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. `(3 sqrt(3))/(8)mg`
B. `(mg)/(2)`
C. `(sqrt(3))/(2)mg`
D. zero

Answer

B

Explanation

For `theta = 30^(@) ` and `mu=(3)/(4)`
The pulling force downward (or down the incline)
is `mg sin theta = mg sin(30^(@)) = (mg)/(2)`
But the resisting force is friction
Maximum friction force that can acts is `mu mg cos theta`
`= (3)/(4)mg cos 30^(@) = (3 sqrt(3))/(8) mg`
The frictional force is greater than the pulling force as `(3 sqrt(3))/(8) mg gt (mg)/(2)`
`:.` block does not slide
So the friction force acting on the block will be static friction and is equal to the pulling force
`f = (mg)/(2)`