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Mechanics Part II

  1. Apr 20, 2006 #1
    Mechanics Part II :)

    Sorry, i got stuck again.. :smile:



    Two Forces, P and Q, act on a particle. The force P has a magnitude 5N and the force Q has magnitude 3N. The angle between the directions of P and Q is 40 degrees. The RESULTANT of P and Q is F

    (a) Find, the magnitude of F

    (b) Find, in degrees to one decimal place the angle between the directions of F and P.

    Now, my stab at part (a) was in the form
    F^2 = Fx^2 + Fy^2
    - = (5cos 40)^2 + (3sin 40)^2
    However, it gives me a different answer to the mark scheme...
    who have... (5+ 3cos40)^2 + (3 in 40)^2 .....

    Why the 5 +3 cos40 ... i dont really get it...

    Any help would be greatly appreciated. :confused:

    (b) similarly for this.. my version would be ; tan(theta)= Fy / Fx ..but then for Fx... there would be 5+3cos 40... which i dont really understand how you get....

    THANKS :)
    Last edited: Apr 20, 2006
  2. jcsd
  3. Apr 20, 2006 #2

    Doc Al

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    Staff: Mentor

    Take the direction of P to be along the x-direction. Thus the x-component of F will be Px + Qx. But Px = 5 N and Qx = 3 cos(40) N, so Fx = 5 + 3 cos(40).
  4. Apr 20, 2006 #3
    ah... so you have to take the x component for force Q as well... correct?

    and there will be no y component for force P.. since well.. there isnt one...

  5. Apr 20, 2006 #4

    Doc Al

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    Staff: Mentor

    Of course.

    Right. Because we cleverly chose our x-axis to be parallel to P.
  6. Apr 20, 2006 #5
    last question..

    are you god? :D ....

    Thank you so much.
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