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Two blocks pushed against wall

  1. Apr 13, 2015 #1
    1. The problem statement, all variables and given/known data;Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is
    2BLOCKS.png



    2. Relevant equations:
    f 1(static friction force acting on block A)=20 N=mass of A multiplied by g.


    3. The attempt at a solution: f2.png
    Here f 2=f1+ m g B= 120 N
    The only thing I don't understand is why f1 is acting on block B?
     
  2. jcsd
  3. Apr 13, 2015 #2

    haruspex

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    Because action and reaction are equal and opposite. If B is exerting a force f1 up on A, A must be exerting a force of the same magnitude down on B.
     
  4. Apr 13, 2015 #3

    jbriggs444

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    The coefficient of [static] friction is a limit on how high the frictional force can go before the objects that are in contact start slipping. It may or may not be the applicable limit in this case. First one needs to decide whether the blocks are sliding or are sticking.

    A coefficient of friction [static or dynamic] relates the tangential force of friction to the perpendicular force with which the surfaces are pressing on one another. This perpendicular force is usually called the "normal force". The word "normal" in this case means "perpendicular", not "ordinary".

    With this in mind, what is the normal force between block B and the wall?
     
  5. Apr 13, 2015 #4
    Oh,this I know of course.:smile:
     
  6. Apr 13, 2015 #5
    But ,f1 is not exerted by A alone,it's a force of friction.It is due to contact between both A and B.Newton's third law will be applicable here also?
     
  7. Apr 13, 2015 #6

    haruspex

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    As far as A is concerned, it is just an external force exerted on it. Doesn't matter whether it's friction or, or a nail, or magnetic...
    Similarly for B. Yes the action and reaction law applies.
     
  8. Apr 13, 2015 #7
    Oh,this I didn't know of course.:smile:
     
  9. Apr 13, 2015 #8
    Thanks jbrigs444 and haruspex .
     
  10. Apr 14, 2015 #9
    Then f 1(static friction force acting on block A) should also be =f2 +mA multiplied by g
     
  11. Apr 14, 2015 #10
    But if it is true we will not get the same answer for f2 i.e 120N.
     
  12. Apr 14, 2015 #11

    jbriggs444

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    Why?
     
  13. Apr 14, 2015 #12
    For the same reason i.e Newton's third law.
     
  14. Apr 14, 2015 #13

    jbriggs444

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    Show your work. Why does Newton's third law require that f2 = f1 + mAg?
     
  15. Apr 14, 2015 #14
    Because of newton's third law f2 will also act on block A But in downward direction because f2 is acting on B upward.So for equilibrium of block A.
    So f1=f2+mA g
     
  16. Apr 14, 2015 #15

    jbriggs444

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    That is a completely incorrect understanding of Newton's third law. F2 acts between block B and on the wall. It does not act on block A at all.
     
  17. Apr 14, 2015 #16
    Oh,sorry.I misunderstood f2.My bad.I thought f2 is force of friction between the two blocks experienced by Block A.
    I labeled that f2 =friction between wall and block B but forgot.Really sorry.
     
    Last edited: Apr 14, 2015
  18. Apr 14, 2015 #17
    Even this is wrong.But now I am clear .I know force of friction between the two blocks experienced by Block A is f1 acting in downward direction.
     
  19. Apr 14, 2015 #18
    My ideas and thoughts are all jumbling.I think,I should sleep now.
     
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