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2 forces, net torque

  1. Jul 2, 2014 #1
    1. The problem statement, all variables and given/known data

    Two 59.9N forces are applied in opposite directions to the 28.3cm long figure shown. If the angle θ = 29.6°, what is the net torque about the center of mass?

    https://loncapa.physics.mcmaster.ca/enc/57/e175e7955c9fbe9618cfad45ea8ac77518fe77da941b3d81f2ec13253475e5b47f4b6f4ee2fd3fcaccec028dd206d1f1f6af315c50f34af8819d09fd59b32077424b5007996cb661.png

    Same question as this:https://www.physicsforums.com/showthread.php?t=413599 (Not my question, but same picture just different numbers)
    But it was unresolved?

    2. Relevant equations

    torque = force * radius * sin theta

    3. The attempt at a solution

    Got net torque = torque of force 1 + torque of force 2
    torque 1= 0.283*59.9*sin 119.6
    torque 2= - 0.283*59.9*sin 119.6
    net torque = 0
    my answer is wrong and I have no clue why. :(
    assistance and guidance will be greatly appreciated.
    Thank you!
     
    Last edited: Jul 2, 2014
  2. jcsd
  3. Jul 2, 2014 #2

    Nathanael

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    The torques are in the same direction, so they're either both positive, or both negative

    (Assuming the picture is the same as in the other thread you posted, because I can't see the picture on this thread)
     
  4. Jul 2, 2014 #3
    Hi!
    It is the exact same picture.
    But the force directions are in the opposite direction (?), so I took that into consideration when calculating torque. Could you explain why? Sorry
     
    Last edited: Jul 2, 2014
  5. Jul 2, 2014 #4
    Also, I just re-calculated assuming the forces were in the same direction. I got 2*14.7N*m = 29.5 N*m which is also incorrect?
     
  6. Jul 2, 2014 #5

    Nathanael

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    The forces are indeed in opposite directions, but they both cause the object to rotate in the same direction.

    It isn't only the direction of the force that is important, but the direction of rotation that the force causes (counter-clockwise in your picture).



    (Imagine holding a pencil (in the middle) with two people applying the same force in the directions shown in your picture. The forces are equal and opposite, but is the pencil going to rotate? It will still rotate, because the torques are in the same direction)


    Probably not the best explanation but ask more if you don't understand
     
  7. Jul 2, 2014 #6

    Nathanael

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    When calcuting the torque, you want to multiply the force by the distance from the pivot (in this case the pivot is the center of mass)

    Quick question, is the center of mass in the center of the object? (There are those "diamond shaped" lines in your picture and I'm not sure what they represent?)


    Another question, why did you multiply it by the sine of 119.6?
     
  8. Jul 2, 2014 #7
    Thanks- that was a great explanation.
    So in order for the object to rotate, the torques must not be equal (cancel out).
    However, I've tried finding the angles at various points between the force and the triangle and have gotten 14.74N*m.
    I can't seem to get a proper torque calculation.
     
  9. Jul 2, 2014 #8
    Me neither :/
    No one has clarified. It's an online assignment so all I'm given is the question.
    Apparently, according to the previous thread with the same question, the centre of mass doesn't matter? So I tried coupling the forces but doing so only results in the same torque values of 14.74 again.

    Edit: I multiplied by sin 119.5 because the angle between the radius of rotation and F is given (?)
     
  10. Jul 2, 2014 #9

    Nathanael

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    Yes. For the torque to cancel out it must be equal and opposite (equal in magnitude, opposite in direction).
    If it doesn't cancel out, there will be a net torque


    In your calculation you put 0.283 as the distance from the center of mass, but that can't be right (at least not for both) since 0.283 is the length of the entire object.
     
  11. Jul 2, 2014 #10
    :bugeye: So I just guessed 14.7N*m and it's right!
    I guess I assumed two different points (like the previous post suggested) and calculated the torques separate from one another when they are actually about one centre of mass, meaning one is on the centre of mass thus 0 torque (right?).
     
  12. Jul 2, 2014 #11

    Nathanael

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    Ah ok. I was a bit thrown of because when I looked at the picture, my first thought was to multiply the force by cos(θ)

    But your method is just as good ([itex]cos(29.6°)=sin(119.6°)[/itex] so we get the same answer)


    It actually halfway makes sense that the center of mass doesn't matter (I won't try to explain it since I don't fully understand it yet)


    Exactly.

    And if you pretend that the center of mass is at the geometrical center, then you'll get two torques, but each of them will be half as large (the "length factor" will be divided in half) so you'll get the same answer (which reflects the idea that the center of mass is irrelevant)


    I wish I had a more intuitive understanding of why the center of mass is irrelevant.
    It's because this is a special case (where the torques are symmetrical).
    Why does this make the COM irrelevant? I'm not entirely sure (mathematically it's clear, but it's not intuitively clear to me)


    Sorry that I can't explain that last part to you. Perhaps someone else who understands it better will explain the logic clearly.
     
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