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Help with torque problem please

  1. Nov 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A uniform horizontal rod of mass 2.8 kg and
    length 2.1 m is free to pivot about one end
    as shown. The moment of inertia of the rod
    about an axis perpendicular to the rod and
    through the center of mass is given by I =(m ℓ^2) /12

    If a 4.8 N force at an angle of 63◦
    to the horizontal acts on the rod as shown, what is the
    magnitude of the resulting angular acceleration about the pivot point? The acceleration
    of gravity is 9.8 m/s^2
    .
    Answer in units of rad/s^2


    2. Relevant equations
    Sum of the torques = I * a
    Torque = R * F * sin(theta)

    3. The attempt at a solution

    Here is what I have so far:

    (sum of the torques) = (Torque of force F)
    I * a = R * F * sin(63)
    a = (R * F * sin(63)) / I

    I = (M * L^2)/3 (It is rotating at its endpoint not at its center of mass)

    a = (3 *(R * F * sin(63))) / (M * L^2)
    a = (3 *(2.1 * 4.8 * sin(63))) / (2.8 * 2.1^2)
    a = 2.1821 rad/s^2 <-----But this answer is incorrect. Where did I go wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 20, 2013 #2
    Sorry if you came to this but I just figured it out. I forgot the torque due to gravity
     
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