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Homework Help: Minimum force needed

  1. Oct 7, 2015 #1
    1. The problem statement, all variables and given/known data
    is part C lack of information ? can someone give me a hint how to start ?

    for part a ,
    -(1.35-0.95)(200cos60)-2(200sin60)=-386N(anticlockwise)

    for part b , -386.4/2 = 193N (in upward direction)

    2. Relevant equations


    3. The attempt at a solution
     

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  2. jcsd
  3. Oct 8, 2015 #2

    haruspex

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    There is enough information.
    Suppose you resolve the tension into a force parallel to AB and a force perpendicular to it. What can you say about how these components contribute to the moment about A?
     
  4. Oct 10, 2015 #3
    If i resolve the tension into a force parallel to AB and a force perpendicular to it , then I need to take the perpendicular distance multiply the force to get the moment . But , the angle between the force and x , y -axis is unknown , how to determine the force ? how to know it's minimum ?
     
  5. Oct 10, 2015 #4

    haruspex

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    What is the moment about A of the component parallel to AB?
    There is enough geometrical information to find the angle ABF.
     
  6. Oct 10, 2015 #5
    why angle ABF is required ?
     
  7. Oct 10, 2015 #6

    haruspex

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    You are asked to find the smallest force that could be applied at B in order to produce the same momemt about A that the force along BF has. I suggested resolving the force along BF into a component parallel to AB and a force perpendicular to AB. In order to do that you need to find the angle which BF makes to AB.
     
  8. Oct 12, 2015 #7
    well, i still dont knw what do you mean here, btw , i managed to get the angle of ABF = arc(tan 0.4/2) = 108.7
     
  9. Oct 12, 2015 #8
    What direction must the minimum force at B have in relation to line AB?
     
  10. Oct 12, 2015 #9
    Sorry, I am nt sure
     
  11. Oct 12, 2015 #10
    Exercise: Take a 1m bar and apply a 10N force to one end. In what direction will the force produce the maximum moment at the other end of the bar?
     
  12. Oct 12, 2015 #11

    haruspex

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    Ok, so now you know how to resolve the applied force F into components parallel to AB (call this FAB) and perpendicular to AB (FP). What moment does FAB have about A?
     
  13. Oct 13, 2015 #12
    max moment will produce when the force is applied exactly 1m from the point .
     
  14. Oct 13, 2015 #13
    this looks weird, this is the first time i resolve the force with the angle more than 90 degree
     

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  15. Oct 13, 2015 #14

    haruspex

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    That's the right diagram, but you did not answer my question. What is the moment of the FAB component about A?
     
  16. Oct 13, 2015 #15
    That is correct, but I was asking about the direction of the force at that point. For example, in this drawing, which direction of the 10N force (A, B, or C) produces the maximum moment at point O?: Moments.jpg
     
  17. Oct 13, 2015 #16
    B
     
  18. Oct 13, 2015 #17
    Correct! So, what angle to any moment arm would give the minimum force to produce a given moment?
     
  19. Oct 13, 2015 #18
    90 degree is to produce max force , how to determine the min force ?
     
  20. Oct 13, 2015 #19

    haruspex

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    No, 90 degrees produces the maximum moment for a given magnitude of force; which is the same as saying it requires the minimum magnitude of force for a given moment.
     
  21. Oct 13, 2015 #20
    ok , i understand it . but , how to solve the question ?
     
  22. Oct 14, 2015 #21

    haruspex

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    Use the diagram you posted at #13.
    What I was trying to get you to see was that the component of F parallel to AB has no moment about A, so you might as well set it to zero. That leaves only the component perpendicular to AB. You know what moment this needs to have about A, and you know the distance AB, so what is the magnitude of this force?
     
  23. Oct 14, 2015 #22
    so , i have found out that the AB = 2.3m , this is due to sin11.3= 0.45/AB
    the angle between the line AB and the horizontal line = 11.3 degree becoz i take 180 degree - 60 degree-108.7 degree , so i gt 11.3 degree

    so i have found out the force = 386.4/ 2.3 = 168 N ... is it correct ?
     
  24. Oct 14, 2015 #23
    why the distance AB is required ?
     
  25. Oct 14, 2015 #24

    haruspex

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    The 0.45 is wrong, leading to the wrong value for the minimum force.
     
  26. Oct 14, 2015 #25
    why the distance AB is required ? if i substituted 0.45 with 0.4 , then my ans would be correct ?
     
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