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The moment about a point on a post

  1. Sep 22, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the moment about point A

    Prob.4-109.jpg

    2. Relevant equations



    3. The attempt at a solution
    Calling counter-clockwise positive

    (300N)(1m) + 500cos(30)(.2) - 250(4/5)(.5) = 286.6Nm

    Not the correct answer. What am I doing wrong?
     
  2. jcsd
  3. Sep 22, 2014 #2

    Simon Bridge

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    The moment arm should be measured from point A.
    Is the point of action of the 500N force only 20cm from point A?
     
  4. Sep 22, 2014 #3
    I thought it could be measured to anywhere on the line of action of the component of the force that creates the torque?

    Thanks for the quick answer.
     
  5. Sep 22, 2014 #4
    300(1) + 500(cos30)2 - 250(4/5)(3.041) = 557.8Nm

    Also incorrect :/.
     
  6. Sep 22, 2014 #5

    Simon Bridge

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    ... but that distance is different for different positions on the line of action - so which do you choose?
    Anyway - there is no distance between A and the line of action of the 500N force than is 0.2m.

    The 500N force is not 2m away from A either.
    You are also using the wrong angles.

    Note: there are two ways of doing this problem.
    It is important not to get them mixed up.

    One way is to take the distance between the pivot (A) and the point of action of the force and multiply that by the component of the force perpendicular to this moment arm, the other is to use the whole magnitude of the force multiplied by the perpendicular distance between the line of action of the force and a parallel line through the pivot.

    The actual formula is ##\tau = \vec r \times \vec F## where ##\vec r## is the vector pointing from the pivot to the point of action of the force.
    So you can always just formulate the whole thing as vectors and use the cross-product formula.
     
  7. Sep 22, 2014 #6
    Thank you Simon. I used the cross-product and it worked. I'm still kind of curious as to why my original method didn't really work. You don't happen to have a graphical explanation?
     
  8. Sep 22, 2014 #7

    Simon Bridge

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    You simply did not apply the definitions correctly.
    It's like making a pie with peaches and wondering why you don't have apple pie.

    I mean: 500sin(30)(0.2) would be the moment from the 500N force about the point 2m directly above point A ... but 500cos(30)(0.2) is not anything useful ... apples and peaches.

    Note: In terms of magnitudes: ##\tau = rF\sin\theta = r(F\sin\theta)=(r\sin\theta)F## where ##\theta## is the interior angle between the vectors. Notice how you can associate the sine with either the F or the r and get the same answer? However, the thought process is quite different:
    ... ##F\sin\theta## would be the component of ##\vec F## that is perpendicular to ##\vec r##, while ##r\sin\theta## would be the perpendicular distance between ##\vec F## and the parallel line through the pivot.
    Sometimes one way is easier than the other.

    Your problem, all the magnitudes were to 1 sig fig ... which could mean that you can make some approximations safely.
     
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