Chapter 8.5, Problem 8CFU

To determine

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

Answer to Problem 8CFU

Magnitude: 74 N

Direction: $253°\text{ in respect to }\left(+x\right)$

Explanation of Solution

Given information:

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

  $\begin{array}{c}\overrightarrow{F_1}=33\text{ N}\\ {\theta }_1=90°\\ \overrightarrow{F_2}=44\text{ N}\\ {\theta }_2=60°\end{array}$

The third force produces equilibrium on the object.

Calculation:

Now resolve the components as shown below.

  $\begin{array}{c}\overrightarrow{F_1}=\left(33\cos 90°\right)\overrightarrow{i}+\left(33\sin 90°\right)\overrightarrow{j}\\ =33\overrightarrow{i}\\ \overrightarrow{F_2}=\left(44\cos 60°\right)\overrightarrow{i}+\left(44\sin 60°\right)\overrightarrow{j}\\ =22\overrightarrow{i}+22\sqrt{3}\overrightarrow{j}\end{array}$

The magnitude of the resultant vector is evaluated as follows:

  $\begin{array}{c}\left|\overrightarrow{F_1}+\overrightarrow{F_2}\right|=\sqrt{{\left(22\right)}^2+{\left(33+22\sqrt{3}\right)}^2}\\ \approx 74\text{ N}\end{array}$

The direction of the resultant vector is evaluated as follows:

  $\begin{array}{c}\tan \theta =\frac{33+22\sqrt{3}}{22}\\ \theta ={\tan }^{-1}\left(\frac{3+2\sqrt{3}}{2}\right)\\ \approx 73°\end{array}$

Since third force produces equilibrium on the object, so magnitude of this resultant force will be same but the direction will be opposite, i.e. $180°+\theta =253°$ .

Thus the magnitude and direction of resultant vector is 74 N and $253°$ in respect to $\left(+x\right)$ respectively.