Chapter 5, Problem 81A

To determine

Magnitude and direction of third force.

Answer to Problem 81A

  $F_3=74.4\text{N,}253°$

Explanation of Solution

Given:

The given forces are:

  $\begin{array}{l}{F}_1=33\text{N,}\text{90}°\\ F_2=44\text{N,}60°\\ F_3=?\end{array}$

Formula Used:

Vector resolution along x and y axis.

  $\begin{array}{l}{F}_x=F\cos \theta \\ F_y=F\sin \theta \end{array}$

Where, $F$ is a vector with angle $\theta$$F_x$ and $F_y$ are components along x and y axis.

Calculation:

For three forces to be in equilibrium, the magnitude of third force is equal to the vector sum of other two forces and the direction will be opposite of the resultant.

So, resolve and add the two given forces

  $\begin{array}{c}{F}_{1x}=F_1\cos {\theta }_1\\ =33\times \cos 90°\\ =0\text{N}\\ F_{1y}=F_1\sin {\theta }_1\\ =33\times \sin 90°\\ =33\text{N}\\ F_{2x}=F_2\cos {\theta }_2\\ =44\times \cos 60°\\ =22\text{N}\\ F_{2y}=F_1\sin {\theta }_2\\ =44\times \sin 60°\\ =38.1\text{N}\end{array}$

Now,

  $\begin{array}{c}{F}_{3x}=F_{1x}+F_{2x}\\ =0+22\\ =22\text{N}\\ F_{3y}=F_{1y}+F_{2y}\\ =33+38.1\\ =71.1\text{N}\\ F_3=\sqrt{F_{3x}{}^2+F_{3y}{}^2}\\ =\sqrt{{22}^2+{71.1}^2}\\ =74.4\text{N}\end{array}$

The direction is,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{F_{3y}}{F_{3x}}\right)+180°\\ ={\tan }^{-1}\left(\frac{71.1}{22}\right)+180°\\ =253°\end{array}$

Conclusion:

Therefore, the third force is $F_3=74.4\text{N,}253°$ .