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Calculating the net torque! help!

  1. Feb 16, 2008 #1
    Calculate the net torque about point for the two forces applied as in the figure below

    The rod and both forces are in the plane of the page.

    torque= ? Nm


    i know that F2 = 0 since its pointing through point 0, but for F1 I've so far got 16, 0, -16, -24... which were all wrong. I'm using the torque=F x L... which includes sin theta.

    please help me solve this question asap. thank-you
  2. jcsd
  3. Feb 16, 2008 #2
    F2 is acting on the rod at an angle, therefore there is a componant of the force acting through the pivot, and a componant acting vertically upwards.

    So first things first, you need to determine the upward componant of F2.

    Then you will have two forces acting on the rod at two distances, from which you can calculate the overall torque.
  4. Feb 16, 2008 #3
    ok. am i on the right track?
  5. Feb 16, 2008 #4
    Well, you're almost there. I think you flipped the distances round, 12*3 and 8*2. However, F1 is acting at a distance of 5 m (2 + 3) from the pivot.

    Also, you're adding the forces, but they aren't acting in the same direction.

    I think it would help you to follow things through if you were to resolve the F2 force to find it's vertical and horizontal componants, then re-draw the diagram given to you. Once you have this picture in front of you, you will find it a lot easier to see what's going on.
  6. Feb 16, 2008 #5
    i don't know how to find the x and y components to tell you the truth.

    so should my process be:
  7. Feb 16, 2008 #6
    OK, forget F1, F1 is fine as it is because it's only acting vertically, there is no horizontal componant to it.

    For F2 though, you just did find the vertical componant, that's what 12sin30 is. That is the force acting upward at a distance of 2m.

    Now bring back F1, which is a force of 8N acting at a distance of five metres from the pivot.

    If we assume the up direction as positive, that means we have a force of 12sin30 at 2m acting up, and a force of -8N at 5m acting up.
  8. Feb 16, 2008 #7
    oooh so it would be



    (i'm sorry im really not good at this)
  9. Feb 16, 2008 #8
    OK, crash course in resolving forces.

    You have 8N force acting at a 90 degree angle to the horizontal bar. Therefore the vertical componant of that force is 8 * sin90 = 8*1 = 8N. The horizontal componant of that force is 8 * cos90 = 8*0 = 0N. Therefore the full 8N force is acting vertically downwards on that bar.

    You also have a 12N force acting on the bar at a 30 degree angle. That's 12 * sin30 = 6 N vertically and 12 * cos30 = 10.39 N acting horizontally. Since the horizontal force acts through the pivot, it doesn't contribute anything to the torque, so we can ignore it.

    So now the situation is like this:
    Code (Text):
       6 N
              8 N
    Taking the upward direction as positive we can show the diagram like this:
    Code (Text):
       6 N    -8 N
        |      |
        |      |
    Now try calculating the torque from that last diagram.
  10. Nov 19, 2008 #9
    -8N x 5m = -40 Nm
    6N x 2m = 12 Nm
    -40 + 12 = -28Nm
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