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I have three masses $\left(F_\alpha, \, F_\beta , \, \text{and} \,F_g \right)$ 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 $9.81 \, \frac{\mathrm{m}}{\mathrm{s}^{2}}$ (gravity).

$$ \begin{alignat}{7} & F_{\text{wind}} && ~=~ & 60 \phantom{.0} & \, \mathrm{N} \\[2px] & F_{\alpha} && = & 313.9 & \, \mathrm{N} \\[2px] & F_{\beta} && = & 619 \phantom{.0} & \, \mathrm{N} \\[2px] & F_{g} && = & 882.9 & \, \mathrm{N} \\ \end{alignat} $$

I'm required to find the angles for vector $F_\alpha$ and $F_\beta$ as shown in below equations (which is derived from the vector's individual components ($x$ and $y$):

$$ \begin{alignat}{7} F_α \, \cos{\left( α \right)} & \, + \, F_β \, \cos {\left( β \right)} && + F_\text{wind} & ~=~ 0 \tag{1} \\[2px] F_α \, \sin{\left(α\right)} & \, + \, F_β \, \sin {\left( β \right)} && - F_g & ~=~ 0 \tag{2} \end{alignat} $$

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

How do I find the angle α & β from these two equations?

Edit 2:
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I have re-organized the equation and square it as such:
cos²a = (619² cos²β + 60² + 2(619cosβ * 60)) / 313.9²
sin²a = (619² sin²β + 882.9² - 2(619sinβ * 882.9)) / 313.9²

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  • $\begingroup$ Providing a diagram would be really useful here. $\endgroup$ – Gert Feb 29 '16 at 3:01
  • $\begingroup$ @Gert I will edit the post to include a free body diagram and what I've worked so far from below's guide $\endgroup$ – Kai Feb 29 '16 at 3:05
  • $\begingroup$ @Gert I have edited accordingly, though I'm not too sure whether my workings is correct or not. $\endgroup$ – Kai Feb 29 '16 at 3:55
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    $\begingroup$ The squaring of the right hand sides is not correct. You forgot the cross products of the terms. $(x+y)^2 = x^2+y^2+2xy$. I ** strongly ** suggest that you work hard to improve your algebra or you will get totally lost in physics. $\endgroup$ – Bill N Feb 29 '16 at 4:28
  • $\begingroup$ Truth be told my algebra is a little bit rusty. If taken on this context, squaring the right hand side will actually be (-619cosβ - 60)² / (-313.9)² which is then be -619²cos²β - 60² - 2(-619cosβ60)? @BillN $\endgroup$ – Kai Feb 29 '16 at 4:51
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This is a typical solution trick when you have a system involving sine and cosine of the same unknown angle: Re-organize the equations so that you have $$\cos\alpha = stuff $$ and $$\sin\alpha = other stuff.$$ Square these equations and add them. The angle $\alpha$ is eliminated because $$\sin^2\alpha + \cos^2\alpha = 1.$$

In your case, it's nice that the coefficiencts on both $\alpha$ terms are the same, and also on both $\beta$ terms. You can use a double angle formula for the remaining $\beta$ terms to solve for $\beta.$ Then you can solve for $\alpha$. Remember that tool, and teach it to someone else.

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  • $\begingroup$ Hi, I will try to solve using this method, I have a quick question though. By squaring these equations (after I re-organizing) do I square the left hand side of the equation as well? (cosα & sinα) $\endgroup$ – Kai Feb 29 '16 at 2:50
  • $\begingroup$ Of course...do the algebra correctly. I assume you know algebra! $\endgroup$ – Bill N Feb 29 '16 at 4:18
  • $\begingroup$ just basic ones though, not involving cos /sin functions, however would you like to take a look at my edit and see whether am I heading in the right direction? thanks! :) $\endgroup$ – Kai Feb 29 '16 at 4:24
  • $\begingroup$ Also if you divide both sides you will have $\tan \alpha = \frac{\ldots}{\ldots}$ which is solved for $\alpha$. $\endgroup$ – ja72 Feb 29 '16 at 5:58
  • $\begingroup$ @ja72 No, because the right-hand sides contain sums of trig functions of the other angle, $\beta$. $\endgroup$ – Bill N Feb 29 '16 at 12:45
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You can eliminate the angle $\alpha$ from the equations with the trick the other answers give you *. But then you will end up with an equation of the form

$$ A \cos \beta + B \sin \beta + C = 0$$

To solve this do the following transformation

$$ \left. \begin{align} A & = R \cos \psi \\ B & = R \sin \psi \end{align} \right\} \begin{aligned} R & = \sqrt{A^2+B^2} \\ \psi & = \arctan\left( \frac{B}{A} \right) \end{aligned} $$

