Forces ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ act at a point. The magnitude of ${\overrightarrow{F}}_1$ is 9.00 N, and its direction is ${60}^{\circ }$ above the x-axis in the second quadrant. The magnitude of ${\overrightarrow{F}}_2$is 6.00 N, and its direction is below the x-axis in the third quadrant.

(a) What are the x- and y-components of the resultant force?

(b) What is the magnitude of the resultant force?

The magnitude of ${\overrightarrow{F}}_1$ is 9.00 N, and its direction is $60°$ above the x-axis in the second quadrant.

The magnitude of ${\overrightarrow{F}}_2$ is 6.00 N, and its direction is $53.1°$ below the x-axis in the third quadrant.

The X and Y components of a vector $\overrightarrow{A}$making an angle $\theta$ with the X axis can be summarized as

role="math" localid="1659392035683" $\overrightarrow{A}=Acos\theta \hat{i}+Asin\theta \hat{j}.....\left(1\right)$

The magnitude of a vector $\overrightarrow{A}=A_x\hat{i}+A_y\hat{j}$ is

$A=\sqrt{A_x^2+A_y^2}.....\left(2\right)$

From equation (1), ${\overrightarrow{F}}_1$ can be written as

$\begin{array}{rcl}{\overrightarrow{F}}_1& =& \left\{-9\cos \left(60°\right)\hat{i}+9\sin \left(60°\right)\hat{j}\right\}\text{N}\\ & =& \left\{-4.5\hat{i}+7.8\hat{j}\right\}\text{N}\\ & & \end{array}$

From equation (1), ${\overrightarrow{F}}_2$ can be written as

$\begin{array}{rcl}{\overrightarrow{F}}_2& =& \left\{-6\cos \left(53.1°\right)\hat{i}-6\sin \left(53.1°\right)\hat{j}\right\}\text{N}\\ & =& \left\{-3.6\hat{i}-4.8\hat{j}\right\}\text{N}\\ & & \end{array}$

Thus, the resultant force is

$\begin{array}{rcl}\overrightarrow{F}& =& {\overrightarrow{F}}_1+{\overrightarrow{F}}_2\\ & =& \left\{-4.5\hat{i}+7.8\hat{j}\right\}\text{N}+\left\{-3.6\hat{i}-4.8\hat{j}\right\}\text{N}\\ & =& \left\{\left(-4.5-3.6\right)\hat{i}+\left(7.8-4.8\right)\hat{j}\right\}\text{N}\\ & =& \left\{-8.1\hat{i}+3\hat{j}\right\}\text{N}\\ & & \end{array}$

From equation (2), the magnitude of the resultant force is

$$\begin{gathered} F=\sqrt{{8.1}^2+3^2}\text{N} \\ =8.64\text{N} \end{gathered}$$

Thus, the resultant force is $8.64\text{N}$