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Question About Question

  1. Feb 11, 2007 #1
    1. The problem statement, all variables and given/known data
    A box of mass m rests on the flat floor of a truck. The coefficients of friction between the box and floor are mu_s and mu_k. The truck stops at a stop sign and then starts to move with an acceleration of a.

    If the box is a distance x from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck?

    2. Relevant equations
    F = ma


    3. The attempt at a solution
    I have a few questions about the problem, mainly because I don't understand the situation. (Well I have started on it, but it turned out nonsense so I decided that I really need to understand the whole picture first.) These might be really stupid questions but it isn't clear for me....

    First of all, the truck is driving, and THEN it stops right? It didn't start out at the stop sign and then start to accelerate?

    Second, the box on the truck, that's static friction, isn't it? Because the box is not moving. It takes static friction first to get it to fall off the truck?
     
  2. jcsd
  3. Feb 11, 2007 #2

    AlephZero

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    1 You don't need to worry about what happened while the truck was stopping. After it stopped, the box was distance x from the rear of the truck. It doesn't matter how the box got to be at that position.

    2 Yes, you have apply a force bigger than static friction before the box moves at all. Then when it is moving, the resistance is dynamic friction.
     
  4. Feb 11, 2007 #3

    cepheid

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    Sure, but the motion of the truck before reaching the stop sign is irrelevant to this problem. You know this, because 'x', the distance of the box from the end of the truck bed, is given at the instant the truck begins to move (from the stop sign). So, the only reason for introducing the stop sign is to let you know that at the beginning of this problem, everything is at rest. I agree that the wording could have been made much clearer by simply stating that the truck is initially stopped at a stop sign, rather than saying it stops at a stop sign. Oh well...
     
  5. Feb 11, 2007 #4
    Thanks for the help guys.

    Sorry, I'm still confused about this problem. So far, I have the FBD: n upwards, mg downwards, and fs to the right. So summing all the forces together, I have--

    y axis: n = mg
    x axis: f_s = u_s*mg

    and then when it starts to move:

    f_k = ma
    u_k*mg = ma
    a = u_k*g

    But then I don't know where to go from here....I'm thinking of the eqn of motion 2, but I don't know how to integrate that in there. I could plug in the a that I got from above to find the time, but then what about the acceleration of the truck? How do I put it all together?
     
  6. Feb 12, 2007 #5

    AlephZero

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    You seem to be using "a" for two different things, the accleration of the truck and the accleration of the box.

    Suppose the truck is accelerating from left to right.

    First think about the case where the truck accelerates slowly and the box doesn't slip. The truck has acceleration a. The box also has accleration a. To accelerate the box, the horizontal force on the box (from friction on the floor of the truck) = ma.

    If the force ma is less than u_s*mg, the box doesn't slip relative to the truck, and they both have the same accleration forwards.

    If the truck accelerates fast, and ma > u_s*mg, then the box will slip. When it starts to slip, the force on the box (forwards) is u_d*mg.

    So, the box is being acclerated forwards by the force u_d*mg, but the truck is acclerating faster that the box, with accleration a.

    When the truck has travelled a distance d and the box has travelled a distance d-x, then the box falls off the back of he truck.
     
  7. Feb 12, 2007 #6
    I'm so confused. I mean, I understand that...but to apply it, I'm having troubles.

    Okay, so they do have different acceleration-- a > u_sg

    So is it like this:

    x = 1/2at^2 for the truck, then I'd solve for time and get (2x/u_sg)^(1/2). Then I plug it into the eqn d - x = 1/2at^2 to get x?

    What about u_k? If I use that, then u_kg should equal a right?
     
  8. Feb 12, 2007 #7

    Doc Al

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    Some comments:
    (1) The truck acceleration (with respect to the road) is given as "a".
    (2) The only force acting on the box while it slides is the kinetic friction force; use that (and Newton's 2nd law) to find the acceleration of the box with respect to the road.
    (3) Note that you are given the distance that the box moves with respect to the truck. So first find the acceleration of the box with respect to the truck. (Big hint.)
     
  9. Feb 12, 2007 #8
    Thanks. I sorta get it now.

    Edit: Nvm, I got the answer already, thank you so much! :)

    Edit2: Erm stupid question and I went over this chapter already but how do you tell whether it's accelerating with respect to truck or to the road?
     
    Last edited: Feb 12, 2007
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