# Angles in a static equilibrium

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:

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

• Providing a diagram would be really useful here.
– Gert
Feb 29 '16 at 3:01
• @Gert I will edit the post to include a free body diagram and what I've worked so far from below's guide
– Kai
Feb 29 '16 at 3:05
• @Gert I have edited accordingly, though I'm not too sure whether my workings is correct or not.
– Kai
Feb 29 '16 at 3:55
• 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. Feb 29 '16 at 4:28
• 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
– Kai
Feb 29 '16 at 4:51

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.

• Hi! May I know how did you derived the first 2 equation: cosα = acosβ+cx & sinα = −asinβ+cy from?
– Kai
Feb 29 '16 at 7:23
• In the OPs case, $A$ and $B$ are the same making the solution much easier than the general case you have outlined. Feb 29 '16 at 12:52
• Hi I have to retract back as using this I didn't get the correct answer, α = 55.58°, β = 79.07°
– Kai
Mar 1 '16 at 0:41
• The methods above are correct. The mistake must be somewhere else. Mar 1 '16 at 12:35
• 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. Mar 2 '16 at 6:14

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.

• 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)
– Kai
Feb 29 '16 at 2:50
• Of course...do the algebra correctly. I assume you know algebra! Feb 29 '16 at 4:18
• 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! :)
– Kai
Feb 29 '16 at 4:24
• Also if you divide both sides you will have $\tan \alpha = \frac{\ldots}{\ldots}$ which is solved for $\alpha$. Feb 29 '16 at 5:58
• @ja72 No, because the right-hand sides contain sums of trig functions of the other angle, $\beta$. Feb 29 '16 at 12:45