IIT JEE , AIEEE Forum - Physics , Mechanics

umang

Solve -

A block of mass m is placed on a smooth wedge of inclination theta . The whole system is accelerated horizontally so that the block does not slip on the wedge . Find the magnitude of the force exerted by the wedge on the block .

Pls give the detailed solution !!!

sen_holmes

as system is moving with acc a hor;

therefore a pseudo force a will act on the block

now actual normal force =mg*cos

and pseudo force=ma*sin

therefore net force on block due to wedge i.e. normal force=mgcos

+masin

yahiyafirdous

Since the mass remain static on the wdge, hence there is a force exerted on the mass by the wedge in horizontal direction.

Hence mg Sin

=F Cos

F=mg tan

If this is correct, then plz rate me

pearlypeeble

i will be solving the question from ground frame
since the body is having accl in horizontal direction only,
thus balancing forces in vertical direction gives
N cos(theta)=mg
N=mg sec(theta)
this is the contact force that should be acting

umang

Hey pearlypeeble , yes , this is the right answer . But , i want to know why cant we resolve weight force ???Why only normal force ?

Pls explain as soon as possible !!!

nikhilwalani

we can resolve weight force along the normal but the component of accl. will also come in that direction. since the accl. is in horizontal direction therefore balancing along vertical direction will directly give the answerand let me tell u that this answer is same as that of sen_holmes

umang

Hey nikhil ,

the correct answer is mg/cos(theta) . Also , the 'a' (acceleration) used by sen_holmes is not provided in the question !!!!!!

nikhilwalani

u can eliminate accl. by resolving forces acting along the inclined plane

Ask & Discuss Questions with Community & Experts