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Homework Help: Friction between objects

  1. Aug 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Please see attached

    2. Relevant equations


    3. The attempt at a solution
    (a) The blocks can have the same acceleration due to static friction.
    consider the 2 blocks on the left: Ff=fmg

    then consider the 2 blocks on the right: Ff is also fmg? I am rather confused at this point... how does M come into the equation? because the answer for part (a) is 2(M+m)/(M+2m)f. There is no g anywhere in the equation, though from the equation, friction is to the normal, which is mg?

    THank you!

    Attached Files:

  2. jcsd
  3. Aug 24, 2012 #2
    The frictional force is given as fs. It is equivalent to mgμ.
    It is a system of 4 objects in unison, one identical acceleration.
  4. Aug 24, 2012 #3
    yeah okay, but how to do the question? Can give me a head start? Thanks!
  5. Aug 24, 2012 #4
    The static friction force created by small mass m on top of the right side will pull both itself and the 2 blocks behind it.
    As a body of 4 blocks there is only one the accelerations value.
  6. Aug 24, 2012 #5
    sorry, I don't really get what you mean, please elaborate... So for the small mass on the right, f value must be (m+M)f?
  7. Aug 24, 2012 #6


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    Gold Member

    If the blocks are all to move with the same acceleration then the small mass [itex] m [/itex] relative to the large mass [itex] M [/itex] must be stationary. I suggest drawing a free body diagram for the [itex] m, M [/itex] system, being careful to get the direction of the frictional force correct.
  8. Aug 24, 2012 #7
    Net external force applied on a system always equals to total mass of system into its acceleration .
  9. Aug 24, 2012 #8
    F=2(M+m)a .........(1)
    fs=(2m+M)a ........(2)
  10. Aug 24, 2012 #9
    Okay, I think I know...

    So the whole thing F=2(M+m)a .........(1) (as stated by azizlwl)
    the greatest friction is between the blocks on the left, so fs=(2m+M)a ........(2) (thanks azizlwl!), because the bottom block on the right need not be considered. So F/fs=2(M+m)/(2m+M), and F=2(M+m)f/(2m+M)

    Then for the 2nd part, F=2(M+m)a .........(1) still,
    then f is only for the block M on the right, thus f=Ma,
    by the same reasoning, F=2(M+m)f/M.

    Yay thanks guys! Any tips for force analysis (and consider which few objects as a system?)
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