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Homework Help: Block on Block Problem

  1. Dec 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Two masses M1 = 4.00 kg and M2 = 7.10 kg are stacked on top of each other as shown in the figure. The static coefficient of friction between M1 and M2 is μs = 0.430. There is no friction between M2 and the surface below it.

    What is the maximum horizontal force that can be applied to M2 without M1 sliding relative to M2?

    2. Relevant equations


    Ffriction = μFnormal

    3. The attempt at a solution

    For the 4 kg box:

    Fnet = ma
    μmg = ma

    (0.43)(9.81) = a = 4.23 m/s/s

    I understand that this must be the acceleration for the whole system, but when I calculate the max force, do I do it for the whole system's mass or for the 7.10 kg box only? Why or why not?

    Attached Files:

    • IMG.png
      File size:
      3.8 KB
  2. jcsd
  3. Dec 24, 2013 #2
    Draw a free body diagram for the other box as well. Also, write the net force equations in the x and y directions.
  4. Dec 24, 2013 #3

    For the y-direction:

    ∑forces = 0 ; since the object is not accelerating in that direction

    For the x-direction:

    Fnet= Fapp

    Fapp=ma= (7.10)(4.23) = 16.88 Newtons

    But, when we are applying the force to the lower box, aren't we essentially pushing the whole system? Therefore, shouldn't the mass of the whole system be considered?

    Attached Files:

  5. Dec 24, 2013 #4
    You are missing one force on the FBD you just drew. And instead of F normal of the system, I suggest that you think it as F12 (i.e. force of mass 1 onto mass 2) Also the friction force on the FBD of your initial post should be pointing in the other direction.
  6. Dec 24, 2013 #5
    Is it the static friction of the top box?
  7. Dec 24, 2013 #6
  8. Dec 24, 2013 #7

    Fapp-Ff = ma

    Fapp - μmg = ma

    Fapp = ma + μmg

    = (7.1)(4.23) + (0.43)(4)(9.81) = 46.91 Newtons ?
  9. Dec 24, 2013 #8
    I don't know how to approach problems that deal with multiple masses. Frustrating.
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