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Stacked blocks

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Imagine that there is a slab (call it block 1) on top of which sits a block (call it block 2). The slab connects to a cable that passed through a pulley and attaches to another block (call it block 3), hanging freely. Assuming that the coefficient of kinetic friction is the same for both blocks (and the same with static friction):

    a) Find the acceleration of the system.
    b) Verify the familiar limits.
    c) Under what conditions does block 2 stay on top of block 1?

    2. Relevant equations/The attempt at a solution

    If blocks 1 and 2 move together, then all three blocks have the same acceleration:

    [tex]a=\frac{m_3g-(m_1+m_2)g\mu_k}{(m_1+m_2+m_3}[/tex]
    If block 2 begins to slip, then we still take all the accelerations as the same (block 2 is just beginning to slip, so it is still pretty much moving with block 1 (i.e. same acceleration) and it's frictional coefficient will be static)

    TO MAKE A LONG STORY SHORT: DID I SET UP NEWTON'S 2ND LAW CORRECTLY?

    then Newtons 2nd Law for block 1, block 2 and block 3(respectfully) are:
    [tex]
    \sum(1,x): T+F_{f,s}-F_{f,k}=(m_1+m_2)a
    [/tex]
    [tex]
    \sum(1,y): F_{n,1}-(m_1+m_2)g=0
    [/tex]

    [tex]
    \sum(2,x): F_{f,s}=m_2a[/tex]
    [tex]\sum(2,y): F_{n,2}-m_2g=0[/tex]

    [tex]\sum(3,x): 0=0[/tex]
    [tex]\sum(3,y): m_3g-T=m_3a[/tex]

    Where the frictional forces take on the familiar value of the coefficient times the normal. Solving, I get:

    [tex]a=
    \frac{m_3g+\mu_sm_2g-\mu_kg(m_1+m_2)}{m_1+m_1+m_3}[/tex]


    For part b) while testing the familiar limits, i found that as [tex]m_3 \rightarrow \infty[/tex] then [tex]a=g[/tex], which makes sense. I found that if [tex]m_2=0[/tex], then we get what we expect. And when I calculate when [tex]m_2 \rightarrow \infty[/tex], [tex]a=0[/tex]. BUT, when I try to calculate as [tex]m_1\rightarrow \infty[/tex], [tex]a=\frac{-\infty}{\infty}[/tex], which doesn't make sense. Did I screw up Newtons 2nd law?

    For part c) all I did was use newton's 2nd law for the second body and solved for [tex]a[/tex], the maximum acceleration: [tex]a=\mu_sg[/tex], which doesn't make sense either since it should depend on mass.

    Thoughts?!?!
     
  2. jcsd
  3. Sep 9, 2009 #2

    kuruman

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    Gold Member

    Part (a) is OK. It is the same result as if blocks 1 and 2 were glued together to form a single block of mass m1+m2.

    Part (b) is not OK. The force that accelerates m2 (net force) is the force of static friction, fs. Newton's 2nd Law says that it must be equal to m2 times the acceleration from part (a). As you increase m3, you increase the acceleration which means that fs must increase accordingly. But fs cannot increase forever. At the point where fs can no longer increase, and ony then, can you say fs = μsN. You have all the ingredients to you can put it together from this point on.
     
  4. Sep 9, 2009 #3
    Thanks for the reply kuruman.

    So my equation for the acceleration is correct then? You say that I cannot include fs = μsN until the point in which the static friction can not keep up with the acceleration of blocks 1 and 3, so then it shouldn't be in my equation for the acceleration?

    What about part c)? Under what condition does block 2 stay on top of block 1? I understand that this only happens if fs = μsN, eg if μs=fs/N (or less), but in this equation what is fs?
     
  5. Sep 9, 2009 #4

    kuruman

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    The force of static friction fs is whatever is necessary to provide the observed acceleration. Because it is the net force on mass m2, you can write

    [tex]f_{s}=m_{2}a=m_{2}\frac{m_{3}g-\mu_{k}g(m_{1}+m_{2})}{m_{1}+m_{2}+m_{3}}[/tex]

    That is always true unless the top mass starts sliding. At the threshold of when sliding is just about going to happen, the force of static friction has reached its maximum value and can no longer "provide the observed acceleration" past that value. What does the above equation become then? What is the maximum value of the force of static friction?
     
  6. Sep 9, 2009 #5
    When:

    [tex]
    \mu_s\geq \frac{F_{f,s}}{F_n}=\frac{am_2}{g}\left(\frac{m_{3}g-\mu_{k}g(m_{1}+m_{2})}{m_{1}+m_{2}+m_{3}}\right)
    [/tex]

    ?

    Also what about the problem with negative acceleration when m_2 -> infinity?
     
  7. Sep 9, 2009 #6

    kuruman

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    Gold Member

    Start with

    [tex]f_{s}=m_{2}a \leq \mu_{s}F_{n}[/tex]

    Replace a and Fn with the appropriate quantities.
    What about it? Do you really think that if you make mass m2 large enough, say the size of a battleship, the masses will accelerate in the opposite direction? I didn't think so. What is more likely to happen instead? Remember, the acceleration you have has been derived assuming that the masses accelerate so that the hanging mass drops.
     
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