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Weight of a torque question

  1. Jun 9, 2013 #1
    I know how to complete an actual torque equation and have done so, my question is more of what sign do I address the force of the problem. One problem I have is:

    F1=50N, d1=9cm, d2 = 10cm and d3=8cm
    F1 is situated between d2 and d3. I used the equation t=f(la) so my equation was (50N)(0.1m)sin90° which equals 5Nm.

    I just want to be clear, in this equation, since f1 is pointing downward (basically on the y-axis) that I still put it in the equation as a positive number.
     
  2. jcsd
  3. Jun 9, 2013 #2

    CWatters

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    I couldn't follow your description. Any chance of a drawing?

    A problem might involve several forces and/or torques pointing in different directions. Unless stated in the problem it's upto you which direction(s) you define as "positive" but you should do so at the start and stick to it.

    Then at the end when you have worked out the sign of the answer it will tell you the direction.
     
  4. Jun 9, 2013 #3
    this is a basic example. I have other problems where there are two forces both going in that downward direction. I just want to make sure I am completing these problems correctly so I have a guide for the rest of my work down the line.
     

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  5. Jun 9, 2013 #4

    CWatters

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    the magnitude of your answer is correct (5Nm) but you should state that it's in the "5Nm clockwise direction".

    Suppose the problem had another force F2 = 100N acting downwards at d1.

    Then you would start by defining which direction you consider positive. Lets say clockwise is positive. Then you can write..

    T = F1*d2 - F2*d1

    It's -F2*d1 because F2 acts in the anticlockwise direction. Both forces are downwards but one acts clockwise and the other anticlockwise.

    Substituting the values gives...

    T = 50*0.1 - 100*0.09
    = 5 - 9
    = -4Nm

    The answer is -ve so the total torque is 4Nm in the anticlockwise direction.

    Suppose we had choosen anticlockwise as +ve at the start. Then we would have written..

    T = -F1*d2 + F2*d1

    and

    T = -50*0.1 + 100*0.09
    = 4Nm

    This time the answer is +ve but because we chose anticlockwise as positive at the start the answer is still 4Nm in the anticlockwise direction.

    So which ever direction you choose as positive does not matter as long as you are consistent and refer back to your choice at the end.

    Sorry I've edited this reply a few times. All done now.
     
  6. Jun 9, 2013 #5
    that definitely helped, thank you so much! I went to use this rule and apply it to another problem I had.

    The question for this one was if F1=50N and F2=10N and d1=9cm, and the distance between F2 and a is 2 times the distance between F1 and a, what is the torque about a?

    I set up total equation as:

    T= (-50N)(0.03m)sin90° + (10N)(0.06m)sin90° = -0.9Nm in the clockwise direction
     

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  7. Jun 9, 2013 #6

    CWatters

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    Yes that's correct, although personally I would have written "0.9Nm in the anticlockwise direction". Same thing really.

    If the exam asks to show your working it's essential to spell out which direction you are defining as +ve at the start. That's because in some exams you get some or even most of the marks for your working even if you get the final answer wrong due to a trivial arithmetic error. Don't make the examiner work out which direction you assumed to be +ve.
     
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