Laws of MotionHard
Question
A triangular block of mass M rests on a smooth surface as shown in figure. A cubical block of mass m rests on the inclined surface. If all surfaces are frictionless, the force that must be applied to M so as to keep m stationary relative to M is :
Options
A.Mg tan 30o
B.mg tan 30o
C.(M + m)g tan 30o
D.(M + m)g cos 30o
Solution
F.B.D. of wedge is w.r.t. ground and
F.B.D. of block is w.r.t. wedge.
Let a is the acceleration of wedge due to force F.
FP is pseudo force on block
mg sin 30o − ma cos 30o = 0 [Equilibrium of block in x direction w.r.t. wedge]
a = g tan 30o
F = ( M + m)a [Newtons II law for the system of block and wedge in horizontal direction]
⇒ F = (M + m) g tan 30o
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