Question

A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle threta = 30 degree with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface?

Answer 1

mg sin 15 = k mg cos 15 (where k = constant or coefficient of friction)

K = tan 15 = tan (45 – 30)

K = (tan 45 – tan 30)/ 1 + tan 45 tan 30 = (31/2 – 1)/ (31/2 + 1).

Answer 2

$\tan( \alpha ) = coefficient \: of \: friction \\ \tan( \frac{\pi}{6} ) = coefficient \\ \frac{1}{ \sqrt{3} } = coefficient.$