A crate is resting on a ramp that is inclined at an angleabove
the horizontal. As the angle is incrased the crateremains in place
until the angle reaches a value of 38degrees. Then the crate begins
to slide down the slope. (a) Determine the coefficient of static
friction between the crateand the ramp surface. (b) The coefficient
of kinetic frictionbetween the crate and the ramps surface is
0.600. Find theacceleration of the moving crate.

The angle of repose or the angle of sliding is defined as the angle of inclined plane with the horizontal at which an object placed on the plane is just start to slide down.

The relation between the angle of friction and the coefficient of static friction is given by

[tex]\mu _{s}=tan\theta[/tex]

[tex]\mu _{s}=tan38=0.78[/tex]

Thus, the coefficient of static friction is 0.78.

(b) As the crate is moving down, then the friction force is kinetic friction.

The force acting along the plane downward = mg Sinθ

The normal reaction, N = mg cosθ

The friction force acting upward along the plane, f = μk N = μk mg Cosθ

The net force acting along the plane downward

Fnet = mg Sinθ - μk mg Cosθ

According to the newton's second law, Fnet = mass x acceleration

so, m x a = mg Sinθ - μk mg Cosθ

a = g Sinθ - μk g Cosθ

here, μk = 0.6 and θ = 38°, g = 9.8 m/s^2

By substituting the value, we get

a = 9.8 ( Sin38 - 0.6 x Cos 38)

a = 9.8 (0.6156 - 0.6 x 0.788)

a = 1.4 m/s^2