Ncert Exercises & Solutions

A body of mass \(10 ~\text{Kg}\) is acted upon by two perpendicular forces, \(6 ~\text{N}\) and \(8 ~\text{N}.\) The resultant acceleration of the body is:

(a)	\(1~\text{ms}^{-2}\) at an angle of \(\text {tan}^{-1} \left(\dfrac{4}{3}\right ) \) w.r.t. \(6 ~\text{N}\) force.
(b)	\(0.2~\text{ms}^{-2}\) at an angle of \(\text {tan}^{-1} \left(\dfrac{3}{4}\right ) \) w.r.t. \(8 ~\text{N}\) force.
(c)	\(1~\text{ms}^{-2}\) at an angle of \(\text {tan}^{-1} \left(\dfrac{3}{4}\right ) \) w.r.t. \(8 ~\text{N}\) force.
(d)	\(0.2~\text{ms}^{-2}\) at an angle of \(\text {tan}^{-1} \left(\dfrac{3}{4}\right ) \) w.r.t. \(6 ~\text{N}\) force.

Choose the correct option from the given ones:

1.	(a) and (c)	2.	(b) and (c)
3.	(c) and (d)	4.	(a), (b) and (c)

Hint: \(F_{net}= \sqrt{F_x^2 +F_y^2}\)

Step 1: Find the resultant force and the magnitude of the acceleration.
Given, mass \(m=10 ~\text{Kg}\)
Let \(F_1=6 ~\text{N}\) and \(F_2=8 ~\text{N}\) be the forces.
Since the forces are perpendicular, the net force on the body is given by;
\(\Rightarrow F_{net} =\sqrt{F_1^2 +F_2^2} =\sqrt{6^2 +8^2}\)
\(\Rightarrow F_{net} =\sqrt{36+64} =10~\text{N}\)
Using \(F=ma\) the acceleration of the body is;
\(\Rightarrow a= \frac{F}{m} = \frac{10}{10} =1~\text {ms}^{-2}\)

Step 2: Find the direction of the acceleration.
Let \(\theta_1\) be the angle between the \(F_{net}\) and \(F_1\) and \(\theta_2\) be the angle between the \(F_{net}\) and \(F_2.\)

From the vector diagram of the forces, we can state that;
\(\Rightarrow \tan \theta_1 =\frac{F_2}{F_1}\) and \(\tan \theta_2 =\frac{F_1}{F_2}\)
\(\Rightarrow \tan \theta_1 =\frac{8}{6}\) and \(\tan \theta_2 =\frac{6}{8}\)
\(\Rightarrow \theta_1 =\tan ^{-1}\left(\frac{4}{3} \right)\) w.r.t \(F_1\) and \( \theta_2 =\tan ^{-1}\left(\frac{3}{4} \right)\) w.r.t \(F_2.\)
Therefore, the body accelerates at \(1~\text{ms}^{-2}\) an angle of \(\text {tan}^{-1} \left(\dfrac{4}{3}\right ) \) w.r.t. \(6 ~\text{N}\) force.
Hence, option (1) is the correct answer.

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