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

Answer

B

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.
Shear force `=mu mg cos theta`
`=(3)/(4)xxmgxx(sqrt(3))/(2)=(3sqrt(3))/(8)mg =0.06mg`
But, pulling force`=mg sin theta =mg sin 30^(@)=0.5mg lt f_(smax). :. ` block does not slide.
Hence frictional force (shear forc) between the block of the plane at this situation will be `=mg sin 30^(@)=(mg)/(2)("not" (3sqrt(3))/(8)mg)`
Alternate Sol.
`tan theta=(1)/(sqrt(3))=(sqrt(3))/(3)=(1.73)/(3)=0.58ltmu`
`:. ` block does not slide . `:. f_(s)=mg sin 30^(@)`