Ncert Exercises & Solutions

A body of mass $10$ kg is acted upon by two perpendicular forces, $6$ N and $8$ N. The resultant acceleration of the body is:

(a)	$1~\text{ms}^{-2}$ at an angle of $tan^{-1}\Big(\frac43\Big)$ w.r.t. $6$ N force
(b)	$0.2~\text{ms}^{-2}$ at an angle of $tan^{-1}\Big(\frac34\Big)$ w.r.t. $8$ N force
(c)	$1~\text{ms}^{-2}$ at an angle of $tan^{-1}\Big(\frac34\Big)$ w.r.t. $8$ N force
(d)	$0.2~\text{ms}^{-2}$ at an angle of $tan^{-1}\Big(\frac34\Big)$ w.r.t. $6$ N force

$\begin{array}{l} \end{array}$
Choose the correct option:

(1) Hint: The resultant acceleration will be in the direction of the resultant force.

Step 1: Find the resultant force and the magnitude of the acceleration.

Consider the adjacent diagram
Given, mass = m = 10 kg

\begin{array}{l} Resultant force = F = \sqrt{F_{1}^{2} + F_{2}^{2}} = \sqrt{36 + 64} \\ = 10 N \\ a = \frac{F}{m} = \frac{10}{10} = 1 m / s^{2}; along R \end{array}

Step 2: Find the direction of the acceleration.

\begin{array}{l} Let θ_{1} be angle between R and F_{1} \\ \begin{matrix} \tan θ_{1} = \frac{8}{6} = \frac{4}{3} \\ θ_{1} = \tan^{- 1} (4 / 3) w . r . t . F_{1} = 6 N \end{matrix} \end{array}

Let θ_{2} be angle between F and F_{2}

\tan θ_{2} = \frac{6}{8} = \frac{3}{4}

θ_{2} = \tan^{- 1} (\frac{3}{4}) w . r . t . F_{2} = 8 N

(a)	\(1~\text{ms}^{-2}\) at an angle of \(tan^{-1}\Big(\frac43\Big)\) w.r.t. \(6\) N force
(b)	\(0.2~\text{ms}^{-2}\) at an angle of \(tan^{-1}\Big(\frac34\Big)\) w.r.t. \(8\) N force
(c)	\(1~\text{ms}^{-2}\) at an angle of \(tan^{-1}\Big(\frac34\Big)\) w.r.t. \(8\) N force
(d)	\(0.2~\text{ms}^{-2}\) at an angle of \(tan^{-1}\Big(\frac34\Big)\) w.r.t. \(6\) N force