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Newton's Law conceptual question

  1. Jun 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Two boxes of masses m and 3m are stacked.
    The surface between the more massive box and
    the horizontal surface is smooth and the surface between the boxes is rough. If the less massive box does not slide
    on the more massive box, what is t
    he static friction force on the less massive box?
    A) F
    B) F/2
    C) F/3
    D) F/4

    The answer is D.

    2. Relevant equations

    So F= (F/3)

    3. The attempt at a solution

    So F= (F/3)

    I keep getting C as the answer. However, the solutions manual claims that the answer is D (I also asked my professor and he too said it is) Can anyone explain to me why it is D? Thank you.
  2. jcsd
  3. Jun 16, 2014 #2
    You are applying force "F" on which block ? Is it not mentioned in the problem ?
  4. Jun 16, 2014 #3
  5. Jun 16, 2014 #4


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    Hello Steven, and welcome to PF.
    You forgot to tell us what F is, but I can guess.
    Did you make a drawing and a free body diagram for
    1) the assembly of the two boxes
    2) the less massive box on top by itself ?

    If you want to, you can also make one for the large box by itself. In the horizontal direction there is not only F and ma but yet another force in play !
    (And in the vertical direction there are two mg forces and N, the normal force).
  6. Jun 16, 2014 #5
  7. Jun 16, 2014 #6
    Isn't it just a=FM? So if the mass is 3M is it A=F/3M? Or do you do A=F/(Small box + Big box) so A=F/(3M+1M)?
  8. Jun 16, 2014 #7
    I said acceleration of the whole system. Thus we have,

    F=(3m+m)a => a=F/4m

    Note that even i apply force F anywhere, it will still be net force on whole system. (Just treat two blocks as one single block.)


    We switch to frame of small block kept on larger one. What is the pseudo force acting on that ?
  9. Jun 16, 2014 #8


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    If I push 2 boxes of mass m and 3m with an acceleration of 1 m/s^2 and they move as one, how much force am I exerting on the system?
  10. Jun 16, 2014 #9
    The static friction?
  11. Jun 16, 2014 #10
    Yes, the pseudo force will be equal and opposite to static friction in the frame of smaller block. Once you find the pseudo force, you have your answer. :)
  12. Jun 16, 2014 #11
    This is where I'm confused. How do I find the pseudo force?
  13. Jun 16, 2014 #12
    Ok, if you observe from ground you see that acceleration of both the block is F/4m. Now question states that smaller block does not slip on the bigger one. Hence if you stand on the bigger block, you'll see that you are in rest with respect to smaller one. But, there is a net force acting on it right, because it is accelerated w.r.t ground. So in the frame of small block you apply force "mass of smaller block times acceleration in it" opposite to the force F because in that frame it is at rest.
    The force which you apply just for the sake of frame change is the pseudo force.
  14. Jun 16, 2014 #13
    So what is the acceleration in it? Is it just 3 because of the bigger box?
  15. Jun 16, 2014 #14
    The acceleration in both the block is same and that is F/4m. Now pseudo force in the smaller box is Fp = m(mass of smaller block) * F/4m(acceleration)...

    This should equal and opposite to static friction if in this frame small block is not slipping. Will it not be ?
    Last edited: Jun 16, 2014
  16. Jun 16, 2014 #15
    Thank makes sense! Thank you very much!
  17. Jun 16, 2014 #16

    Andrew Mason

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    More to the point, what are you calling a pseudo force? The large mass exerts a real force (through static friction) on the smaller (this force = ma). The smaller mass exerts a real force on the larger mass of -ma. This reduces the net force on the larger mass to 4ma - ma = 3ma.

    There is no reason to analyse this in the accelerating frame of the masses. The forces should be analysed in an inertial frame of reference.

    Last edited: Jun 16, 2014
  18. Jun 16, 2014 #17


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    I don't see anything particularly complicated about this (but, then, I am not a physicist!). There is a total mass of 4m and a force of F is applied to it. The acceleration of the entire body is F/(4m). Since the object of mass m does not slip, it accelerates at F/4m. To accelerate a body of mass m and a= F/(4m) we must have a force of F= ma= mF/(4m)= F/4. There is no external force on the body of mass m so that must come from friction.

    As a check, in order to accelerate just the 3m object at a= F/4m, there must be a total force 3m(F/4m)= (3/4)F on it. That is, in fact, the original force, F, minus the F/4 friction force the upper object, of mass m, is applying (opposite to the F force) on it.

    (Now that I review, that is what Sankalpmittal said.)
    Last edited by a moderator: Jun 16, 2014
  19. Jun 16, 2014 #18
    Andrew Mason and HallsofIvy you both are correct.

    Andrew Mason, if you see a thing from the ground, you'll see that the total mass(the system) accelerates by F/4m and in the frame of the smaller block it is at rest relatively. You can also solve this problem by inertial frame (of what I did from accelerating frame). I think that is what you did.

    No one is WRONG here.

    P.S. I am not a physicist as such. I am just an ordinary high school pass out.
  20. Jun 16, 2014 #19


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    Pretty good for a high school pass out! As a general rule , though, pseudo forces should be used sparingly at the Intro level . Why use the pseudo force concept of F_net - ma = 0 when
    the simple F_net = ma will suffice here. They say the same thing, but sooner or later you'll use the pseudo force concept once too often, and it will sting you. Ouch!
  21. Jun 17, 2014 #20
    Ummm thanks for the first statement..

    And why will it sting ? Umm curious...
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