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Friction Force on 2 blocks on a frictionless surface...

  1. Jan 23, 2017 #1
    1. The problem statement, all variables and given/known data
    The two blocks ##(m=16 kg)## and ##(M=88 kg )## , are not attached to each other.The coefficient of static friction between the blocks is ##μ=0.33##, but the surface beneath the larger block is frictionless.What is the minimum magnitude of the horizontal force ##\vec F## required to keep the smaller block from slipping down the larger block ?

    2. Relevant equations
    ##\vec {F_t}=m\vec a##
    ##F_f=μN##


    3. The attempt at a solution

    I made a free body diagram for m and M.Picture in attached files.
    So I cannot find ##F_{mM}## or ##F_{Mm}## And ıf this was true M should move right but we dont know acceleration of ##M##. Is there some other force that I dont know ?

    Thanks
     

    Attached Files:

    Last edited by a moderator: Jan 23, 2017
  2. jcsd
  3. Jan 23, 2017 #2

    PeroK

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    1) If you apply a horizontal force to ##m##, what will happen to the system?

    2) What would happen if the larger block were fixed to the ground? Could you calculate the required horizontal force on ##m## in this case?
     
  4. Jan 23, 2017 #3
    What information do you know when the small block is right at the point of slipping - but does not slip downward?

    Edit: Sorry PeroK. (Stupid autocorrect.) I should have left it alone but didn't notice you posted until after I posted.
     
  5. Jan 23, 2017 #4
    1)I guess system will move ?

    2)Required horizontal force for what ? I didnt understand ohh If u mean what I mean then ##F=\frac {mg} {μ}##
     
  6. Jan 23, 2017 #5

    PeroK

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    Can you put those two things together?

    Hint: in the case where ##M## moves why is most of the applied force transmitted though ##m## to ##M##?
     
  7. Jan 23, 2017 #6
    Well the friction force should be equal to Weight.The friction force depends the ##N## and that depends the ##F## and I guess ##F_{Mm}##
     
  8. Jan 23, 2017 #7
    Cause M has bigger mass ?

    well I can I guess but I need acceleration I guess which I dont have that info ##(M+m)a=\frac {mg} {μ}##
     
  9. Jan 23, 2017 #8

    PeroK

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    Even if a number in not given in the problem, you can still use that quantity. In this case, you can still consider the acceleration. It might tell you something about the forces involved. So:

    ##a = ?##
     
  10. Jan 23, 2017 #9
    I guess I see your point wait I ll solve I ll find a then from there I ll find ##F_{Mm}## then I ll find ##F## right ?
     
  11. Jan 23, 2017 #10

    PeroK

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    That's the idea.
     
  12. Jan 23, 2017 #11
    I solved it thanks
     
  13. Jan 23, 2017 #12
    Or wait I have to ask something.I am so tired I thought I can figure it out but I have to ask.

    ##F_{mM}=Ma## isnt this correct cause according to the question and my tryings it should be ##ma##.

    Why ##ma## but not ##Ma## ?
     
  14. Jan 23, 2017 #13

    PeroK

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    There's more to it than that.

    One observation is that you seem to be trying to jump straight to the answer from half way through. This is not an easy problem and it has maybe 4-5 calculations in it. There's no way to jump to the answer.

    Perhaps you've started to do physics now at a slightly higher level. One aspect of harder problems is that you have a longer chain of calculations to get the answer.

    Anyway, in this case, the first step is to express ##a## in terms of ##F, M## and ##m##.
     
  15. Jan 23, 2017 #14
    I made the all steps,like making a referance frame drawing forces and finding a and then N ok I ll type them just wait a sec
     
  16. Jan 23, 2017 #15
    I guess theres something wrong in our equation.

    we said ##(M+m)a=\frac {mg} {μ}## This is also equal ##F## (that we said),so we can find easily a here cause we know M and m and g and μ.
    And If F=(M+m)a which its something like 476 which thats not the answer ??

    I am so tired.I dont know this is true or not but I ll go to sleep
     
  17. Jan 23, 2017 #16

    Doc Al

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    I would attack this problem in this order:
    (1) What friction force is required between the blocks?
    (2) What normal force (between the blocks) is thus needed?
    (3) What acceleration is thus produced?

    And so on.
     
  18. Jan 24, 2017 #17
    (1)##F=μN##
    (2)well I guess ##ma=F-N-F_{Mm}## from there ##N## equals to zero I guess but its wrong so ##a##should be zero I guess ??
    or ##N=F-F_{Mm}## but this means theres no acceleration.
    (3)##F=(M+m)a##
     
  19. Jan 24, 2017 #18

    Doc Al

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    What must the friction force equal so that the block does not slip down?
     
  20. Jan 24, 2017 #19
    weight which is ##mg##
     
  21. Jan 24, 2017 #20

    Doc Al

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    Right. Now answer my question #2.
     
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