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Statics: Magnitudes/Resultant force

  1. Jul 18, 2012 #1
    1. The problem statement, all variables and given/known data
    I'll attatch it.

    3. The attempt at a solution
    Alright, so I work out the angle of the right side...

    cos-1(12/13) = 22.6°

    Find the components of the right side...

    F1x = 800cos(40)
    = 612.84
    F1y = 800sin(40)
    = 514.23

    F2x = 600cos(22.6)
    = 553.92
    F2y = 600sin(22.6)
    = 230.58

    I'm not sure where to go from here, not even sure if the angles right. Looking forward to some help thanks.

    Attached Files:

  2. jcsd
  3. Jul 18, 2012 #2


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    Add them up remembering to be consistent with the sign (especially for Total Fx).

    eg Total Fy = F1y+F2y

    Then check the answer is in the format requested.
  4. Jul 18, 2012 #3
    Total Fy = 744.81
    Total Fx = 58.92

    Cause one of the x's is going in the opposite direction right?

    But the answers meant to be:
    F = 747N
    Angle = 85.5degrees
  5. Jul 18, 2012 #4


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    Reconstruct FTotal using pythagorous..

    FTotal = SQRT (Fy2+Fx2)

    likewise the angle
  6. Jul 18, 2012 #5
    FTotal = 747N

    but with the angle what would I need to use? Sine rule or something?
  7. Jul 18, 2012 #6
    Still struggling to find out how to find the angle of the triangle.
  8. Jul 18, 2012 #7
    Soh - sin (theta) = opposite/hypotenuse
    Cah - cos (theta) = adjacent/hypotenuse
    Toa - tan (theta) = opposite/adjacent
  9. Jul 18, 2012 #8


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    You know the length of all three sides. Draw it on your diagram. Apply basic trig. eg
    Angle = Cos-1(.... you know the rest.
  10. Jul 18, 2012 #9
    the diagram isn't a right angled triangle though? how can i use them trig function?
  11. Jul 19, 2012 #10
    I got the answer... from cos^-1(58.92/744.8)
    = 85.46
    = 85.5

    But I'm struggling to get my head around the diagram. Where's the angle that I'm meant to be finding and what does it mean by "it's direction compared to the x axis clockwise."
  12. Jul 19, 2012 #11


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    You chose +ve to be to the the right and since Total Fx is positive the Total vector points to the right of vertical.

    Total Fy = 744.81
    Total Fx = 58.92

    Since Total Fy >> Total Fx the resulting vector is almost vertical.

    So by inspection the resulting vector is almost but not quite vertical and points/leans slightly to the right. You calculated about 85.5 degrees to the horizontal which confirms this. Draw it on the diagram.

    The question asked for the angle relative to the x-axis anticlockwise. Presumably they mean the anti clockwise direction is to be considered positive. So this also 85.5 degree angle you calculated.
  13. Jul 19, 2012 #12


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    Tell you what, why not draw vectors representing

    Total Fy
    Total Fx
    and the resulting vector Total F

    on the diagram and post it again.
  14. Jul 24, 2012 #13
    Hi guys I'm back with another one, this ones doing my head in I can't seem to solve it...

    Determine the magnitude of resultant force and its direction, measured counterclock wise from the positive x axis; If Fa= 770N, Fb= 240N and Theta= 35 degrees.

    So I get the components...

    Fax = 770*cos(36.87) = 616
    Fay = 770*sin(36.87) = 462
    Fbx = 240*cos(35) = 196.6
    Fby = 240*sin(35) = 136.7

    Fx = 812.6
    Fy = 592.7

    sqrt(812.6^2 + 592.7^2) = 100.9.336N

    to get angle:

    tan-1(812.6/592.7) = 53.9°

    Answers meant to be:
    R= 732 N
    Theta= 55.04 degrees
  15. Jul 24, 2012 #14
    noticed that the picture didn't attach, here it is..

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