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.

Options

A. `a = g(tanθ) ` , `N = (mg)/(cosθ)`
B. `a = g(sinθ) ` , `N = (g)/(cosθ)`
C. `a = -g(tanθ) ` , `N = (2mg)/(cosθ)`
D. `a = -g(tanθ) ` , `N = (3mg)/(2cosθ)`

Answer

A

Explanation

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.

As m is at rest, moving with same acceleration as wedge in horizontal direction but in vertical direction. the block is at rest.
`Sigma vec(F)_(y)=0implies N cos theta =mg`...(i)
`Sigma vec(F)_(x) = Sigma m_(i)vec(a)_(i) implies N sin theta =mA` ...(ii)
on solving (i) and (ii) we get `A=g tan theta and N=(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`
`implies A=g tan theta`
Also: `N=mg cos theta+mA sin theta`
`=mg cos theta+mg tan theta sin theta =(mg)/(cos theta)`.