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.


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