When a maximum force of 3 N is applied on a body kept on rough inclined plane of shown in the figure, then body remains stationary. The maximum external force up the inclined plane that does not move the block is 12 N. The coefficient of static friction between the block and the plane is [Take, g = 10 m/`s^(2)`]

Options

A. `(sqrt(3))/(5)`
B. `(1)/(sqrt(3))`
C. `(3sqrt(3))/(5)`
D. `(5)/(3sqrt(3))`

Answer

A

Explanation

Block does not move upto a maximum applied force of 3 N down the inclined plane. Equation of motion is 3 + mg sin `30^(@)` = F
`3 + mg sin 30^(@) = mu mg cos 30^(@)`

Similarly, block also does not move upto a maximum applied force of 12 N up the plane. Now, equation of motion is
mg sin `30^(@) + F =12 `
or `mg sin 30^(@) + mu mg cos 30^(@) =12 `
Solving Eqs. (i) and (ii), we have
`mg sin 30^(@) = (9)/(2) ` and `mu "mg cos " 30^(@) = (15)/(2)`
Hence dividing
`:. (1)/(mu) tan 30^(@) = (9//2)/(15//2)=(3)/(2) implies mu = (5)/(3sqrt(3))`