Question

A body of mass $5$ kg is acted upon by two perpendicular forces $8$ N and $6$ N. Given the magnitude and direction of the acceleration of the body.

Answer 1

$\text{Step 1: Drawing FBD}$ $\text{[Refer Fig. 1]}$

$\text{Step 2: Finding Resultant Force}$ $\text{[Refer Fig. 2]}$

$\text{Magnitude of Resultant:}$

Since $F_1$ and $F_2$ are perpendicular

$\therefore$ Magnitude of Resultant force on body is given by

$R=\sqrt{F_1^2+F_2^2}$

$=\sqrt{8^2+6^2}N$ $=10N$

$\text{Direction of Resultant:}$

$\theta ={\tan }^{-1}\frac{F_2}{F_1}$ $\text{[Refer Fig. 2]}$

$={\tan }^{-1}\frac{6}{8}$ $={36.9}^o$ $\approx {37}^o$

$\text{Step 3: Applying Newton's second Law for finding acceleration}$ $\text{[Refer Fig. 3]}$

$F_{net}=ma$

$\Rightarrow$ $10=5a$

$\Rightarrow$ $|\overrightarrow{a}|=2m/s^2$

Hence magnitudes of acceleration is $2m/s^2$ directed at ${37}^o$ from the 8 N force.