1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.

    ....______ ________

    2. Relevant equations


    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




    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]A=487.7 N

    Final answer 487.7 N

  2. jcsd
  3. Sep 26, 2012 #2


    User Avatar
    Homework Helper

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

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook