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Finding 2 of 3 Forces Given 1 and a Resultant

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  1. Aug 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Force A has a magnitude of 200lb and points 35° N of W. Force B points 40° E of N. Force C points 30° W of S. The resultant of the three forces has a magnitude of 260lb and points 85° S of W.
    (a) What is the magnitude of Force B?
    (B) What is the magnitude of Force C?

    2. Relevant equations
    F = √(Fx2 + Fy2)
    Fx = FcosΘ
    Fy = FsinΘ

    where

    F = Force
    Θ =Angle

    3. The attempt at a solution
    Having drawn these vectors, I've been trying to use components which could've been easier if 2 forces were given.

    Resultant's X = Rx = 260cos85°
    Resultant's Y = Ry = 260sin85°
    Force A's X = Ax = 200cos35°
    Force A's Y = Ay = 200sin35°
    Bx = Bsin40°
    By = Bcos40°
    Cx = Csin30°
    Cy = Ccos30°
    And formed equations

    R = A + B + C
    Rx = Acos35° + Bx + Cx
    Ry = Asin35° + By + Cy
    With these, I don't really know because 2 are missing while having no other equations for them like A+B=1 or stuff like that that give hints. Seems like something is missing, but i don't know because others have solved it. Don't know what is next.
     
  2. jcsd
  3. Aug 22, 2015 #2

    BvU

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    Hello cad, welcome to PF :smile: !

    You have two unknows (B and C) and two equations.
    You are making life difficult for yourself by dragging in Bx, y and Cx, y.

    [edit]By the way, I hope you don't mean R = A + B + C but ##\vec R = \vec A + \vec B + \vec C## :rolleyes:
     
    Last edited: Aug 22, 2015
  4. Aug 22, 2015 #3
    So 260 = 200 + ##\vec B + \vec C## would be the first equation and i'll have to find the other equation
    By any chance, does the 2nd equation contain angles in it? Or anything?

    Ah yes, sorry about that
     
  5. Aug 23, 2015 #4

    BvU

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    260 and 200 are not vectors. I take it you do know about vector addition ?

    You have two equations

    with two unknowns: the magnitudes B and C.​

    Perhaps I confused you with my blunt comment
    but the intention was that you would substitue the magnitudes of B and C
     
  6. Aug 23, 2015 #5
    I see. I came up with estimated answers B = 1780.15 and C = 2006.18. I also checked and resulted to 259.99... Am I doing this right?
     
  7. Aug 23, 2015 #6

    BvU

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    Does it match with the drawing you mentioned in post #1 ?
     
  8. Aug 24, 2015 #7

    BvU

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    Check your angles. Do they have the same reference ?
    In math and physics, east is zero angle, north ##\pi/2##
     
  9. Aug 24, 2015 #8

    BvU

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    Never mind, it matches. Note that finding the magnitude of ##\vec A + \vec B + \vec C## is 260 is not enough to ensure ##\vec R = \vec A + \vec B + \vec C##; the direction has to be the same too.
    But from you answer I expect you did the right thing and equated x and y components. WEll done.
     
  10. Aug 27, 2015 #9
    Hi, im a newbie here... can someone explain how did he got the values of B and C??
     
  11. Aug 27, 2015 #10
    how did you got B and C??
     
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