# Homework Help: Angles in a Static Equilibrium

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1. Feb 29, 2016

### kaienx

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):

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)

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. Feb 29, 2016

### ehild

Bring the terms to the common denominator, and note that you can simplify by replacing the sum of the square terms by 6192 .

3. Feb 29, 2016

### kaienx

could you elaborate further on (simplify by replacing the sum of the square terms by 6192) ?

4. Feb 29, 2016

### ehild

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
5. Feb 29, 2016

### kaienx

I've done some simplifying and adding up based on your guidance and got:

783112.41 + 619²(sin²β + cos²β) + 74280 (cosβ + sinβ) / 313.9².

6. Feb 29, 2016

### ehild

I just noticed that your second squared equation is wrong. Correct. And I do not understand what your last formula is.

7. Feb 29, 2016

### kaienx

my last formula? meaning sin2α+cos2α=1? is it?

8. Feb 29, 2016

### ehild

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

9. Feb 29, 2016

### kaienx

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.

10. Feb 29, 2016

### ehild

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?