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Force required to keep block from sliding down (check)

  1. Sep 26, 2012 #1
    1. The problem statement, all variables and given/known data
    The two blocks (with m=16kg and M=88kg) shown in Fig. 6-56 are not attached. The coefficient of static friction between the blocks is [itex]\mu[/itex]stat=0.38, but the surface beneath the larger block is frictionless. what is the minimum magnitude of the horizontal force [itex]\vec{F}[/itex] required to keep the smaller block from slipping down the larger block?

    Fig. 6-56 looks approximately like this.

    ....______ ________
    ....|.......||...........|
    [itex]\vec{F}[/itex].|..m...||...........|
    ....|_____||.....M...|
    ..............|...........|
    ..............|_______|
    _________________________________
    FRICTIONLESS GROUND

    2. Relevant equations
    [itex]\sum[/itex][itex]\vec{F}[/itex]=m[itex]\vec{a}[/itex]

    [itex]\vec{F}[/itex]fric=[itex]\mu[/itex]stat[itex]\vec{N}[/itex]


    3. The attempt at a solution
    I denote [itex]\vec{F}[/itex] from the problem text as [itex]\vec{F}[/itex]A

    I denote the force of block m and block M, [itex]\vec{F}[/itex]mM

    Likewise, the force of block M and block m is denoted [itex]\vec{F}[/itex]Mm. I treat this as the "normal" force for the friction equation, but I'm not sure if this is correct.

    Fgravm=mg=156.8 N

    [itex]\vec{a}[/itex]=[itex]\vec{F}[/itex]A/104


    [itex]\vec{F}[/itex]mM=[itex]\vec{F}[/itex]Mm=[itex]\frac{88}{104}[/itex][itex]\vec{F}[/itex]A

    [itex]\vec{F}[/itex]fric=[itex]\mu[/itex][itex]\vec{F}[/itex]Mm=.38*[itex]\frac{88}{104}[/itex][itex]\vec{F}[/itex]A=.322[itex]\vec{F}[/itex]A

    To prevent downward motion [itex]\vec{F}[/itex]fric must cancel out (i.e. be at least) [itex]\vec{F}[/itex]gravm

    [itex]\vec{F}[/itex]fric=[itex]\vec{F}[/itex]gravm
    .322[itex]\vec{F}[/itex]A=156.8N
    [itex]\vec{F}[/itex]A=487.7 N

    Final answer 487.7 N

    Thanks
     
  2. jcsd
  3. Sep 26, 2012 #2

    ehild

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    The result is correct. But in case of static friction, Ffric≤μFMm.
    At the same time Ffric=mg.

    ehild
     
  4. Sep 26, 2012 #3
    correct
     
  5. Sep 26, 2012 #4

    PhanthomJay

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    Looks good! Might want to round your answer to 490 N. I am not particularly thrilled about using [itex] F_mM = m/(m+M)F_A [/itex], although it is correct. Free body diagrams and the use of Newton's laws avoid having to use this equation.
     
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