I'm getting two different answers for the below problem depending on how I represent the relevant vector. My guess is the inverse sine function is affecting things, but I don't understand how I need to compensate.
Problem: A weight W is suspended from two lines. The force on each line are A and B at angles a, b. Solve for angle a.
Let $-W = -50\ \text{N},\ A = 35\ \text{N},\ b = 60^\circ.$ (The answer is $a = 74.42^\circ$)
\begin{align}
\pmb{A}
&= A<\cos c, \sin c> \text{OR}\\
&= A<-\cos a, \sin a>\\\\
\pmb{B}
&= B<\cos b, \sin b>\\\\
\pmb{W}
&= W<0, -1>\\
\end{align}
Here's my work for getting $\pmb{A}$:
I verified that both of these forms make the same vector:
\begin{align}
\pmb{A}
&= 35<-\cos 74, \sin 74>\\
&= 35<\cos(180-74), \sin(170-74)>\\
&= 35<-0.28, 33.6>
\end{align}
Below is the approach I used to find the correct value of $a$. It uses $\pmb{A} = A<\cos c, \sin c>\text{, where }c = 180 – a$:
x direction:
\begin{align}
0 &= A\cos c + B\cos b\\
B &= -A\frac{\cos c}{\cos b}
\end{align}
y direction:
\begin{align}
0 &= -W + A\sin c + B\sin b.\ \text{(Substitute for B.)}\\\\
W &= A\sin c -A\frac{\cos c}{\cos b}\sin b\\\\
\frac{W}{A} &= \sin c – \frac{\cos c}{\cos b}\sin b\\\\
\frac{W}{A}\cos b &= \sin c\cos b – \cos c\sin b\\
&= \sin(c – b)\\\\
c – b &= \arcsin(\frac{W}{A}\cos b)\\
180 – a – b &= \arcsin(\frac{W}{A}\cos b)\\\\
a &= 180 – b – \arcsin(\frac{W}{A}\cos b)\\
&= 180 – 60 – \arcsin(\frac{50}{35}\cos 60)\\
&= 74.42^\circ
\end{align}
Now, here is the approach yielding the incorrect value of $a$ where I began with $a$ directly. $\pmb{A} = A<-\cos a, \sin a>$:
x direction:
\begin{align}
0 &= -A\cos a + B\cos b\\
B &= A\frac{\cos a}{\cos b}
\end{align}
y direction:
\begin{align}
0 &= -W + A\sin a + B\sin b.\ \text{(Substitute for B.)}\\\\
W &= A\sin c + A\frac{\cos a}{\cos b}\sin b\\\\
\frac{W}{A} &= \sin a + \frac{\cos a}{\cos b}\sin b\\\\
\frac{W}{A}\cos b &= \sin a\cos b + \cos a\sin b\\
&= \sin(a + b)\\\\
a + b &= \arcsin(\frac{W}{A}\cos b)\\
a &= – b + \arcsin(\frac{W}{A}\cos b)\\
&= -60 + \arcsin(\frac{50}{35}\cos 60)\\
&= -14.42^\circ
\end{align}
These two answers, $a_1 = 74$ and $a_2 = -14$, make $60^\circ$ when added together. Again, my guess is the $\arcsin$ is causing issues, but I don't see what the logical error is. In an effort to see what's happening or if a compensating term would mean something to me, I wrote the equation for adding these two answers:
\begin{align}
\text{Let}\ X &= \arcsin(\frac{W}{A}\cos b)\\
a_1 + a_2 &= (180 – b – X) + (-b + X)\\
&= 180 – 2b\\
a_1 &= 180 – 2b – a_2\\
\end{align}
I'm left with the odd sensation where, looking at the algebra, it seems clear that the angles I'm getting would be different, but at the same time I would also expect that starting with an equivalent vector $\pmb{A}$, I ought to get the same final answer for the angle $a$. I've heard of compensating by 180, but $180 – 2b$ was unexpected.
Lastly, does the $2b$ suggest the discrepancy is related to the double-angle property? I looked into it and didn't feel it was leading anywhere.
Best Answer
Yes, arc sine is a tricky tool. It takes a function that produces the same value infinitely many times, and supposes that you can invert that function. The way we get this to work usually is we only count on it when the result should be between $-90$ and $90$ degrees ($-\frac\pi2$ to $\frac\pi2$ radians).
Given $b = 60$ degrees, the only way we could possibly have $a + b = \arcsin(\text{anything})$ is if $a$ is between $-150$ and $30$ degrees. It isn't, so this doesn't work.
The "correct" method is fraught with danger too. If the actual correct result had put $a < 30$ degrees, that would imply $c > 150$ degrees, which would imply $c - b > 90$ degrees, so $c - b = \arcsin(\text{whatever})$ would be impossible.
I would not use either method. I would construct the force diagram as shown below, where the vector sum comes out to $\vec A + \vec B + \vec W = 0$ and where the vectors $-\vec B$ and $\vec W$ make an angle of $30$ degrees. We then want to find the angle between $\vec A$ and $\vec W$ so that $A = 35.$ Based only on that information, there would be two possible solutions for $\vec A$ as shown in the figure below, but in this case only one of those solutions would have a positive vertical component (which is required by the setup of the problem), so that is the solution to choose.
To work this out, let $\theta$ be the angle between $\vec A$ and $\vec B.$ Then by working out the components of $\vec W$ and $\vec A$ perpendicular to $\vec B,$ we get $$ 35 \sin\theta = A\sin\theta = W \sin 30^\circ = 50 \sin 30^\circ = 25 $$ and therefore $$ \sin\theta = \frac{25}{35} = \frac 57.$$ One solution is $\theta_1 = \arcsin \frac57 \approx 45.58^\circ.$ Another solution is $\theta_2 = 180 - \arcsin \frac57 \approx 134.42^\circ.$ Now, realizing that $a,$ $b,$ and $\theta$ are the three interior angles of the triangle in the original diagram, we have $$ a = 180^\circ - b - \theta = 120^\circ - \theta, $$ which leads to the two answers \begin{align} a_1 &= 120^\circ - \theta_1 = \phantom{-}74.42^\circ, \\ a_2 &= 120^\circ - \theta_2 = -14.42^\circ. \\ \end{align} So the two answers you found correspond to the two solutions of the triangular force diagram. Which method found which solution was simply a matter of which solution would put the quantity in the sin function (either $a + b$ or $c - b$) into the range $[90^\circ,90^\circ].$
Also notice that because $\theta_1$ and $\theta_2$ are the two solutions of $\arcsin\frac57$ that fall in the range $[0,180^\circ],$ we have $\theta_1+\theta_2=180^\circ.$ So if if we add $a_1$ and $a_2,$ calculated according to the equations above, We get \begin{align} a_1 + a_2 &= (120^\circ - \theta_1) + (120^\circ - \theta_1) \\ &= 240^\circ - (\theta_1 + \theta_2) \\ &= 240^\circ - 180^\circ \\ &= 60^\circ. \end{align} The angles that actually get "compensated" by $180^\circ$ are $\theta_1$ and $\theta_2$, because those are the angles whose sine is actually taken.
By the way, if we try to generalize the problem to any given values of $b,$ $A,$ and $W,$ we can end up with no solution, or we can end up with two valid solutions to the same problem. For example, if we merely change $A$ to $27,$ while keeping $b=60^\circ$ and $W=50,$ then $a_1$ and $a_2$ will both be positive and will represent two equally possible configurations of the lines holding up the weight.