Chapter 8, Problem 44SGA

To determine

To find: the magnitude and direction of the resultant vector for the given diagram.

Answer to Problem 44SGA

  $\begin{array}{c}\text{Magnitude}=41\text{ m/s}\\ \text{Direction}=108.65°\end{array}$

Explanation of Solution

Given information:

  

Calculation:

Let $\overrightarrow{r}$ be the resultant vector, $x$ be the horizontal component of vector and $y$ be the vertical component of vector. The vectors are as follows:

  $\begin{array}{c}\overrightarrow{v_1}=12\overrightarrow{j}\\ \overrightarrow{v_2}=x\overrightarrow{i}+y\overrightarrow{j}\\ =\left(30\cos 116°\right)\overrightarrow{i}+\left(30sin116°\right)\overrightarrow{j}\end{array}$

The magnitude of $\overrightarrow{r}$ will be evaluated as follows:

  $\begin{array}{c}\left|\overrightarrow{r}\right|=\sqrt{\left|\overrightarrow{v_1}+\overrightarrow{v_2}\right|}\\ =\sqrt{{\left(30\cos 116°\right)}^2+{\left(12+30sin116°\right)}^2}\\ \approx 41\text{ m/s}\end{array}$

Thus the magnitude of resultant vector is $41\text{ m/s}$ .

The direction of $\overrightarrow{r}$ will be evaluated by using the law of tangent as follows:

  $\begin{array}{c}\tan \theta =\frac{y}{x}\\ =\frac{12+30\sin 116°}{30\cos 116°}\\ =\frac{12+30\left(0.898\right)}{30\left(-0.438\right)}\\ =\frac{12+26.94}{-13.15}\\ =-\frac{38.94}{13.15}\\ \theta ={\tan }^{-1}\left(-\frac{38.94}{13.15}\right)\\ \approx 108.65°\end{array}$

Thus the direction of resultant vector is $108.65°$ .