Chapter 8.4, Problem 2MCQ

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 2MCQ

  $\begin{array}{l}h=11.4\text{ mm}\\ v=24.9\text{ mm}\end{array}$

Explanation of Solution

Given information:

The magnitude and direction of given vector is as follows:

  $\begin{array}{l}\text{Magnitude}=27\text{ mm}\\ \text{Direction}=245°\end{array}$

Calculation:

The figure of given vector for 2.7 mm magnitude and angle direction $245°$ is shown below.

The exterior angle of the triangle is evaluated as follows:

  $360°-245°=115°$

The acute angles $\left(x\text{ and }y\right)$ of triangle can be evaluated as follows:

  $\begin{array}{c}y=180°-115°\left[\text{Linear angle property}\right]\\ =65°\\ \\ 90°+y+x=180°\left[\text{Angle }\text{}\text{sum property of triangles}\right]\\ 90°+65°+x=180°\\ x=180°-\left(90°+65°\right)\\ =25°\end{array}$

Thus the direction of the vector is $65°$ .

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

The horizontal component of the vector is evaluated as follows:

  $\begin{array}{c}h=\text{magnitude}\times \text{cos}\left(\text{direction}\right)\\ =27\cos \left(65°\right)\left[\text{Put the value of magnitude and direction}\right]\\ =11.4\text{ mm}\end{array}$

The vertical component of the vector is evaluated as follows:

  $\begin{array}{c}v=\text{magnitude}\times \sin \left(\text{direction}\right)\\ =27\sin \left(65°\right)\left[\text{Put the value of magnitude and direction}\right]\\ =24.9\text{ mm}\end{array}$

Thus the horizontal and vertical component of the given vector is $11.4\text{ mm}$ and $24.9\text{ mm}$ respectively.