Chapter 8.5, Problem 21E

To determine

To find: the magnitude and direction of the third force.

Answer to Problem 21E

Magnitude: 39.8 N

Direction: $270°$

Explanation of Solution

Given information:

Let the components be $\overrightarrow{F_1}\text{ and }\overrightarrow{F_2}$ .

  $\begin{array}{c}\overrightarrow{F_1}=\overrightarrow{F_2}=23\text{ N}\\ {\theta }_1=60°\\ {\theta }_2=120°\end{array}$

The third force produces equilibrium.

Calculation:

Now resolve the components as shown below.

  $\begin{array}{c}\overrightarrow{F_1}=\left(23\cos 60°\right)\overrightarrow{i}+\left(23\sin 60°\right)\overrightarrow{j}\\ =11.5\overrightarrow{i}+11.5\sqrt{3}\overrightarrow{j}\\ \overrightarrow{F_2}=\left(23\cos 120°\right)\overrightarrow{i}+\left(23\sin 120°\right)\overrightarrow{j}\\ =-11.5\overrightarrow{i}+11.5\sqrt{3}\overrightarrow{j}\end{array}$

The magnitude of the resultant vector is evaluated as follows:

  $\begin{array}{c}\left|\overrightarrow{F_1}+\overrightarrow{F_2}\right|=\sqrt{{\left(11.5-11.5\right)}^2+{\left(11.5\sqrt{3}+11.5\sqrt{3}\right)}^2}\\ =\sqrt{{\left(0\right)}^2+{\left(23\sqrt{3}\right)}^2}\\ =23\sqrt{3}\\ \approx 39.8\text{ N}\end{array}$

The direction of the resultant vector is evaluated as follows:

  $\begin{array}{c}\tan \theta =\frac{11.5\sqrt{3}+11.5\sqrt{3}}{11.5-11.5}\\ \tan \theta =\frac{23\sqrt{3}}{0}\\ =\text{undefined}\end{array}$

The value of $\tan \theta$ is undefined but the vertical component is positive, so $\theta =90°$ but in opposite direction i.e. $180°+\theta =270°$ .

Thus the magnitude and direction of resultant vector is 39.8 N and $270°$ respectively.