11mgcosθ?
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 The force exerted by the wedge on the block has a manitude is the ans (mg) or( mg cos theta)
-
UP 0 DOWN 0 0 3
3 Answers
3if its smooth ...only normal force is applied by the wedge on the block ....mgcos(theta)
1well actually if the horizontal acceleration of the system is 'a'
then the net force acting on the block is ma in that direction as the block is at rest w.r.t the wedge
now , there are two forces acting on the block
(i)force by gravity mg downwards
(ii)force by wedge (let it be F at angle of theta to the reverse direction of mg
then as the net force exerted is ma along horizontal
Fcosθ=mg
and Fsinθ=ma
answer will be mg when a=0 or when you want the vertical component of the force