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B Force acting on a seesaw?

  1. Jul 22, 2017 #1
    Here's a question that I've been trying to solve for a while but keep on running into dead ends, and I can't seem to find any info on the internet to help me. Anyways I was wondering what the integral of torque is? For my specific example I have a rod that is not equally balanced on a fulcrum (as shown on in the picture below)

    EKtjN.png
    Now gravity would of course pull the rod down but my question is with what force?? I know that Tnet = Iα

    But my problem here is finding α, since α=a/r could I substitute that in and say that Tnet = mr2 (a/r) = mar , and then just integrate as if I were finding the moment of inertia ( a∫r dm). I know this involves the density of the rod, but the example shown in the picture isn't all that important, I'm looking for an explanation on how to find the net torque caused by gravity. Thanks!
     
  2. jcsd
  3. Jul 22, 2017 #2
    Find the torque produced by gravity on a differential segment of the rod, then integrate from end to end.
     
  4. Jul 27, 2017 #3

    CWatters

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    Science Advisor
    Homework Helper

    Lots of ways to analyse this. One way is to model the rod as two short rods one either side of the pivot. Then model each rod as a mass acting at that rods centre of gravity. Sum the torques to give the net torque. Calculate the moment of inertia and hence the angular acceleration.
     
  5. Jul 27, 2017 #4

    Doc Al

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    Staff: Mentor

    Finding the net torque should be easy, since gravity can be treated as acting at the rod's center of mass. (You can also, as CWatters suggests, model it as two short rods. Same answer, of course.)
     
  6. Jul 29, 2017 #5
    I guess the easiest way to find the torque is to first find the center of gravity, if the density and shape is uniform, than that's just the middle point. Then just apply torque T = F*r, whereby F the gravitational force on the rod (mass times g) and r the distance between the fulcrum and the center of gravity. Then you can compute alfa by T/I...
     
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