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Force and motion

  1. Sep 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A small box (mass m1) is placed upon a large box (mass m2); the large box i placed on a horizontal table. The boxes begin a rest on the table. The lower box is hit by a horizontal blow at the time t=0 resulting in a initial velocity v0. The frictioncoefficient between all surfaces are µk
    2. Relevant equations
    a) Draw force-diagrams for both boxes as the move relative to each other. Indicate all Newtons-Third-Law-Pairs.

    b) Determine all velocities and accelerations of the boxes as the move relative to each other. At what time T do the boxes not move relative to each other anymore?

    c) Determine the acceleration of the boxes when they do not move relative to each other. What length L do the boxes move in this fase? The initial velocity can be concidered know as v1

    3. The attempt at a solution
    I started drawing the force diagrams. No problem there. It was easy to see that the normal force in both cases were pairing up with the gravitational force.

    For question b) I came up with this equation:
    vbox1=v0-a*t​
    Applied on the second box (the one underneath the other):

    The acceleration must be the kinetic friction force divided by the mass:
    vbox1=v0-(fk)/m*t​
    Naturally I got:
    vbox1=v0k*g*t​

    For the box on top:
    vbox2=-μk*g*t​

    For the acceleration I differentiate both:

    abox1=(vbox2)'=-g*µk
    abox2=(vbox2)'=-g*µk

    It is now that I am lost. I do not know how to find the time by which the boxes are not moving relatively to each other anymore...
    Picture of scenario:
    Udklip3.png
    Force Diagrams:
    Udklip2.png
    Udklip.png
     
  2. jcsd
  3. Sep 27, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    What are fk1 and fk2? The friction forces need more thought here.
    If you call it box 1, why do you say "second box"? Its acceleration is wrong because you are missing a force acting on it.

    For the box on top, be careful with the sign. In which direction will it accelerate?
     
  4. Sep 27, 2015 #3

    Doc Al

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    Staff: Mentor

    Careful: The normal force and the gravitational force are not 3rd-law pairs.
     
  5. Sep 27, 2015 #4
    Oh, yeah! That's right! They are just equal and opposite...


    Do you mean that box 2 has a velocity of 0 at t=0 and it therefore because of the friction will accelerate in the same direction as the velocity of box 1?

    I understand that it is a bit confusing when refered to as both box 1 and 'second box' but they are the same

    fk1 and fk2 should be the same - I realise that now
     
  6. Sep 27, 2015 #5

    mfb

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    2016 Award

    Staff: Mentor

    For example, yes.

    They are not, no matter what they represent (which is still undefined).
     
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