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QUICK torque and static equilibrium

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A uniform beam of length 1.0m and mass 10 kg is attached to a wall by a cable (the cable makes a 30 degree angle with the beam). The beam is free to pivot at the point where it attaches to the wall. What is the tension in the cable?

    2. Relevant equations
    F(net) = F(beam) - F(cable) = 0
    torque = [F(beam) x d(beam)]+[F(cable) x d(cable)]

    3. The attempt at a solution
    I think these equations are correct. I was able to figure out the torque of the beam (10kg x 9.8m/s2 x 1.0m) but I am having trouble figuring out what to do with the cable. How and where do I use the angle? Do I use it to find the distance of the cable itself? or its distance up the wall?
  2. jcsd
  3. Nov 12, 2008 #2
    Ey mate hope this helps havent touched this for a while now

    your equations look correct

    F(net) = F(beam) - F(cable) = 0
    torque = [F(beam) x d(beam)]+[F(cable) x d(cable)]

    Now by the looks you can work out what to do once you have your torque due to the weight of the beam.

    First off (you didnt mention where the centre of mass of the beam was*, generally its in the middle). Now what you have done is correct as long as your center of mass is at the end of the beam which is unlikely.

    You can work out the force(beam), which is its weight, the distance at which it acts on the beam from the pivit point. Now **You** want the torque around this point to be zero** thus let torque equal zero and solve using your torque equation.

    Now im assuming the cable is attached to the end of the beam, thus you know all your variables besides F(cable) pulling up. Thus you should be able to solve for this force and find the required value.

    Once you have this the rest should fall into place

    KEY*** torque is equal to zero about the pivit point thus you can solve for F(wire) knowing the other varibales

    Hope this helps Trent (sorry about it being long, justifying it to myself along the way)
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