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Finding the Resultant Force, help needed.

  1. Nov 4, 2009 #1
    Hi, I need some guidance regarding this exercise, I attached a visual representation of the exercise

    1. The problem statement, all variables and given/known data

    Find the Resultant Force for the following system of forces shown


    FORCES SHOWN on a Vector:


    F1= 5 kips on the X quadrant in a 60 degree angle

    F2= 5 kips straight down on the Y axis

    F3= 8 kips on the X quadrant in a 30 degree angle



    2. Relevant equations

    These right?

    Rx= [tex]\Sigma[/tex]Fx= F1 + F2 + F3
    Ry=[tex]\Sigma[/tex]Fy= F1 + F2+ F3


    (theta)Qx= Tan- Ry/Rx


    3. The attempt at a solution


    I want to know if I am going in the right direction, if not please guide me, this is what I tried.


    Rx= [tex]\Sigma[/tex]Fx= F1 + F2 + F3

    =+5(sin 60 )- 5 + 8(cos 30 )

    = -6.16 kips


    Ry=[tex]\Sigma[/tex]Fy= F1 + F2+ F3

    = -5( sin 60)-5 + 8 (cos 30)

    = -8.571 kips
     

    Attached Files:

    Last edited: Nov 4, 2009
  2. jcsd
  3. Nov 4, 2009 #2

    rl.bhat

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    In ΣFx, there is no F2 and F1x = F1*cosθ.
    In ΣFy, F1x is positive, and F3x = F3sinθ and it is negative.
     
  4. Nov 4, 2009 #3
    So like this right?


    Rx= [tex]\Sigma[/tex]Fx= F1 + F3

    =+5(cos 60 ) x 8(cos 30 )

    = answer here


    Ry=[tex]\Sigma[/tex]Fy= F1 + F2+ F3

    = 5( sin 60)-5 + 8 (sin 30)

    = answer here
     
  5. Nov 4, 2009 #4

    rl.bhat

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    Note down whether the components are on the +ve or -ve axis.
     
  6. Nov 4, 2009 #5
    This is the next steps right?


    R= [tex]\sqrt[]{}[/tex] Rx+ Ry

    = answer here


    then...


    (theta)Qx= TAN-Ry/Rx

    = answer here
     
    Last edited: Nov 4, 2009
  7. Nov 4, 2009 #6
    What do you mean?, my english understanding isn't that great can you say that in another way.

    I just want assurance that I am on the right path, thanks for the help btw I appreciate it.
     
    Last edited: Nov 4, 2009
  8. Nov 4, 2009 #7

    rl.bhat

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    =+5(cos 60 ) x 8(cos 30 ) This step should be

    =-5(cos 60 ) + 8(cos 30 )
    5( sin 60)-5 + 8 (sin 30) This step should be
    5( sin 60)-5 - 8 (sin 30)
     
  9. Nov 4, 2009 #8
    I had it like that then got confused by your advise that said " *cos " and I took that as you saying multiply then I added an X, sorry lol.

    Thanks for clearing that up, so the next steps I showed you in post#5 above is the right path correct?, that would pretty much wrap this exercise up, I will proceed once you tell me.
     
    Last edited: Nov 4, 2009
  10. Nov 4, 2009 #9

    rl.bhat

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    Carry on with post5#
     
  11. Nov 4, 2009 #10
    Alright thanks, I appreciate the help and guidance :)
     
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