Chapter 5.1, Problem 13SSC

To determine

To Calculate: The sum of three vectors.

Explanation of Solution

Given:

Vector K $=-4.0$

Vector M $=5.0\left(37.0°\right)$ above horizontal

Vector L $=6.0$

Formula Used:

For addition of vectors, add the components of each vector in same direction and then find the resultant.

Horizontal and vertical components of a vector A can be obtained by:

  $\begin{array}{l}{A}_x=A\cos \theta \\ A_y=A\sin \theta \end{array}$

Resultant can be obtained by:

  $R=\sqrt{R_x{}^2+R_y{}^2}$

The angle of the resultant from the horizontal can beobtained by:

  $\theta ={\tan }^{-1}\left(\frac{R_y}{R_x}\right)$

Calculation:

Components of vector K :

  $\begin{array}{l}{K}_x=-4.0\\ K_y=0\end{array}$

Components of vector M :

  $\begin{array}{l}{M}_x=5.0\cos 37°=3.99\\ M_y=5.0\sin 37°=3.00\end{array}$

Components of vector L :

  $\begin{array}{l}{L}_x=6.0\\ L_y=0\end{array}$

Component of resultant in x -direction:

  $\begin{array}{c}{R}_x=K_x+L_x+M_x\\ =-4.0+6.0+3.99\\ =5.99\end{array}$

Component of resultant in y -direction:

  $\begin{array}{c}{R}_y=K_y+L_y+M_y\\ =0+0+3.0\\ =3.0\end{array}$

Magnitude of resultant can be calculated as follows:

  $\begin{array}{c}R=\sqrt{R_x{}^2+R_y{}^2}\\ =\sqrt{{5.99}^2+{3.0}^2}\\ =6.7\end{array}$

The angle of the resultant from the horizontal can be calculated as follows:

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{R_y}{R_x}\right)\\ ={\tan }^{-1}\left(\frac{3.0}{5.99}\right)\\ =26.6°\end{array}$

Conclusion:

Thus, the magnitude and direction of the ant’s displacement is $8.32\text{cm}$ and $19.85°$ respectively.