The equation is now $$ cos\beta\cos\psi + \sin\beta \sin\psi = \cos(\beta-\psi) = -\frac{C}{R} $$

which is solved for

$$ \begin{split} \beta & = \arccos\left( -\frac{C}{R} \right) + \psi \\ & = \arccos\left( -\frac{C}{\sqrt{A^2+B^2}} \right) + \arctan\left( \frac{B}{A} \right)\end{split}$$

footnotes:

  • make the equations of this form $$\begin{align} \cos \alpha & = a \cos\beta+c_x \\ \sin \alpha & = -a \sin \beta + c_y \end{align}$$
  • square both sides and add them for $$ 1 = 2 a c_x \cos\beta - 2 a c_y \sin\beta + c_x^2 + c_y^2 +a^2 $$ $$ \left(2 a c_x\right) \cos\beta + \left(- 2 a c_y\right) \sin\beta + \left(c_x^2 + c_y^2 +a^2-1\right) = 0 $$
  • Match the $A$, $B$ and $C$ coefficients.
  • Once $\beta$ is known, then divide the two equations above for $$ \tan \alpha = \frac{c_y - a \sin\beta}{c_x + a \cos\beta} $$

Edit 1

Here is the actual solution:

$$\left. \begin{align} -313.9 \cos(\alpha) + 619 \cos(\beta) + 60 & = 0 \\ 313.9 \sin(\alpha) + 619 \sin(\beta) - 882.9 & = 0 \end{align} \right\} \begin{aligned} 313.9 \cos(\alpha) & = 619 \cos(\beta) + 60 \\ 313.9 \sin(\alpha) & = - 619 \sin(\beta) + 882.9 \end{aligned} $$

Square and add the two equations (on each side) to get

$$ \left. 98533.21 = 74280 \cos(\beta) - 1093030.2 \sin(\beta) + 1166273.41 \right\}\\ 74280 \cos(\beta) - 1093030.2 \sin(\beta) + 1067740.2 = 0 $$

$$ \begin{aligned} \beta & = \arccos\left( -\frac{C}{\sqrt{A^2+B^2}} \right) + \arctan\left( \frac{B}{A} \right) \\ A & = 74280\\ B & = -1093030.2 \\ C & = 1067740.2\\ \beta &= 1.41284652 = 80.9501426° \\ \end{aligned} $$

Finally, $\alpha$ can be solved with the 2nd equation:

$$ \sin(\alpha) = 2.81267919-1.97196559 \sin(\beta) $$ $$ \alpha = 1.04567064 = 59.9125144° $$

Now you can plug the values of $\alpha$ and $\beta$ into the two original equations to confirm it balances the forces.

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  • $\begingroup$ Hi! May I know how did you derived the first 2 equation: cosα = acosβ+cx & sinα = −asinβ+cy from? $\endgroup$ – Kai Feb 29 '16 at 7:23
  • $\begingroup$ In the OPs case, $A$ and $B$ are the same making the solution much easier than the general case you have outlined. $\endgroup$ – Bill N Feb 29 '16 at 12:52
  • $\begingroup$ Hi I have to retract back as using this I didn't get the correct answer, α = 55.58°, β = 79.07° $\endgroup$ – Kai Mar 1 '16 at 0:41
  • $\begingroup$ The methods above are correct. The mistake must be somewhere else. $\endgroup$ – ja72 Mar 1 '16 at 12:35
  • $\begingroup$ Your $\alpha=55.58°$ and $\beta=79.07°$ do not solve the system of equations. What does solve the system of equations is $$ \begin{align} \alpha & = 59.912514710332826° \\ \beta & = 80.950143077864119° \end{align} $$ which I get with my equations above. $\endgroup$ – ja72 Mar 2 '16 at 6:14

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