# Applications of vectors: Net force

Net force. Three forces are applied to an object, as shown in the figure. Find the magnitude and direction of the sum of forces.

So I broke each force into its horizontal and vertical components and then added all the corresponding points to get <-93.7,-29.2>. Then I computed the magnitude and got approximately 98. Then to find the direction I used inverse tangent for the horizontal and vertical component sum and got approximately 17.3 degrees. Now how do I find out what direction that angle is pointing?

• It seems unlikely that the net force is pointing slightly above the horizontal axis, no? – fosho Jan 31 '16 at 21:41
• Components of your vector are both negative, so it must be in the third quadrant. – Aretino Jan 31 '16 at 21:58

The function $\tan(\theta)$ is periodic with period $\_\_\_$, which just means that angles differing by a multiple of $\_\_\_$ have the same tangent.

So, you can always replace $17.3^\circ$ with $17.3^\circ + \ \_\_\_$ (really, you can add any integer multiple of $\_\_\_$ to the $17.3^\circ$) to get an angle with the same tangent. Since your point has negative $x$- and $y$-coordinates, the angle had better be in the third quadrant, between $180^\circ$ and $270^\circ$.

Let the resultant vector be $\rm \vec{F}=F_x\hat{i} + F_y\hat{j}$. ( right = $\rm+x$ , top = $\rm +y$ )

\begin{align}\rm F_x &= \rm |\vec{F_2}| \cos(30^\circ) - |\vec{F_1}|\cos (45^\circ) - |\vec{F_3}| \cos (60^\circ) \\ &= -93.7\end{align}

\begin{align}\rm F_y &= \rm |\vec{F_2}| \sin(30^\circ) + |\vec{F_1}|\sin (45^\circ) - |\vec{F_3}| \sin (60^\circ) \\ &= -29.2\end{align}

$$\therefore \theta \approx \arctan\frac{-29.2}{-93.7} = 17.3^\circ$$

Now, we know $\tan \theta$ is positive in $I$ and $III$ quadrant, or $\theta \in(0,90^\circ)\cup(180^\circ,270^\circ)$. So you can either find $\cos\theta$ or $\sin\theta$ and show that they are negative, so the angle must be in $III$ quadrant, or you can just visualise it graphically: