Question

a wooden block is placed on an inclined plane. the block just begins to slide down when the angle of the inclination is increased at 45degrees what is the coefficient of the friction

Answer 1

Here, the various force involved are:

Weight , mg of the body, vertically downward.

Normal reaction ,R, perpendicular to AB.

Force of friction ,F ,up the plane AB

Now, mg can be resolved into two rectangular component …i.e.- mg cos 45 and mg sin 45

In equilibrium,

F=mg sin 45°…………(1)

R=mg cos 45°…………(2)

On dividing ( 1) by (2), we get

F/R= mg sin 45°/mg cos 45°

or, friction coefficient= tan 45°.

(most probably correct)

Answer 2

Answer:

tan 45* is the answer.... 1 is the answer

Explanation:

Hope it helps