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2D Moments confusion, does d have to be perpendicular?

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data
    http://imageshack.us/a/img812/8836/5tfn.jpg [Broken]

    Uploaded with ImageShack.us

    Find F such that the moment around A will not exceed 1000N*m for any value of θ.
    2. Relevant equations
    M = fd


    3. The attempt at a solution

    I got the right answer, I found the distance of a -> b
    AB= √( (2^2) + (3^2) )
    AB=3.6

    and then I did
    M = fd
    1000 = f (3.6)
    f = 277.8 N

    This is correct according to the answer in the back, but I don't understand why.
    1. I was under the impression for: M = fd , d must be the perpendicular distance. Clearly, the d we are using is not perpendicular to f. Does d not have to be perpendicular to the force?

    2. I thought this was a trick question at first, and logically I thought it made most sense that the greatest moment around A would be caused by a force only in the vertical direction. So I tried doing it where Fx = 0 and Fy is the variable of interest.

    So i tried, M = (Fy)d
    1000 = (Fy)(3)
    but the fy I get is clearly not the answer. Can someone explain this to me.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 21, 2013 #2
    If the force is not perpendicular to the distance, what is its moment?
     
  4. Sep 21, 2013 #3
    I don't understand what you mean, Voko.
     
  5. Sep 21, 2013 #4
    You wrote: "I was under the impression for: M = fd , d must be the perpendicular distance. Clearly, the d we are using is not perpendicular to f."

    Clearly F and d are not perpendicular for every possible angle. So how does one go about computing the moment in such a case?
     
  6. Sep 21, 2013 #5
    Would I break it into components instead?

    Mx = fcosθ * (d)
    My = fsinθ * (d)
     
  7. Sep 21, 2013 #6
    In 2D, moments are not vectors, so they have no components.

    Have you not been given a definition of the moment of force? It should be in your course material/textbook, find it.
     
  8. Sep 21, 2013 #7
    right right right.

    M = fcosθ(3) - fsinθ(2)
    10 000 = fcosθ(3) - fsinθ(2)

    I have no clue what comes next. I've tried reading it, clearly I am confused. Wait, I am confused even more, my solution in the initial post yielded the right answer, was that by complete fluke?

    It says in my book "d is the perpendicular to the line of action of the force" I still can't see how d will ever be the perpendicular on the line of action .
     
  9. Sep 21, 2013 #8
    What that definition really means is that when you compute the moment of force at point B about point A, you first find the component of force perpendicular to AB. Then the moment is the product of that component with AB.
     
  10. Sep 21, 2013 #9
    http://imageshack.us/a/img547/2850/nzd7.jpg [Broken]

    Uploaded with ImageShack.us

    Your not given θ or F, maybe because its 1am, but I swear I can't see what you are trying to help me see.

    I understand F can be broken down into Fc and Fc1. Fc1 isn't important because Fc1 goes through our moment pt A, thus causing no moment.

    I have no clue why Fc = F.
     
    Last edited by a moderator: May 6, 2017
  11. Sep 22, 2013 #10
    So you do understand that ##M = F_c d ##, correct? Now, what is ##F_c## if ##F## is at angle ##\alpha## with AB (note I said ##\alpha##, not ##\theta##)?
     
  12. Sep 22, 2013 #11
    Or let me ask that differently: does ##F_c## ever exceed ##F##? What is the max value of ##F_c##? At what angle with AB?
     
  13. Sep 22, 2013 #12
    alpha = 90
     
  14. Sep 22, 2013 #13
    So, what is the magnitude of ##F_c## at 90 degrees with AB, if the total magnitude of the force is F?
     
  15. Sep 22, 2013 #14
    so wait, is θ just thrown in there to mess with me. Should I have known right from the jump that the largest moment generated around A would be Fc (perpendicular to AB)
     
  16. Sep 22, 2013 #15
    You see, since you are required to consider all possible angles ##\theta##, you might as well consider all possible angles ##\alpha##, which is easier, right?
     
  17. Sep 22, 2013 #16
    Oooooooooohhhhh, Voko, you're JEDI MASTER.
     
  18. Sep 22, 2013 #17
    So, do you understand now why your initial solution was correct?
     
  19. Sep 22, 2013 #18
    Yah, I missed the obvious until you pointed it out.

    I appreciate you takin the time, voko.
     
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