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Friction between two stacked blocks

  1. Nov 25, 2010 #1
    Block A rests on block B, which lies on a flat frictionless surface. There is friction only between the two blocks. If I were to exert on block A a horizontal force of magnitude less than that of the maximum static frictional force between the two blocks so that for block A Fnet,x=0, then block B would receive an equal frictional reaction force in the opposite direction. That block A would remain stationary while block B would accelerate in the direction of the force is clearly counterintuitive. Can someone please point out the flaw in my thinking?

  2. jcsd
  3. Nov 25, 2010 #2
    I'm not exactly sure if this is the answer you're looking for, but I'll give it a go anyways.

    If the force F you exert on A is smaller than the static friction between the two blocks, they will not move relatively to each other.
    Therefore you can consider the two blocks as being one block with the total mass M = m_A + m_B. They will move forward with constant acceleration a = F/M
  4. Nov 25, 2010 #3
    Hi mr.physics

    you have two equations governing the motion parallel to the surface
    [tex]m_A a_A = F - f_{fr}[/tex]
    [tex]m_B a_B = f_{fr}[/tex]
    with an additional constraint - if we are to have static friction- that the two blocks are at rest relatively to each other.
    Assuming they both start at rest, this requirement imposes the same acceleration on them:
    [tex]\frac{F - f_{fr}}{m_A}=\frac{f_{fr}}{m_B}[/tex]
    [tex]F=f_{fr}(1+\frac{m_A}{m_B}) \leq \mu_S m_A g (1+\frac{m_A}{m_B}) [/tex]
    which is the maximum force that can be applied to A for static friction to hold the two blocks together

    Please double-check and let me know if this is convincing
  5. Nov 25, 2010 #4
    To me it looks OK yes...

    It DOES seem a little weird though. Wouldn't you normally just think that the force F applied just has to be smaller than the maximum possible static friction which is f_statis = mu_s * mass_a * g ?
  6. Nov 25, 2010 #5
    I agree this is not very intuitive but the result seems to have the right limit behavior so I guess this is one of those cases were math helps.
  7. Nov 25, 2010 #6
    Hmm well... if you look at block A alone and consider block B as being a table:
    Force equilibrium (at max static friction): F - mu_s * m * g = 0.
    The moment F gets bigger than mu_s * m * g, there is no longer force equilibrium, thus the block A will accelerate relatively to the table (or block B).

    So....?! Both my physical arguments and your mathematical arguments seem to be right. What IS right then :P ?
  8. Nov 25, 2010 #7
    Hi doubleyou

    I am not sure I am getting your counter-example: if the block B is very massive (is that what you mean by table?) from
    [tex]F \leq \mu_S m_A g (1+ m_A / m_B)[/tex]
    one gets the usual result
    [tex]F \leq \mu_S m_A g [/tex]

    What am I missing?
  9. Nov 26, 2010 #8


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    Getting back to the OP:

    Except that Fnet isn't zero. The pair of blocks are accelerating due to the force. For block A, Fneta = massa x acceleration = total_force - friction_force. Friction force between block A and B = total_force - Fneta = massb x acceleration = Fnetb.
    Last edited: Nov 26, 2010
  10. Nov 26, 2010 #9
    what is this adding to the previous replies?
  11. Nov 26, 2010 #10
    It doesn't really matter what the mass of B is. In my mind, block A would still start moving relatively to B, when the applied force F is bigger than the friction between the two blocks f_friction = mu_s * mass_a * g...... independant of the mass of B
  12. Nov 26, 2010 #11
    For the values of F when
    mu_Sm_A g < F < mu_S m_A g (1+(m_A)/(m_B) and a_A is not equal to a_B, so block A should be accelerating relative to block B (kinetic friction) and the blocks are not moving together. I think that from the way you have formulated your equations, a_A = a_B only in the specific scenario when F=mu_S m_A g (1+(m_A)/(m_B).

