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Tipping a crate

  1. Jun 3, 2009 #1
    A worker wants to turn over a uniform 1010 N rectangular crate by pulling at 53.0 degrees on one of its vertical sides. The floor is rough enough to prevent the crate from slipping.

    1. What pull is needed to just start the crate to tip?
    2. How hard does the floor push on the crate?
    3. Find the friction force on the crate.
    4. What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

    1. torque = (F_pull)cos(270-53)(1.5m) = 1010 * 1.5 = 1515 N
    the problem is getting started. i dont think i did this right.
  2. jcsd
  3. Jun 3, 2009 #2


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    Homework Helper

    . torque = (F_pull)cos(270-53)(1.5m) = 1010 * 1.5 = 1515 N
    This is the counterclockwise torque. What is the clockwise torque?
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