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Basic Equilibrium Force

  1. Sep 15, 2016 #1
    1. The problem statement, all variables and given/known data
    000.PNG

    2. Relevant equations
    xcomponent of vector= magnitude*cosine of theta
    ycomponent of vector= magnitude*sine of theta
    Basic trig operations after this. Honestly I think that I am just having trouble with the algebra

    3. The attempt at a solution
    Because the magnitude of the added vectors a and b must equal the magnitude of c, I added the vectors, with the exponent x as the magnitude for force b, and used Pythagorean theorem 000.PNG to come up with the magnitude of the sum. I acted as if force a was acting along the positive x axis, putting the angle of force b at 144. Set this equal to the magnitude of force c. See below

    186=((210cos(0)+xcos(144)^2+(210sin0+xsin(144))^2)^0.5
    34596=(210-.81x)^2+.588x^2
    34596=1.006x^2-340.2x+44100
    0=1.006x^2-340.2x+4504
    x=13.8,324.7
    I took the 324.7 being the magnitude of force b, obviously if this was right I wouldnt be here
    Many thanks for the help
     
  2. jcsd
  3. Sep 15, 2016 #2
    What are the x and y components of Alex's force? Let ##\theta## be the angle that Charles' force makes with the x axis. In terms of ##\theta##, what are the x and y components of Charles' force? What is the force balance in the x direction?
     
  4. Sep 15, 2016 #3
    So the x component of Alex's force is 210 and the y component is 0. x and y components of Charles' force would be 186cosθ and 186sinθ respectively. With their added forces being 180 degrees from 144 degrees or 324°, we can say the arctan of (186sinθ/210+186cosθ)=324. So 186sinθ/(210+186cosθ)=-.727.
     
  5. Sep 15, 2016 #4
    No. Does it look that way to you from the diagram?
    Correct.
    I asked for the force balance in the x direction. Once you know the x components of Alex's force and Charles' force, writing this equation should be easy.
     
  6. Sep 15, 2016 #5
    Equilibrium Force in x= -(-123.43+186cosθ)
     
  7. Sep 15, 2016 #6
    Forgive me, I had rearranged the problem so that alex's force was acting upon the positive x axis, that has been corrected.
     
  8. Sep 15, 2016 #7
    I get $$186\cos{\theta}-123.43 = 0$$Do you see this? From this equation, what is the value of ##\theta##?
     
  9. Sep 15, 2016 #8
    Placing the image on a grid, with Betty's force being at 270°, this would make the balancing force be at 90° making the added x components of Alex's and Charles' force be 0. Making the x component of Force C be 123.43, taking the arccos of the magnitude (186) divided by 123.43 gives an angle of 47.12°. Thanks for the help
     
  10. Sep 15, 2016 #9
    I get 48.4 degrees.

    Now you need to do the force balance in the y direction, which includes Betsy's force.
     
  11. Sep 15, 2016 #10
    or 48.425. my mistake
     
  12. Sep 15, 2016 #11
    Adding those two forces gives a force in the positive y direction of 309 N, with the magnitude of Betty's force being equal in the opposite direction.
     
  13. Sep 15, 2016 #12
    Although I am unsure about "the other possible direction for equilibrium"
     
  14. Sep 15, 2016 #13
    I get 301.
     
  15. Sep 15, 2016 #14
    210sin126=169.9 186sin48.4=139.1 169.9+139.1=309
     
  16. Sep 15, 2016 #15
    Suppose Charles' force was pointing 48.4 degrees below the x axis instead of above the x axis. Wouldn't the x component of his force be the same?
     
  17. Sep 15, 2016 #16
    Oh, I used 200 for some reason.
     
  18. Sep 15, 2016 #17
    Adding the y components using 210N at 126° for Alex and 186N at -48.4° for Charles I get 30.8 N with the equilibrium force having a magnitude equal but in the -y direction
     
  19. Sep 15, 2016 #18
    Nicely done.
     
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