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Angles in a Static Equilibrium

  1. Feb 29, 2016 #1
    1. The problem statement, all variables and given/known data
    I have 3 masses (Fα, Fβ & Fg) with 2 pulleys, and a wind variable which is in static equilibrium. I have already calculated the appropriate forces for the 3 masses by multiplying it with 9.81m/s² (gravity).

    Fwind = 60N
    Fα = 313.9N
    Fβ = 619N
    Fg = 882.9N

    I'm required to find the angles for vector Fα & Fβ as shown in below equations (which is derived from the vector's individual components (x & y):

    lVCR1.jpg

    2. Relevant equations
    - Fαcos α + Fβcos β + Fwind = 0 — (1)
    Fαsin α + Fβsin β - Fg = 0 — (2)

    3. The attempt at a solution
    Replacing these with actual values
    :
    - 313.9cos α + 619cos β + 60 = 0 — (1)
    313.9sin α + 619sin β - 882.9 = 0 — (2)

    I have re-organized the equation:
    cosα = 619cosβ + 60 / 313.9
    sinα = - 619sin β + 882.9 / 313.9
    square it as such:
    cos²α = (619² cos²β + 60² + 2(619cosβ * 60)) / 313.9² — (1)
    sin²α = (619² sin²β + 882.9² - 2(619sinβ * 882.9)) / 313.9² — (2)

    Adding them up as sin2α+cos2α=1

    1 = (619² cos²β + 60² + 2(619cosβ * 60)) / 313.9² + (619² sin²β + 882.9² + 2(619sinβ * 882.9)) / 313.9²

    I'm not too sure if I'm doing this correctly or not.
     
    Last edited: Feb 29, 2016
  2. jcsd
  3. Feb 29, 2016 #2

    ehild

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    Bring the terms to the common denominator, and note that you can simplify by replacing the sum of the square terms by 6192 .
     
  4. Feb 29, 2016 #3
    could you elaborate further on (simplify by replacing the sum of the square terms by 6192) ?
     
  5. Feb 29, 2016 #4

    ehild

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    First check the signs in your second squared equation .
    Using the identity a/c+b/c=(a+b) /c
    you get
    1 =( 619² cos²β + 60² + 2(619cosβ * 60) + 619² sin²β + 882.9² - 2(619sinβ * 60) )/ 313.9²
     
    Last edited: Feb 29, 2016
  6. Feb 29, 2016 #5
    I've done some simplifying and adding up based on your guidance and got:

    783112.41 + 619²(sin²β + cos²β) + 74280 (cosβ + sinβ) / 313.9².
     
  7. Feb 29, 2016 #6

    ehild

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    I just noticed that your second squared equation is wrong. Correct. And I do not understand what your last formula is.
     
  8. Feb 29, 2016 #7
    my last formula? meaning sin2α+cos2α=1? is it?
     
  9. Feb 29, 2016 #8

    ehild

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    sin2α+cos2α=1 is right and should be used. But the whole formula is wrong. It is not even an equation.
    Go back to
    1 =( 619² cos²β + 60² + 2(619cosβ * 60) + 619² sin²β + 882.9² - 2(619sinβ *882.9 60) )/ 313.9²
     
  10. Feb 29, 2016 #9
    I saw my mistake! Handling big numbers really are confusing. I have edited the first post for clarity's sake.
    1 = 619² cos²β + 60² + 2(619cosβ * 60) + 619² sin²β + 882.9² - 2(619sinβ *882.9))/ 313.9²
    From there:
    1 = 619² (cos²β + sin²β) + 60² + 882.9² + 2(619cosβ * 60) - 2(619sinβ *882.9) / 313.9²
    313.9² = 619² (1) + 60² + 882.9² + 74280cosβ - 1093030.2sinβ
    0 = 619² + 60² + 882.9² - 313.9² + 74280cosβ - 1093030.2sinβ
    0 = 1067740.2 + 74280cosβ - 1093030.2sinβ

    And I'm stuck again.
     
  11. Feb 29, 2016 #10

    ehild

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    Yes, it is difficult to work with big numbers. That is, why we prefer to work symbolically, and substitute the data at the end. I do not have time and nerve to check your numbers.
    Divide the whole equation with the constant term, getting an equation of form C sin(β) +D cos(β) - 1=0 . C and D are not big numbers.
    How would you solve it, knowing that sin2(β) + cos2(β)=1?
     
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