A block of mass 2kg rests on a rough inclined plane making an angle of `30^@` with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is

Answer

`9.8 N`

Explanation

As the block is at rest therefore maximum force of static friction
`f_(s) = mu _(s) R` But `R = m g cos theta`
`:. f_(s) = mu_(s) m g cos theta = 0.7 xx 2xx 9.8 cos 30^(@)`
`= 1.4 xx 9.8 xx 0.866 = 11.9 N`
But in this case actual force of static friction cannot exceed the force due to gravity `(mg sin theta)` acting down the plane
`:.` Actual force of static friction
`= mg sin theta = 2 xx 9.8 xx sin 30^(@) = 9.8 N`.