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Couple Moments

  1. Jun 24, 2015 #1
    1. The problem statement, all variables and given/known data
    For F = 70 lb, compute the combined moment of the two forces about (a) point O, (b) point C, (c) point D. The moments are positive if counterclockwise, negative if clockwise.

    2. Relevant equations
    M = F d , M = r x F

    3. The attempt at a solution
    eb7827a6-ea06-4264-b498-46a8d2b9d9da_zpsfrvx7swv.jpg
     
  2. jcsd
  3. Jun 24, 2015 #2
    Shouldt M = Fd here? where F=70 and d=37.014? i got an answer of 2590.95 but its not correct.
     
  4. Jun 24, 2015 #3

    haruspex

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    The image is a bit hard to read, but I assume the two forces are equal and opposite. M=Fd is right, where d is a certain distance. What distance? (It probably isn't the distance AB here.)
     
  5. Jun 24, 2015 #4

    gneill

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    Which part of the question are you trying to answer here? It looks like you're trying find the moment about point B, which isn't listed in the question statement. It asks for moments about points O, C, and D.

    Also note that in order to use the form M = Fd you must be certain that the F is perpendicular to the d. Otherwise you should use the vector dot product or use geometry to find the component of F that is perpendicular to d.
     
  6. Jun 24, 2015 #5
    Yes are equal, opposite, parallel, and colinear.
    Why is it not the distance AB, i thought the distance between couples was the distance for the formula M=Fd.
     
  7. Jun 24, 2015 #6
    if the two forces are parallel then the distance AB is perpendicular because point A and point B are each on the axis of either force
     
  8. Jun 24, 2015 #7

    haruspex

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    Not collinear. If they were collinear there would be no moment.
    It is, but the distance AB is the distance between two arbitrary points in the lines of application of the forces. It is not necessarily the distance between the lines.
     
  9. Jun 24, 2015 #8
    yes sorry, they are not on the same axis non collinear.
     
  10. Jun 24, 2015 #9
    i just tried using M=rxF and summed the two and got -2165.239## \hat k##
    but that was wrong answer
     
  11. Jun 24, 2015 #10
    I thought the moment would be the same for any point? isnt that a properties of couple moments in 2d?
     
  12. Jun 24, 2015 #11
    well they are both points on the lines of application of the forces, i dont see how they are not the distance between the lines.
     
  13. Jun 24, 2015 #12

    haruspex

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    The distance between two parallel lines (or two skew lines ) is the shortest distance between points on them. If you pick two arbitrary points they are not necessarily as close as they could be, in fact they could be arbitrarily far apart.
     
  14. Jun 24, 2015 #13
    Ok i can see that i was assuming incorrectly that the line AB is prependi
    Ok, i was wrong to assume line AB was perpendicular to the couple. So would the best way to solve this would to be use ∑M = rxF from both the forces to point 0, then repeating that for the other two point?
     
  15. Jun 24, 2015 #14

    haruspex

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    Whichever way you do it will involve finding the perpendicular distance from a point to a line, so you might as well stick with your approach. You just need to do a bit of geometry to find that distance.
     
  16. Jun 24, 2015 #15
    ok, but for future reference does the vector r in equation M= r x F have to be perpendicular to F?
     
  17. Jun 24, 2015 #16

    haruspex

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    No, the process of computing the cross product handles that. |r x F| = |r||F| sin(theta), where theta is the angle between the vectors.
     
  18. Jun 24, 2015 #17
    well why didnt my calculations work for M = r F
    20150624_210536_zps6csfxhl7.jpg
     
  19. Jun 24, 2015 #18

    haruspex

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    It'll take me a while to tease through your numbers, but using AB as the distance vector and taking the cross product I get 2520. Most likely you have a sign error somewhere.
     
  20. Jun 24, 2015 #19
    ok thanks that is correct, i just did it the long way by summing the two moments together using ∑M = rF = 2520.227 units
     
  21. Jun 24, 2015 #20
    so the quicker way is to cross rAB with F?
     
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