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A block with mass M rests on a frictionless surface and is connected to a horizontal spring of force constant k, the other end of which is attached to a wall. A second block with mass m rests on top of the first block. The coefficient of static friction between the blocks is mu*s. Find the maximum amplitude of oscillation such that the top block will not slip on the bottom block.

I did the exact same question as this only with numbers and got the right answer subbing my numbers into
A=mu(coefficient of static friction*g/w^2

So, i thought the answer would be
A=(mu*g*m)/k
But I was told that the correct answer does not depend on mu (the coefficient of static friction). So I'm guessing it must get cancelled out of the equation? I'm confused because I needed it in the first question when I actually subbed numbers in?

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asked by Jon

  1. Well, I can tell you the coefficent of friction does enter into the answer. Consider nofriction between the blocks, than any acceleration will make the top block slip. If the coefficent is infinite, as you nailed the two blocks together, then you could have a very large acceleration and they wouldn't slip.

    My concern here is what does Mu*s mean? does that mean the static coefficent of friction? If so, I stick with my above answer. If it means the static coefficent times some value s, then I need to know what s is.

    Let look at it.

    x= Asinwt
    v= Aw coswt
    a= -A w^2 sinwt

    So the max occurs at sinwt=1, or max acceleration is Aw^2

    Now, the max force on the top block is mu*mg, so acceleration <mu*mg/m
    acceleration<mu*g

    or Aw^2<mu*g
    but w is... sqrt (k/(M+m)
    Ak/2M<mu*g
    or A<mu*g*(M+m)/k

    m is the mass of the top box, M the bottom.

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    bobpursley

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