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Variable friction proof

  1. Jun 17, 2009 #1
    A warehouse worker is shoving boxes up a rough plank inclined at an angle B above the horizontal. The plank is partially covered with ice, with more ice near the bottom of the plank than near the top, so that the coefficient of friction increases with the distance x along the plank : mu=Ax, where A is a positive constant and the bottom of the plank is at x=0. (For this plank, the coefficients of kinetic and static friction are equal: mu(k)=mu(s)=mu.) The worker shoves the box up the plank so that it leaves the bottom of the plank moving at speed v0. Show that when the box first comes to rest, it will remain at rest if v02>or=(3gsin2B)/(AcosB)

    I drew a FBD and started the problem by solving for mgsinB = AxmgcosB. Basically that the friction was perfectly counteracted by that component of weight.
    => Ax=tanB

    I don't know where to go from here. I didn't know if I should use kinematics or energy or what?
     
    Last edited: Jun 17, 2009
  2. jcsd
  3. Jun 17, 2009 #2

    cepheid

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    Hmm, well, I'm not sure your first step makes sense. If the box is moving up the plank, then wouldn't the || component of its weight and the friction force both be pointing in the same direction? So they wouldn't counteract each other. They would both work together as forces opposing the motion.

    Edit: I see. You were talking about after the box comes to rest. Okay.
     
  4. Jun 17, 2009 #3
    so I tried substituting for x according to the equation v2= v02 + 2a (x-x0). that gives:
    v02= -2asinB/(AcosB)

    then if I use ma=-mgsinB...

    v02= 2gsin2B/AcosB

    off by a factor of 3/2?
     
  5. Jun 17, 2009 #4

    cepheid

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    Here's one method. Although energy is not conserved, you can still keep track of things because you know exactly how much work is done by friction. Because the friction force is not constant, you have to integrate. If you don't know what integration is, and haven't done calculus, well then this could be difficult to understand. Maybe there is some other way of solving it. Here is what I did:

    We want the height at which the box stops to be ≥ the height at which the box is in static equilibrium. The latter is easy: you have derived it

    Height at which box is in static equilibrium: x = tanB / A

    Height at which box stops = height at which all of initial kinetic energy is used up. I.e.:

    initial kinetic energy = work done by friction + final potential energy

    mv02 = ∫F(x)dx + mg xsinB (xsinB is the height of the box)

    ∫F(x)dx = ∫(AxmgcosB)dx = (A/2)x2mgcosB

    Plugging the result of the integral back into the energy equation, we get:

    v02 = (A/2)x2gcosB + gxsinB

    plugging in x = (tanB)/A , because we want the energy to run out, at least at the height where static equilibrium occurs, so that the block stays there after it stops:

    v02 ≥ (A/2)(sinB/AcosB)2gcosB + g(sinB/(cosB)A)sinB

    When I simplify, I get

    v02 ≥ 3g(sinB)2/2AcosB

    That seems to differ from the result you were supposed to show by a factor of 1/2 though...:confused:
     
  6. Jun 17, 2009 #5

    cepheid

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    Sorry. This equation is only valid if the acceleration is constant. The acceleration is not constant, because the force is not constant. That's why I used integration in my previous post.
     
  7. Jun 17, 2009 #6
    no that's right, it's just that the kinetic energy is 1/2mv^2


    thank you so much this actually makes sense! :D
     
  8. Jun 17, 2009 #7

    cepheid

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    LOL, yeah, good point!


    You're welcome, and I'm glad it does.
     
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