Q5E - University Physics with Modern Physics

Question

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?

 

Step-by-Step Solution

Answer

(a) The X and Y -components of the resultant force are $-8.1 N$ and  3N respectively.

(b) The magnitude of the resultant force is  8.64N

Step 1: Given data

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.

Step 2: Vector components

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

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

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

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

Step 3: Components of the resultant forces

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

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

Step 4: Magnitude of the resultant force

 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  $8.64 \text{N}$