Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question on torue(center of mass being important I think)

  1. Nov 30, 2011 #1
    Question on torque(center of mass being important I think)

    1. The problem statement, all variables and given/known data

    I have a lab report due tomorrow, and I have it all done except one calculation, but it confuses me.

    The center of mass on my meter stick on a stand is 50.1 cm, or .501 meters.

    For part B, we changed the pivot point to 30.0 cm(not the center of mass), or .300 meters. then with 3 mass hanging by a rope are added. I understand that torque is force times radius.

    The question is what OTHER forces are acting on the system besides the weights we added on, and then find the torque, mass, radius, and distance.

    I suppose this is referring to the force of gravity *mass of meter stick with the distance being .501 - .300 = .201 meters. The torque would then be .164675 Nm. Is this right?

    Please help. Thanks.

    2. Relevant equations

    sine = 1
    radius * force = torque

    3. The attempt at a solution
    Last edited: Nov 30, 2011
  2. jcsd
  3. Nov 30, 2011 #2


    User Avatar
    Gold Member

    you didn't give any masses. you should also probably be considering the moment of inertia of the meter stick.
  4. Nov 30, 2011 #3
    Sorry, the mass of the meter stick is 0.0836 kg.

    Well, I don't understand how inertia would tie in with finding the torque of the other forces besides the weights on the tension.

    Wouldn't it just be mass of meter stick, the distance between the pivot and the center of mass, then to find the force, just 0.0836 kg * 9.8 m/s^2, and then torque is radius*force?
  5. Nov 30, 2011 #4


    User Avatar
    Gold Member

    Hrmm... that means... is the system in equilibrium (no motion and not sitting with one end touching the floor)? If so, wouldn't the torque on either side of the fulcrum be the same?
  6. Nov 30, 2011 #5
    its not in static equilibrium, but its not touching the floor because the ruler is on a stand.

    I'm suppose to find the "other forces"(not the weights I added) before I calculate the static equilibrium.

    So I need the torque value 1 + torque value 2 + other forces.
    There are two weights on the system. The center of mass is not the pivot point.
    This would be torque 0.309 + -0.434 + (other forces) <===this is where i need help.
    Last edited: Nov 30, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook