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.

The magnitude of the resultant Z of the two vectors X and Y at an angle θ with each other is given as,

Z=(X² + Y² + 2.X.Y cosθ)^1/2

and the direction of the resultant can be calculated as θ , where from the fig.

tan θ = AC / AB

⇒ θ = tan^-1(AC / AB)

Given,

Mass of the body, m = 5 kg.

Force F₁ along AB = 8 N.

Force F₂ along AC = 6 N

The resultant force F along AD is given as,

Since, the forces are perpendicular, cos θ =0

⇒

The direction of the resultant thus is,

θ = tan^-1 (-6/8)

⇒ θ = -36.87°

The negative sign shows the configuration of the θ with respect to F_2.

The Newton’s second law of motion, the force can be defined as:

F = m× a

⇒ a = F / m

Where,

m is the mass of the body

a is the acceleration of the body.

Thus, the acceleration of the body is,

Ans: The magnitude of the acceleration is 2 m/s² and is 36.87° in the clockwise direction from the force taken on the horizontal axis i.e. F₂.