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Moments - I am so confused please help

  1. Jun 29, 2009 #1
    1. The problem statement, all variables and given/known data

    "In raising the pole from the position show, the tension T in the cable must supply a moment about O of 72 Knm. Determine T."


    2. Relevant equations

    M = Fd

    3. The attempt at a solution

    d = 12sin(30) = 6m
    T = 72/d
    ==> 12m

    I do not know what I am doing wrong here. I'm assuming there is a problem with "12sin(30)" but I checked atleast 3 times and I still get the same eq. Please assist in any manner possible. Thank You in advance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jun 29, 2009 #2

    PhanthomJay

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    The moment of a force about a point O is the force times the perpendicular distance from the line of action of that force to the point O (M=F(d_perp)). Alternatively, the moment of a force about point O is the force times the distance from the point of the application of the force to point O, times the sine of the included angle between the two (M = rF sin theta = 30Tsin theta in this example). This becomes a geometrry/trig problem, where you must determine theta using the appropraite trig functions (as in using the law of sines and cosines, for example).
     
  4. Jun 29, 2009 #3
    Now I'm not too sure if I understand you correctly here, but what I make out of your response is that the angle between, I believe it to be 30 degrees; you take the sin of the angle and then multiply that by the force, in this case T.

    Doing that I arrive at the conclusion that 30Tsin(30) = 72 and solving for T results a 4.8 knm

    I see the answer is wrong and am still unable to make out how to correctly solve this problem. Unfortunately I have a quiz on this material tomorrow evening at 6 and I need your help in trying to understand how to do this problem, for which the author of the book decided not to give the EDIT: workto.

    The final answer is 8.65 knm. Could you kindly walk me step by step through the problem so that I can clearly understand how it is done? (I agree my trig is horrible)
     
    Last edited: Jun 29, 2009
  5. Jun 29, 2009 #4

    tiny-tim

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    Hi zrome! :smile:
    Where do you get 30º from? :confused:

    Hint: draw the vertical line from the top of the cable to the ground … so you have a right-angled triangle, and you want to know the sine of one of the angles. :wink:
     
  6. Jun 29, 2009 #5

    nvn

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    That's a good hint by tiny-tim, to get the height of the vertical line. And, you can compute the horizontal projection of the pole on the ground, right? After that... Hint: Use definition of tan to compute an angle.
     
  7. Jun 30, 2009 #6

    tiny-tim

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    Hi nvn! :smile:

    Actually, we don't need to know the angle …

    we only want its sine, whch we can get from the sides of the right-angled triangle (and Pythagoras) :wink:
     
  8. Jul 1, 2009 #7
    Ya computed the vertical line and was able to see the big triangle I was missing. Thank you for all the help.
     
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