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Block Friction

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data

    A block of mass [tex] M_1[/tex] rests on a block of mass [tex] M_2[/tex] which lies on a frictionless table. The coefficient of friction between the blocks is [tex] \mu [/tex]. What is the maximum horizontal force which can be applied to the lower ([tex] M_2 [/tex]) block for the blocks to accelerate without slipping on one another?

    2. Relevant equations


    3. The attempt at a solution
    The acceleration of the two blocks (assuming they're not slipping) is [tex] a = \frac{F}{M_1 + M_2} [/tex], and you want the upper block ([tex]M_1[/tex]) to not slip, that is, the acceleration times [tex] M_1 [/tex] must be less than or equal to the frictional force. When the blocks start slipping, [tex] M_1 a = \mu M_1 g [/tex] where the frictional force holding the upper block is [tex] f = \mu M_1 g [/tex]. This means that [tex] a = \frac{F_{max}}{M_1 + M_2} = \mu g [/tex], or [tex]F_{max} = \mu g (M_1 + M_2) [/tex].

    I'm not sure if this answer is right, but it makes intuitive sense when looking at the final equation. Thanks for any help!
     
  2. jcsd
  3. Jun 16, 2009 #2
    That is correct. :) Just to set your mind at ease, here's what you said in a more formal form:

    The maximum force that static friction can enact on the upper block, and the corresponding acceleration at which it would travel are:
    fs max = µs*M1*g --> amaxsg

    Now we have our maximal acceleration. We see what the form of the acceleration is for the two blocks moving in tandem, equate the two, and find out Fmax.

    Since they are moving in tandem, we can refer to them as one block of mass M1+M2

    a = F/(M1+M2) = amax = µsg

    Fmax = µsg(M1+M2) //
     
  4. Jun 17, 2009 #3
    Thank you RoyalCat. Here's the catch: The problem originally came in two parts, the first part being that a force F is applied to the upper block, all other components are the same. Intuitively I saw [tex]F_{max} = \mu M_1 g [/tex] in this new situation (both blocks sitting on a frictionless table, recall).

    The text gives [tex] M_1 = 4, M_2 = 5 [/tex] (M1 on top), and says that a force of 27 N applied to the lower block causes them to slip. Thus, from the original formula [tex] F_{max} = \mu g (M_1 + M_2) [/tex] I calculated the appropriate [tex] \mu = 0.306 [/tex]. Then it asks for the maximum force which applied to the upper block, so plugging in [tex] \mu = 0.306 [/tex] should give the text's answer. [tex] F_{max} = (0.306)(9.8)(4) = 12 N [/tex] but the "correct answer" is 21.6 N.

    Either my formula for the top-block force is wrong, we're both wrong for the bottom formula, or the text's answer is wrong, or a combination! What do you think?
     
  5. Jun 17, 2009 #4
    Oh, I see it now.
    If you're pushing the top block at just enough force so that it doesn't slip, what forces are acting on the bottom block, and in what acceleration do they result?
    Now remember that the acceleration of the bottom block is shared by the top block, and since they are moving in tandem, the force acting on them needs to provide the acceleration for both of them.
     
  6. Jun 17, 2009 #5
    Right, that was pretty silly of me. Thanks again - answer makes sense now.

    [tex] F_{max} = (M_1 + M_2)a [/tex]
    [tex] M_2 a = \mu g M_1 \implies a = \mu g \frac{M_1}{M_2} [/tex]
    [tex] F_{max} = (M_1 + M_2) \mu g \frac{M_1}{M_2} = 21.6N [/tex]
     
  7. Jun 17, 2009 #6
    You're welcome. :) It didn't come intuitively to me either, heh.
     
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