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Board pulled

  1. Oct 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A small box of mass 1 is sitting on a board of mass 2 and length L . The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box is Us . The coefficient of kinetic friction between the board and the box is, as usual, less than Us.

    Find the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).


    2. Relevant equations

    F=ma

    3. The attempt at a solution

    I have a couple of questions:
    1. The acceleration does not go to 0 in the x direction because the board is pulled with a constant force and therefore a constant acceleration because force and acceleration are both constant. Is this right? Or do they go to zero because they were at rest?(Inertia, newton's first law)

    [url=http://www.freeimagehosting.net/aavu4][PLAIN]http://www.freeimagehosting.net/t/aavu4.jpg[/url][/PLAIN]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 18, 2012 #2

    SammyS

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    There are two objects involved, each with potentially different acceleration. Which object's acceleration are you asking about?
     
  4. Oct 19, 2012 #3
    I meant that the board is pulled with a constant force and therefore a constant acceleration because force and acceleration are both proportional. Well for the box its zero right? But for the board, is the acceleration constant too? it is being pulled. or is the acceleration for the box and board zero because they were at rest?

    also are my body diagrams correct?
     
  5. Oct 19, 2012 #4

    SammyS

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    Yes.

    Isn't there a net force (on the box) to the right?
    Your diagrams look good enough to answer this question. I'm not sure what all of your symbols mean.
     
    Last edited: Oct 19, 2012
  6. Oct 19, 2012 #5
    Sorry, I labeled the box A with mass m1 and board is B with mass m2. N is my normal force, w is my weight, fs is my static friction force in the opposite direction of motion, and F is my force towards motion. When I do the body diagram for the board do i have to account the weight of the box? or is that only if i do a system body diagram? I trying to solve this by doing individual body diagrams.
     
    Last edited: Oct 19, 2012
  7. Oct 19, 2012 #6

    SammyS

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    What you have in your diagrams looks correct to me.

    Added in Edit:
    What is the maximum force which can be exerted on the Box?
     
    Last edited: Oct 19, 2012
  8. Oct 19, 2012 #7
    Ʃ
    Well I get the correct answer when I do not include my normal force of a on b for the board. But I can't figure it out when I am including it. Why?

    Here is my work:

    the box is just m1=1
    board is m2=2

    For the box
    ƩFy: N(2on1) -W(1)=0 nothing going on in the verical
    ƩFx: fs(2on1) =m(1)*ax

    This turns out to be us*g=ax I substituted my n from my summation of forces in my y to my n from my summation of forces of x on fs(2on1)...fs=us*n


    For the board
    ƩFy: N(2)-W(2)-N(1on2)=0 Nothing going on in the vertical
    ƩFx: F-fs(1on2)=m2*ax

    What am I doing wrong here?
     
  9. Oct 19, 2012 #8

    SammyS

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    You're doing nothing wrong if you're analyzing the situation of the maximum force, F, which can be applied without slipping.

    For a force, F+, just very slightly greater than this F, the frictional force (kinetic friction) will be μkm1g, and a1 ≠ a2 . μk < μs .
     
  10. Oct 19, 2012 #9
    So i am doing is finding the max force but I need the minimum force. How would I change things differentl?. I know fs is less than or equal to us*n but how would I apply that in this case?
     
  11. Oct 19, 2012 #10
    I am only stuck in the part where I need to substite my n in my fs=us*n for the board.. I want to know why is it that when I take n1on2 out of my body diagram i can get the answer. But it should be there because of newtons third law.
     
  12. Oct 19, 2012 #11

    SammyS

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    Well, you found the greatest force, for which the book doesn't slide on the board. the minimum force for which it will slide on the board is infinitesimally larger than the max force with no slipping --- so they are essentially the same value.

    The big difference is that the frictional force produced by kinetic friction will be slightly less than the maximum that can be produced by static friction. So the box will have a little bit smaller acceleration.

    What about the acceleration of the board?
     
  13. Oct 19, 2012 #12
    That is where I get stuck. So i am not sure if I have my body diagram correct. I have this for the board: F-fs1on2=m*ax
    F-us*n=m*ax. What would go in my n according to my body diagram? I have n2-n1on2-w2=0 along my y forces.

    as far as what you said about kinetic friction, do I include that in there? Since the board will be sliding or no?
     
  14. Oct 19, 2012 #13
    i got most of it now, thanks. But why is it that you don't need kinetic friction when it slides?
     
  15. Oct 19, 2012 #14

    SammyS

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    Well, yes the box slides on the board, and you should be using kinetic friction.

    But acceleration, a1, for the box, is not the same as acceleration, a2, for the board.
     
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