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The arm of a crane at a construction site is 15.0 m long

  1. Oct 19, 2011 #1
    1. The problem statement, all variables and given/known data
    The arm of a crane at a construction site is
    15.0 m long, and it makes an angle of 15.8
    with the horizontal. Assume that the max-
    imum load the crane can handle is limited
    by the amount of torque the load produces
    around the base of the arm.
    What maximum torque can the crane with-
    stand if the maximum load the crane can
    handle is 779 N?
    Answer in units of N · m

    2. Relevant equations
    torque = rFsin θ

    3. The attempt at a solution
    torque = radius * Force of gravity * sin θ
    = 15 m * 779 N * sin 15.8
    I don't get what is wrong...
     
  2. jcsd
  3. Oct 19, 2011 #2

    Delphi51

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    Re: Torque

    "torque = 15 m * 779 N * sin 15.8"

    Surely it should be cosine rather than sine?
     
  4. Oct 19, 2011 #3
    Re: Torque

    uhhh can i ask why it's cosine?
     
  5. Oct 19, 2011 #4

    Delphi51

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    Homework Helper

    Re: Torque

    Torque is the vertical force times the horizontal distance from the pivot. The angle of the crane with horizontal is given. So 15*cos(15.8) is the horizontal component of the crane's length.
     
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