91Ó°ÊÓ

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

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

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

The X and Y components of a vector $\stackrel{→}{A}$ making an angle $\mathrm{θ}$ with the X- axis can be summarized as

$\stackrel{→}{A}=Acos\mathrm{θ}\hat{i}+Asin\mathrm{θ}\hat{j}.....\left(1\right)$

The magnitude and direction of a vector $\stackrel{→}{A}=A_x\hat{i}+A_y\hat{j}$ is

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

Here, $\mathrm{θ}$ is the angle made with the X axis

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

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

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

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

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

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

The vector sum of the three forces are

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

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

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

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

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