Q2E - University Physics with Modern Physics

Question

To extricate an SUV stuck in the mud, workmen use three horizontal ropes, producing the force vectors shown in Fig. E4.2.

 

(a) Find the x- and y-components of each of the three pulls.

 

(b) Use the components to find the magnitude and direction of the resultant of the three pulls.

Step-by-Step Solution

Answer

(a) The X- and  Y-components of each of the three pulls are$\left\{-417.58\hat{i}+668.26\hat{j}\right\} N$ , $\left\{844.3\hat{i}+507.3\hat{j}\right\} N$  and $\left\{-247.35\hat{i}-328.24\hat{j}\right\} N$ .

(b) The magnitude of the resultant of the three pulls is $866.1 N$  which makes an angle  $78°$ with the   X-axis.

Step 1: Given data

The force from the first workman is$788 \text{N}$making an angle of ${\theta }_A=32°$  with the  Y axis.

The force from the second workman is 985N  making an angle of   ${\theta }_B=31°$with the  X axis.

The force from the first workman is  411N  making an angle of  ${\theta }_C=53°$ with the   X-axis.

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

  $$\begin{gathered} A=\sqrt{A_x^2+A_y^2} .....\left(2\right) \\ tan\theta =\frac{A_y}{A_x} .....\left(3\right) \end{gathered}$$

Here, $\theta$  is the angle made with the  X axis

Step 3: Components of the forces

From equation (1), the force from the first workman can be written as

 $$\begin{gathered} \overrightarrow{A}=\left\{-788\cos \left(90°-32°\right)\hat{i}+788\sin \left(90°-32°\right)\hat{j}\right\} \text{N} \\ =\left\{-417.58\hat{i}+668.26\hat{j}\right\} \text{N} \end{gathered}$$

From equation (1), the force from the second workman can be written as

 $$\begin{gathered} \overrightarrow{B}=\left\{985\cos \left(31°\right)\hat{i}+985\sin \left(31°\right)\hat{j}\right\} \text{N} \\ =\left\{844.3\hat{i}+507.3\hat{j}\right\} \text{N} \end{gathered}$$

From equation (1), the force from the third workman can be written as

 $$\begin{gathered} \overrightarrow{C}=\left\{-411\cos \left(53°\right)\hat{i}-411\sin \left(53°\right)\hat{j}\right\} \text{N} \\ =\left\{-247.35\hat{i}-328.24\hat{j}\right\} \text{N} \end{gathered}$$

Step 4: Resultant force

 

The vector sum of the three forces are

  $\begin{array}{rcl}\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}& =& \left\{-417.58\hat{i}+668.26\hat{j}\right\} \text{N}+\left\{844.3\hat{i}+507.3\hat{j}\right\} \text{N}+\left\{-247.35\hat{i}-328.24\hat{j}\right\} \text{N}\\ & =& \left\{\left(-417.58+844.3-247.35\right)\hat{i}+\left(668.26+507.3-328.24\right)\hat{j}\right\} \text{N}\\ & & \text{ = }\left\{179.37\hat{i}-847.32\hat{j}\right\} \text{N}\\ & & \end{array}$

From equations (2) and (3), the magnitude and direction of the resultant force is

  $$\begin{gathered} F=\sqrt{{179.37}^2+{\left(-847.32\right)}^2} \text{N} \\ =866.1 \text{N} \end{gathered}$$

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{847.32}{179.37}\right) \\ =78° \end{gathered}$$

  

Thus, the resultant force is $866.1 \text{N}$  and makes an angle $78°$   with the X  axis.