Chapter 8.4, Problem 1MCQ

To determine

To find: the magnitude of horizontal and vertical components by first drawing the given vector with the help of ruler and protector.

Answer to Problem 1MCQ

  $\begin{array}{l}h=1.6\text{ cm}\\ v=1.7\text{ cm}\end{array}$

Explanation of Solution

Given information:

The magnitude and direction of given vector is as follows:

  $\begin{array}{l}\text{Magnitude}=2.3\text{ cm}\\ \text{Direction}=46°\end{array}$

Calculation:

The figure of given vector for 2.3 cm magnitude and angle direction $46°$ is shown below.

Let the horizontal and vertical component be $h$ and $v$ respectively.

The horizontal component of the vector is evaluated as follows:

  $\begin{array}{c}h=\text{magnitude}\times \text{cos}\left(\text{direction}\right)\\ =2.3\cos \left(46°\right)\left[\text{Put the value of magnitude and direction}\right]\\ =1.6\text{ cm}\end{array}$

The vertical component of the vector is evaluated as follows:

  $\begin{array}{c}v=\text{magnitude}\times \sin \left(\text{direction}\right)\\ =2.3\sin \left(46°\right)\left[\text{Put the value of magnitude and direction}\right]\\ =1.7\text{ cm}\end{array}$

Thus the horizontal and vertical component of the given vector is $1.6\text{ cm}$ and $1.7\text{ cm}$ respectively.