Vector Components - Superposition of Forces

Question

I understand that by inspection of the given figure, $F_{1,x}$ (the x-component of $F_1$) must $< 0$, so $F_{1,x} = 250\color{red}{\cos{53^{\circ} }}$ can't be right. But I don't see how $F_{1,x} = (250)\color{red}{\cos{127^{\circ} }}$?

By definition, $F_x = F\cos\theta$ where $F$ is the hypotenuse of a right-angle triangle with $F_x, F_y$ as sides. Working with $127^{\circ}$ does not yield a right-angle triangle. Thank you very much!

---

@Maesumi: Thanks for your response. I understand that $127^{\circ}$ measured from the positive $x$-axis is the same as $-53^{\circ}$ measured from the negative $x$-axis. But knowing this, I still have my questions above.

---

@Maesumi: Thanks for your 2nd response. I understand reference angles but I don't see why $F_{1,x} = (250)\color{red}{\cos{127^{\circ} }} $ makes sense graphically. From my picture, $\cos{127^{\circ}}$ can be found using Cosine Law, but it doesn't involve the pink line, $F_{1,x}$, which is what we want.

Answer 1

If you measure the angle $\theta′ = 180 - θ$ from negative side of $x$-axis then you should use

$$F_x=−F\cosθ′=F\cos(180 - θ).$$ This angle is called the reference angle and one uses the supplementary angle identity $$ \cos θ′= \cos(180 - θ) = -\cosθ $$.

Answer 2

Let's start with abstract vector notation, then we will use rectangular and polar coordinates as convenient.

In vector notation, we have $F_{net}=\sum F_i=F_1+F_2+F_3$.

Since addition of vectors works better in rectangular coordinates, and the data is give to us in polar coordinates, that directs us to our first step.

$F_{net}=F_{net,x}+F_{net,y}$. Let's work on the $x$ component first.

$F_{net,x}=\sum F_{i,x}=250\cos(127)+50+0\approx-100$.

$F_{net,y}=\sum F_{i,y}=250\sin(127)+0-120\approx 80$.

At this point, we have the $x$ and $y$ components of $F_{net}$, and we may now tackle the problem of converting $F_{net}$ to polar coordinates.