Question

A block of mass m is placed on an inclined plane. With what acceleration A towards right should the system move on a horizontal surface so that m does not slide on the surface of inclined plane? Also calculate the force supplied by wedge on the block. Assume all surfaces are smooth.

Answer 1

Method 1 : Analysis of forces on m relative to ground

IF the motion of m is analyzed from ground, its acceleration is A and the forces acting on it are its weight mg and normal reaction N. Asm is at rest, moving with same acceleration as wedge in horizontal direction but in vertical direction, the block is at rest.

$\sum {\overrightarrow{F}}_y=0\Rightarrow N\cos \theta =mg$........(i)

$\sum {\overrightarrow{F}}_x=\sum m_i{\overrightarrow{a}}_i\Rightarrow N\sin \theta =mA$.........(ii)

On solving (i) and (ii), we get A=g

$\tan \theta$ and $N=\frac{mg}{\cos \theta }$

Method 2 : Analysis of forces on m relative to the inclined plane

If the motion of m is analyzed from the view point of an observer standing on the inclined plane (i.e., relative to the plane), its acceleration is 0 and the forces acting on it are : its weight. the normal reaction and a pseudo force of magnitude mA towards left.

$mA\sin \theta =mA\cos \theta$

$\Rightarrow$ $A=g\tan \theta$

Also: $N=mg\cos \theta +mA\sin \theta$

$mg\cos \theta +mg\tan \theta \sin \theta =\frac{mg}{\cos \theta }$