A block of mass `m` is placed on a rough inclined plane. The coefficient of friction between the block and the plane is `mu` and the inclination of the plane is `theta`. Initially `theat-0` and the block will certain stationary on the plane. Now the inclination `theta` is gradually increased. The block presses the inclined plane with a force `mg cos theta`. So welding strength between the block and inclined is `mu mgcos theta`, and the pulling forces is `mg sin theta`. As soon as the pulling force is greater than the welding strength, the welding breaks and the block starts sliding, the angle `theta` for which the block starts sliding is called angle of repose `(lambda)`. During the contact, two contact forces are acting between the block and the inclined plane. The pressing reaction (Normal reaction) and the shear reaction `(` frictional force`)` and the shear reaction (frictional force) . The net contact force will be resultant of both.

If the entire system, were accelerated upward with acceleration `'a'`, the angle of repose , would `:`

Options

A. increases
B. decreases
C. remain same
D. increase of `agtg`

Answer

C

Explanation

Angle `(theta')` fo repose `:`
`m(g+a)sintheta'=F`
`m(g+a)cos theta'=R`
`:. (F)/(R)=tantheta'`
`theta'=tan^(-1)((F)/(R))=alpha`
Hence angle of repose does not change.