Question

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

Answer 1

Given Data

Mass of the body, \(m\)	\(= 5\ \text{kg}\)
Force \(F_1\)	\(= 8\ \text{N}\)
Force \(F_2\)	\(= 6\ \text{N}\)
Angle between forces	\(= 90^\circ\) (perpendicular)

Step 1: Resultant Force

Since the two forces are perpendicular, the magnitude of the resultant force \(F_R\) is:

\[ F_R = \sqrt{F_1^2 + F_2^2} \] \[ F_R = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\ \text{N} \]

Step 2: Magnitude of Acceleration

Using Newton’s second law:

\[ a = \frac{F_R}{m} = \frac{10}{5} = 2\ \text{m s}^{-2} \]

Step 3: Direction of Acceleration

Let \(\theta\) be the angle that the resultant (and hence acceleration) makes with the direction of the 8 N force.

\[ \tan\theta = \frac{F_2}{F_1} = \frac{6}{8} = \frac{3}{4} \] \[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.9^\circ \]

So the acceleration is at an angle of about \(37^\circ\) to the 8 N force, towards the 6 N force.

Final Answer

Magnitude of acceleration: \[ a = 2\ \text{m s}^{-2} \] Direction of acceleration: at \[ \theta \approx 37^\circ \] with the 8 N force, towards the 6 N force.