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Friction problem of two blocks

  1. Nov 1, 2004 #1


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    Figure 6-36: http://www.webassign.net/walker/06-36alt.gif

    5. [Walker2 6.P.071.] Two blocks, stacked one on top of the other, slide on a frictionless, horizontal surface (Figure 6-36), where M = 5.0 kg. The surface between the two blocks is rough, however, with a coefficient of static friction equal to 0.47.

    If a horizontal force F is applied to the bottom block, what is the maximum value F can have before the 2.0 kg top block begins to slip?

    The answer in the back of the book is 32 N. But that's not what I get. I get the same wrong answer 2 different ways:

    friction = mu * mg
    friction = 0.47 * 2 * 9.81
    friction = 9.2214 N

    at this point, I'm stuck.

    I'd like to say that if pushing the 2 kg block with a force of 9.2214 will move it, then how much harder would I have to push a 5 kg block to make its push equivalent to the 2 kg push.

    9.2214 * (5/2) = 23.0535 which is the wrong answer.

    Or I could do it like this:

    9.2214 = 2 * a

    a = 4.6107

    now use this for a for the 5 kg block

    F = ma
    F = 5 * 4.6107
    F = 23.0535 which is exactly what I got before and it is wrong

    Any thoughts...??
  2. jcsd
  3. Nov 1, 2004 #2

    Doc Al

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    You are very close. What is the maximum acceleration that the top block can have? (You already figured that out--that's when the frictional force on it is at maximum.) Now treat the two blocks as a single object (why not?). What force F is required to accelerate both blocks to that value?

    (One of your mistakes was using F = ma, but not using the net force on the object. If you treat the bottom block by itself -- nothing wrong with that! -- don't forget that there are two horizontal forces on it. Recall Newton's 3rd law.)
  4. Nov 1, 2004 #3


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    Thanks, Doc (again!)
    lol... I feel so stupid for not realizing that I'm pushing 7kg, and not 5 !!! I stared at this for an hour.
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