    For values of F when
    [tex] F \leq\ mu_Sm_A g [/tex], [tex] f_{fr} = F [/tex] so your equations indicate that [tex]m_A a_A = F - f_{fr} =0 [/tex] and if [tex] a_A [/tex] and [tex] a_B [/tex] are forcibly set equal because the friction is static, you should arrive at [tex]\frac{o}{m_A}=\frac{f_{fr}}{m_B}[/tex], an inconsistency.

    So I'm not sure [tex]F_{max} [/tex] <= [tex] mu_S m_A g (1+\frac{m_A}{m_B}) [/tex] are the correct values of F for which the two blocks don't slide relative to each other.
    Last edited: Nov 26, 2010
  13. Nov 27, 2010 #12
    The only conditions required for static friction to exist is
    - to have the interfaces moving at the same speed (be at rest wrt to each other)
    - for the friction force to be LESS THAN (not equal to) the product of static friction coefficient and normal reaction of the interface

    My equations simply answer the following question:
    under what conditions are the two blocks moving together (same acceleration, same initial conditions)? The answer of Newton's equations is
    [tex]F \leq \mu_S m_A g (1+\frac{m_A}{m_B}) [/tex]
    even if this is not intuitive, it is correct so you will have to attune your intuition to it
  14. Nov 27, 2010 #13
    Hi dgonphysics

    Sorry, I must still be misunderstanding something. If [tex]F \leq \mu_S m_A g (1+\frac{m_A}{m_B}) [/tex] there will be some values of F when it is greater than "the product of static friction coefficient and normal reaction of the interface". Right?
  15. Nov 27, 2010 #14


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    Correct. The maximum force applied to block A can be greater than the maximum static friction force since the friction force equals the maximum force minus the force it takes to accelerate block A. Using the OP terminology, as long as total_force - fneta is < maximum_static_friction_force, there won't be any slippage.
  16. Nov 27, 2010 #15
    Yes, using my symbols:
    - F is not a friction force but the force applied to block A
    - F can be larger than the maximum value of static friction, [tex] \mu_S m_A g [/tex]
    - [tex]f_{fr}[/tex] (the friction force) cannot be larger than [tex] \mu_S m_A g [/tex](and will not be as long as the quoted constraint on F is respected)
    - F can be larger than [tex]f_{fr}[/tex] because of acceleration
  17. Nov 27, 2010 #16
    But when F is greater than F(fr) doesn't that mean the friction is no longer static?
  18. Nov 27, 2010 #17
    No the maximum value for static friction (that is [tex]\mu_S N[/tex]) applies only to the friction force (the tangential component of the constraint force) not to the force applied to move the object
    If the object is not moving the applied force (or better its component parallel to the friction direction) and friction have the same magnitude but if the object is moving -as in your case since friction occurs with another moving body- they will differ by ma
  19. Nov 27, 2010 #18
    I think i'm beginning to understand what you are saying. If either of you has time can you post up an example in which you solve for the Fneta, Fnetb, and the friction force given some F greater than the maximum_static_friction_force and less than mu_S m_A g (1+(m_A)/(m_B)? Thanks
  20. Nov 27, 2010 #19
    So we have Newton's equations
    [tex]m_A a_A = F - f_{fr}[/tex]
    [tex]m_B a_B = f_{fr}[/tex]
    with the constraints
    [tex] a_A= a_B [/tex]
    [tex]f_{fr}\leq \mu_S m_A g[/tex]
    This solves to
    [tex]f_{fr}=\frac{F}{1+m_A/m_B} [/tex]
    with the constraint
    [tex]F \leq \mu_S m_A g (1+m_A/m_B) [/tex]
    The net forces are simply the force side of Newton's equation:
    [tex]F_{net,A} = F - f_{fr} = F \frac{m_A}{m_A+m_B}[/tex]
    [tex]F_{net,B} = f_{fr} = F \frac{m_B}{m_A+m_B}[/tex]

    F is a known force in the problem, just replace the numbers as you see fit and you will get your example. If you want to avoid numbers you can just use something like the average between the two extremes you want to explore:
    i.e. [tex]F= \mu_S m_A g (1 + \frac{m_A}{2 m_B})[/tex]
    and solve for whichever quantity you wish to calculate
    Last edited: Nov 27, 2010
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