Question

Refer to figure (2 − E1). Find (a) the magnitude, (b) x and y component and (c) the angle with the X-axis of the resultant of \[\overrightarrow{OA}, \overrightarrow{BC} \text { and } \overrightarrow{DE}\].

Answer 1

First, let us find the components of the vectors along the x and y-axes. Then we will find the resultant x and y-components. 
x-component of \[\overrightarrow{OA} = 2 \cos 30^\circ= \sqrt{3}\]

x-component of \[\overrightarrow{BC}\] = 1.5 cos120°

\[= - \frac{\left( 1 . 5 \right)}{2} = - 7 . 5\]

x-component of \[\overrightarrow{DE}\] = 1 cos 270°

  = 1 × 0 = 0 m 

y-component of \[\overrightarrow {OA}\] = 2 sin 30° = 1
y-component of \[\overrightarrow{BC}\] = 1.5 sin 120°

\[= \frac{\left( \sqrt{3} \times 1 . 5 \right)}{2} = 1 . 3\]

y-component of \[\overrightarrow{DE}\] = 1 sin 270° = −1

x-component of resultant \[R_x = \sqrt{3} - 0 . 75 + 0 = 0 . 98 m\]

y-component of resultant R_y = 1 + 1.3 − 1 = 1.3 m

\[\therefore \text { Resultant, R }= \sqrt{\left( R_x \right)^2 + \left( R_y \right)^2}\]

\[ = \sqrt{\left( 0 . 98 \right)^2 + \left( 1 . 3 \right)^2}\]

\[ = \sqrt{0 . 96 + 1 . 69}\]

\[ = \sqrt{2 . 65}\]

\[ = 1 . 6 m\]

If it makes an angle α with the positive x-axis, then

\[\tan \alpha = \frac{\text { y -component }}{\text { x - component }}\]

\[ = \frac{1 . 3}{0 . 98} = 1 . 332\]

∴ α = tan⁻¹ (1.32)