a block of mass  'm'  is placed over a smooth wedge of inclination  @ . the  whole system is accelerated horizontally such that the block does not slip on the wedge .  the force exerted by the wedge on the block has a magnitude?

ans=  mg/cos@

the system is accelerated horizontally, so net force in the vertical direction will be 0

so Ncos@ = MG => N = mg/cos@

thats the force exerted by the wedge

rel. to the wedge  also the psuedo force has no vertical component. so it wont affect the normal reaction offerd by the wedge

 

rate me plz.

when the wedge is moved with accl^n a the body of mass m experieces a psuedo force ma towards opposite direction of the motion of wedge (as shown in figure).

Now resolving the forces as shown in figure...

we get ( for the body to remain at rest)

ma.cos = mg.sin

or a = g.tan ____________(1)

This is the accl^n to be provided to the wedge so that the block doesnot fall.

So again from figure we get that the force which the block gives the incline perpendicularly = ma.sin + mg.cos 

This force = force given by incline on the block....( from Newton's 3rd Law)

So reqd. force = ma.sin + mg.cos

                     = mg.tan.sin + mg.cos .........[ by (1) ]

                     = mg.(sin²/cos) + mg.cos

                     = mg.(sin² + cos²) / cos

                     = mg / cos...