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Quick Resultant Forces Question

  1. Oct 24, 2016 #1
    1. The problem statement, all variables and given/known data
    s7rMLoN.png

    So we have 65N @ 30*, 30N @ 180* and 20N at 250*

    2. Relevant equations

    Ncos(angle)i + Nsin(angle)j


    3. The attempt at a solution

    Using this video



    I did the following.

    F1(65N) = 65cos(30)i+65sin30i

    Then using the radian circle I saw that 30* has x,y coordinates of sqrt3/2 and 1/2 respectively. So as shown in the video I did

    65(root3/2)i+65(1/2)j

    = 56.3i + 32.5j

    I did this for the left one too (30N).

    But for the one at the bottom (20N) I get the degree of it to be 250. (270-20) = 250.

    250* is not on the radian circle so how can I write 250* in terms of x,y coordinates?
    circle-unit-304560.gif

    If we look at this radian circle. We see that 240* = (-1/2, -sqrt3/2) and then it goes to 270* = (0,-1). How can I find the x,y coordinates for 250* and use it in the equation?

    Thanks in advance.
     
  2. jcsd
  3. Oct 24, 2016 #2

    Doc Al

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    Staff: Mentor

    I don't believe that trig functions of that particular angle can be simply expressed. (Like the others can.) You'll need a calculator.
     
  4. Oct 24, 2016 #3
    Ah yes, I just found this out. Thanks!
     
  5. Oct 24, 2016 #4

    haruspex

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    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If it's a matter of expressing it in surds, you can use the usual expansion of sin(3x), with x=20o, and apply the formula for solving cubics.
     
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