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 F2.
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.