How to Find Angles in a Static Equilibrium System with Pulleys?

[kaienx]

Homework Statement

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):

Homework Equations

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

The Attempt at a Solution

Replacing these with actual values[/B]:
- 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.

 

[ehild]

Homework Statement

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):

Homework Equations

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

The Attempt at a Solution

Replacing these with actual values[/B]:
- 313.9cos α + 619cos β + 60 = 0 — (1)
313.9sin α + 619sin β - 882.9 = 0 — (2)

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

I'm not too sure if I'm doing this correctly or not.

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

 

[kaienx]

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

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

 

[ehild]

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

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

 

[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.

 

[ehild]

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

 

[kaienx]

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

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

 

[ehild]

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

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

 

[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.

 

[ehild]

I saw my mistake! Handling big numbers really are confusing.
0 = 1067740.2 + 74280cosβ - 1093030.2sinβ

And I'm stuck again.

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?