A body of mass 2.5 kg is subjected to the forces shown in figure. Find the acceleration of the centre of maas.

Explanation

Take the X andy axes as shown in theure. The x-component of the resultant force is
`F_x=-6N+(5N)cos37^0+(6N)cos53^0+(4N)cos 60^0`
`=-6N+(5N).(4/5)+(6N).(3/5)+(4N).(1/2)=3.6N`
similarly the y component of the resultant force is
` F_y=5Nsin37^0-(6N)sin53^0+4Nsin60^0`
`=(5N).(3/5)-(6N).(4/5)+(4N).(sqrt3/2)=1.7N`
The magnitude of the resultant force is
`F=sqrt(F_x^2+F_y^2)=sqrt((3.6N)^+(1.7N)^2)~~4.0N`
The direction of the resultant force makes an angle `theta`
with the X-axis where
tantheta=F_y/F_x=1.7/3.6=0.47`
`The acceleration of the centre of mass is
`a_(CM)=F/M=(4.0N)/(2.5kg)=1.6m/s^2`
in the direction of the resultant force