Chapter 8, Problem 58SGA

To determine

To find: the magnitude and direction of the resultant of the given forces.

Answer to Problem 58SGA

Magnitude: 154.63 N

Direction: $13.08°$

Explanation of Solution

Given information:

Let the components be $\overrightarrow{F_1}\text{ and }\overrightarrow{F_2}$ . The force 90 N $\left(\overrightarrow{F_1}\right)$ is acting along x -axis and force 70 N $\left(\overrightarrow{F_2}\right)$ is making $30°$ angle with one another.

Calculation:

Resolve the components as shown below.

  $\begin{array}{l}\overrightarrow{F_1}=90\overrightarrow{i}\\ \overrightarrow{F_2}=\left(70\cos 30°\right)\overrightarrow{i}+\left(70\sin 30°\right)\overrightarrow{j}\end{array}$

The magnitude of the resultant vector is evaluated as follows:

  $\begin{array}{c}\left|\overrightarrow{F_1}+\overrightarrow{F_2}\right|=\sqrt{{\left(90+70\cos 30°\right)}^2+{\left(70\sin 30°\right)}^2}\\ =\sqrt{{\left(150.62\right)}^2+{\left(35\right)}^2}\\ =\sqrt{23911.28}\\ \approx 154.63\text{ N}\end{array}$

The direction of the resultant vector is evaluated as follows:

  $\begin{array}{c}\tan \theta =\frac{70\sin 30°}{90+70\cos 30°}\\ \theta ={\tan }^{-1}\left(\frac{70\sin 30°}{90+70\cos 30°}\right)\\ ={\tan }^{-1}\left(0.23237\right)\\ \approx 13.08°\end{array}$

Thus the magnitude and direction of resultant vector is 154.63 N and $13.08°$ respectively.