetic friction

    3. The attempt at a solution

    I've come up with two free body diagrams. The y direction seems to be negligible since mg=n and they cancel out. In the x direction, I have F_belt pointing in the positive x-direction. I am confused about how to place f_k. I think that it too points in the positive x-direction, and heres my logic: it is not moving faster than the conveyor belt, so its motion would appear opposite the conveyor belt until it too moves at 1.8 m/s.

    Also, there would be a free body diagram for when the suitcase moves at the velocity of the conveyor belt. This diagram would have F_belt pointing in the positive x-direction and f_s pointing in the negative x-direction.

    F_x=8kg(0m/s^2)=F_belt - f_s
    F_belt=f_s
    F_belt= .4 (8kg) (9.8m/s^2)
    F_belt=31.4 N

    then,

    F_x= 8kg*a = F_belt + f_k
    = 31.4N + .6(8kg)(9.8m/s^2) = 78.44 N
    78.44/8= 9.8 m/s^2 = acceleration of suitcase

    kinematics:
    1.8m/s=0m/s + 9.8 m/s^2(delta t)
    delta t= .18 seconds

    x_f=x_i+v_i*t+.5at^2
    x_f=0+0+.5(9.8m/s^2)(.18^2)
    x_f=.16m


    This isn't the answer, according to my book*.
    *I changed the numbers, but this is the method I used with the numbers I had and I didn't get the right answer.
     
    Last edited: Oct 4, 2008
  2. jcsd
  3. Oct 4, 2008 #2
    The force of friction always opposes the direction of the movement.
    So you will have to change you signs where you're calculating acceleration of suitcase.
     
  4. Oct 4, 2008 #3
    I don't understand...was my depiction correct? I know that friction opposes motion, but relative to the point on the belt where the suitcase is dropped, I would think the suitcase is moving away from that point (negative direction if conveyor belt is moving in the positive direction), making the direction of the force of kinetic friction point in the positive direction.

    If this is true, I would have to have positive acceleration, I think.
     
  5. Oct 4, 2008 #4

    PhanthomJay

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    There are far too many forces noted here. The force of the belt acting on the suitcase is one and the same as the force of friction between the belt and suitcase acting on the suitcase. It is not an additional force. The kinetic friction force accelerates the suitcase forward (FBD #1) until it reaches the same speed as the belt (FBD #2), at which point, there is no longer any net force acting on it. So the force of static friction at this point is what?
    Rake-MC: to clarify, the friction force always opposes the direction of relative motion (or pending motion) between the 2 objects in contact.
     
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