1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Friction between objects

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

    Please see attached

    2. Relevant equations

    Ff=μFn,
    F=ma

    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

    CAF123

    User Avatar
    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?)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook