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Frictional force

  1. May 5, 2006 #1
    An 18.8kg box is released on a 38.0o incline and accelerates down the incline at 0.281m/s2. What is the magnitude of the friction force impeding its motion.

    F = m*g*cos(38)
    sum forces parallel to the plane
    m*g*sin(38) - (mu)*(F) = m*a, or
    m*g*sin(38) - (mu)*m*g*cos(38) = m*a
    masses cancel out
    [9.81*sin(38)-(.281)]/[9.81*cos(38)] = .75 = mu

    but it says i'm wrong. i cant be.

    i only have one try left, and that's it.
     
  2. jcsd
  3. May 5, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    All you are asked to find is the friction force. No need to compute the normal force (what you call F, for some reason) or find mu. Rewrite your equation for forces parallel to the plane using "F" to represent the friction force.
     
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