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Determine net torque

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data

    A 17m long ladder is mounted on a fire truck. The ladder itself has mass 130kg , and at the top is a 37kg basket holding a 83kg firefighter.

    If the ladder makes a 60∘ angle with the horizontal, what's the magnitude of the net torque about the ladder's base?

    2. Relevant equations

    I think center of mass (X1M1+X2M2+X3M3)/(m1+m2+m3)
    Torue=rFsinθ

    3. The attempt at a solution

    I get a huge number when i try it out. But since is is asking about the ladders base, would it the sine change to cosine?

    I take the torque from the mass at the end and the torque of the ladder to get the net torque. Kind of stuck.
     
    Last edited: Oct 16, 2013
  2. jcsd
  3. Oct 16, 2013 #2

    rude man

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    Never mind "center of mass of all the forces."

    You have the center of mass of 130 kg of the ladder at 8.5m from the pivot axis. You have 120 kg of mass at 17m from the pivot point. You have a 60 deg. angle with the horizontal.

    Compute the torques due to the ladder mass and the basket cum firefighter.
     
  4. Oct 16, 2013 #3
    Ok then so it is just

    Tnet=(8.5)(9.8)(130)Sin(60)+(17)(9.8)(120)Sin(60)
     
  5. Oct 27, 2013 #4

    rude man

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    Why sin?
     
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