A block of mass m is placed on a rough inclined plane. The corfficient of friction between the block and the plane is `mu` and the inclination of the plane is `theta`.Initially `theta=0` and the block will remain stationary on the plane. Now the inclination `theta` is gradually increased . The block presses theinclined plane with a force `mgcostheta`. So welding strength between the block and inclined is `mumgcostheta`, and the pulling forces is `mgsintheta`. As soon as the pulling force is greater than the welding strength, the welding breaks and the blocks starts sliding, the angle `theta` for which the block start sliding is called angle of repose `(lamda)`. 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). The net contact force will be resultant of both.
Answer the following questions based on above comprehension:
Q. If `mu=(3)/(4)` then what will be frictional force (shear force) acting between the block and inclined plane when `theta=30^@`:

Options

A. increase
B. decrease
C. remain same
D. increase of `a gt g`

Answer

C

Explanation

Initially when the system is not accelerating for the block to just start sliding its acceleration should be zero.
Equation along the incline (for just sliding)
`mg sin theta = mu N` ….(1)
Equation perpendicular the incline)
`N = mg cos theta`...(2)

Dividing equation (1) by equation (2)
`tan theta = mu`
`:. theta = tan_(-1) (mu)` = angle of repose
When the system is accelerating upwards with a acceleration 'a'.
Equation along the incline
`mu N - mg sin alpha = ma sin alpha` ....(3)
Equation perpendicular the incline
`N - mg cos alpha = mg a cos alpha`....(4)
Dividing equation (3) by
(4) and eliminating N we get
`mu = (m(g+a)sin alpha)/(m(g+a)cos alpha)`
`tan alpha = mu`
`:. alpha = tan_(-1) (mu) `= angle of repose
The value of `theta` and `alpha` are same. Hence the angle of repose will remain same.