# Vectors of force and their angles

This is a homework problem which has me stumped. I have just begun Calculus 3 and this is the first section introducing vectors. .

The embedded picture states the problem and shows some of my hen's scratching in an effort to solve it. After studying the example problem from the same sub-chapter, I understand that the two vectors given, when summed, is the vector that is acting in opposition to the vector pulling the weight down. That is $\mathbf{F_1} + \mathbf{F_2} = \langle 0,50 \rangle$. This leads to the following system of equations:

$$\begin{array}{rcl} -\left|\mathbf{F_1}\right|cos(\alpha) + \left|\mathbf{F_2}\right|cos(60^\circ) & = & 0 \\ \left|\mathbf{F_1}\right|sin(\alpha) + \left|\mathbf{F_2}\right|sin(60^\circ) & = & 50 \\ \end{array}$$

Basically, I'm having troubles figuring out how to eliminate one of the two unknowns: $\alpha$ and $\left|\mathbf{F_2}\right|$. Using the system of equations, I can tell you that $\left|\mathbf{F_2}\right|$ is as follows:

$$\begin{array}{rcl} -\left|\mathbf{F_1}\right|cos(\alpha) + \left|\mathbf{F_2}\right|cos(60^\circ) & = & 0 \\ -35cos(\alpha) + \left|\mathbf{F_2}\right|\frac{1}{2} & = & 0 \\ -70cos(\alpha) + \left|\mathbf{F_2}\right| & = & 0 \\ \left|\mathbf{F_2}\right| & = & 70cos(\alpha) \\ \end{array}$$

Also, knowing that the interior angles of a triangle must sum $180^\circ$, it's also quite a simple matter to demonstrate that $\alpha$ is as follows (assigning $\beta$ to the unnamed angle):

$$\begin{array}{rcl} 180 & = & 60 + \alpha + \beta \\ \alpha & = & 120 - \beta \\ \end{array}$$

From the scratching notes you can see in the picture, you can see that I've tried to draw in the unseen vector $\langle 0,50 \rangle$ in the hopes of making a right triangle to get myself closer to the answer. This didn't help me. I had the inspiration that some of the trigonometric functions and identities I learned (too many years ago now) would be of use. So I pulled out my pre-calc book to look up things like SSA and so forth. However, none seem to be the answer. For example, the SSA only works if you know 2 of the three sides and one angle. Well, I've got to of the three required criteria.

I also refreshed myself on the law of sines. However, this doesn't bring me closer to (what I think are) the solutions. So, one again, I can tell you this much:

$$\text{The law of sines} \\ \frac{sin(A)}{a} = \frac{sin(B)}{b} = \frac{sin(C)}{c} \\ \text{I don't know C, or c, but this shouldn't matter} \\ \Rightarrow \frac{sin(60)}{35} = \frac{sin(\alpha)}{\left|\mathbf{F_2}\right|}$$

However, it's pretty obvious that, though I can calculate the left side, it doesn't bring me close enough to get either $\alpha$ or $\left| \mathbf{F_2} \right|$

So, if someone could kindly point me in the right direction, I'd be very grateful. Thanks.

• Why don't you use $|F_2|=70\cos(\alpha)$ in $|F_1|sin(α)+|F_2|\frac{\sqrt3}{2} = 50$ then you have one equation with one var :) Commented Jan 19, 2015 at 0:58
• @Lu_kors I had thought of that, but quickly rejected it because I still had two variables: right? I don't know what $\alpha$ is and thus don't know $sin(\alpha)$ or $cos(\alpha)$ in addition to not knowing $\left| \mathbf{F_2} \right|$. What am I forgetting that you obviously see? Commented Jan 19, 2015 at 18:06

Using Lami's there

$\frac{F_2}{sin (90+\alpha)}=2F_1=\frac{50}{sin (120-\alpha)}$

$\frac{F_2}{cos \alpha}=2F_1=\frac{50}{sin (120-\alpha)}$

$\frac{F_2}{cos \alpha}=\frac{50}{sin 120cos\alpha-cos 120sin\alpha}$

$\frac{F_2}{cos \alpha}=\frac{50}{\frac{\sqrt3}{2}cos\alpha+\frac{1}{2}sin\alpha}$

$\frac{F_2}{cos \alpha}=\frac{100}{\sqrt3cos\alpha+sin\alpha}$

Substituting your answer above where you got $F_2=70 cos \alpha$,

$70=\frac{100}{\sqrt3cos\alpha+sin\alpha}$

$\sqrt3cos\alpha+sin\alpha=\frac{10}{7}$

By using double angle,

R=2, $\theta=60$

$\sqrt3cos\alpha+sin\alpha=2sin(\alpha+60)$

$2sin(\alpha+60)=\frac{10}{7}$

$sin(\alpha+60)=\frac{5}{7}$

$\alpha+60=45.584$

$\alpha=-14.4 Since$60\le \alpha \le 120, \alpha=-14.4F_2=67.79$. I am getting the angle to be -ve, meaning sin is -ve and cos is +ve in the 4th quadrant. Angles in the 3rd quadrant too keep the sin -ve exceeds 120. It sounds a bit improbable to have a negative angle in real life, but according to the question, it makes sense. Good luck. Use$|F_2|=70\cos(\alpha)$in$|F_1|sin(α)+|F_2|\frac{\sqrt3}{2} = 50$then you have: $$35(\sin{\alpha} + \sqrt{3} \cos\alpha)=0$$ Divide by$35\cos\alpha$$$\tan\alpha = -\sqrt3$$ So you can find$\alpha = -60°$No responsibility for possible calc mistakes :P • I'm sorry but I don't see where you get$35(sin\alpha + \sqrt{3} \cdot cos\alpha) = 0\$ Commented Jan 19, 2015 at 22:40
• Ok I see where the left-hand side comes from, but where do you get 0 on the right? Commented Jan 19, 2015 at 22:47
• jup sry has to be 50, will edit it, asap i have a bit more time Commented Jan 19, 2015 at 23:14
• I should have marked this as answered back then. Sorry, and thanks for the help. Commented Jan 13, 2020 at 19:15