IIT JEE , AIEEE Forum - Physics , Mechanics

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(i m gonna use 'u' as the friction coefficient)
See, the 40 N force you have applied will act as the normal reaction for the body. So you get frictional force = u x 40 = 0.5 x 40 = 20 N

Thus the weight(downward) must nullify the frictional force(upward) if the body is to remain at rest. So you get weight = 20 N (which is again verified by the fact that mg = 2 x 10 = 20 N )

So on application of the 15N force as shown in figure, the net force acting on the body becomes

( 20² + 15² ) = 25 N

This 25 N force will be opposed by a frictional force = u x 40 = 20 N again as the 15 N force will have no effect on the normal reaction received by the body from the wall. This is because 15 N force makes an angle 90⁰ with the plane of the wall and has no component perpendicular to the wall.

Thus net force acting on the body becomes (25 - 20) N = 5 N.

This 5 N will make the body move.....

And this 5 N will act along the 25 N force, since frictional force is just opposite to the 25 N force.

Now let

be the angle made by the 5 N force with the 15 N force.

Thus (from figure) tan

= 20/15

So <theta> = tan^-1 (20/15)

= 53⁰

THE BODY MOVES BY MAKING AN ANGLE 53⁰ WITH THE 15 N FORCE......

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