In the figure given below ,the force F to be applied on triangular block of mass M so that the block of mass m placed on it appears stationary w.r.t wedge is ( coeff of friction is 0 between mass and wedge And between wedge and ground also)
a ) mg tan 'theta' b ) ( M + m ) g tan ' theta'
c ) ( M + m ) g cos 'theta' d ) ( M + m ) g sin 'theta'

Dear Student,

From the free body diagram (FBD)

If there is no force on the wedge the acceleration of the block will be a = gsinθ

If the block is to be at rest with respect to the wedge.

Let the force be F.

F = (M+m)a

When we observe from the wedge

there is a pseudo force on the block in the direction opposite to the motion of the block and wedge system.

Resolving the force along the direction of the wedge and the direction perpendicular

we get: macosθ = mgsin θ

a = gtanθ

Thus, F = (M+m) gtanθ


  • 11
(b) is correct;
from the free body diagram of the smaller block m we observe that ,the normal force f is acting prependicular to it which is provided by the wedge,and mg vertically downwards,
if you take component of the force along vertical and horizontal,then you will find that the component of normal force which is NSIN'theta' ,will pull it forward and therefore;
NSIN'theta'=ma .........(i)      (where ma is the psuedo force acting backwards, a if the acceleration of the whole system,, including block and wedge)
NCOS'theta=mg   .......(ii)    (because the block is in rest)

dividing 1 and 2 ,we get {a=gtan'theta'}
so the force is         [{(M+m)gtan'theta'}]
  • 5
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