Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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 −−→ O a , −−→ B C and −−→ D E . - Physics

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 $\overrightarrow{OA} , \overrightarrow{BC} \text { and } \overrightarrow{DE}$.

#### Solution

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 $\overrightarrow{BC}$ =1.5cos120°

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

x-component of $\overrightarrow{DE}$ = 1cos270°

= 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 Ry = 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)
Concept: What is Physics?
Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 2 Physics and Mathematics
Exercise | Q 5 | Page 29
Share