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Compression Forces

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data
    The board sandwiched between the other two boards in Figure P4.77 weighs 95.5N. If the coefficient of friction between the boards is 0.663, what must be the magnitude of the compression forces (assumed to be horizontal) acting on both sides of the center board to keep it from slipping?

    https://www.physicsforums.com/attachment.php?attachmentid=7118&d=1149969727


    2. Relevant equations
    fs< or = usn
    F=ma


    3. The attempt at a solution

    So far, I've figured out that the friction forces must cancel out the weight of the board so I did:

    2F=95.5 N
    F=47.8N
    and from that I figured out the normal force is 72.1 N:
    47.8 N < or = 0.663n
    72.1 N < or = normal force

    I can't figure out how to solve for the compression forces. I assume there are two (one on each side), but I haven't a clue what my next step would be to solving it. Any input would be greatly appreciated.
     
  2. jcsd
  3. Sep 9, 2007 #2

    Doc Al

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    Staff: Mentor

    The normal force is the compression force.
     
  4. Sep 9, 2007 #3
    Can you explain why though? The normal force is a vertical force, whereas it asked me for compression forces which it told me to assume were horizontal.
     
  5. Sep 9, 2007 #4

    Doc Al

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    Staff: Mentor

    The normal force is perpendicular (normal) to the surfaces. Since the surfaces are vertical, the normal forces are horizontal in this case.

    The friction force is vertical; the normal force is horizontal.
     
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