A block weighing 5 kg is located on an inclined plane with an angle of inclination alpha.

A block weighing 5 kg is located on an inclined plane with an angle of inclination alpha. The reaction force acting on the bar is 40 N. Find the angle alpha.

m = 5 kg.

g = 10 m / s2.

N = 40 N.

∠α -?

According to 1 Newton’s law, the condition for the equilibrium of a bar on an inclined plane is the balancing of all the forces that act on it.

It is acted upon by 3 forces: m * g is the force of gravity directed vertically downward, N is the reaction force of the surface of the inclined plane directed perpendicularly to the plane upward, Ffr is the force of friction along the inclined plane.

m * g + N + Ftr = 0 – equilibrium condition.

ОХ: m * g * sinα – Ftr = 0.

OU: 0 = – m * g * cosα + N.

Ftr = m * g * sinα.

N = m * g * cosα.

cosα = N / m * g.

cosα = 40 N / 5 kg * 10 m / s2 = 0.8.

∠α = arcsin0.8 = 53 °.

Answer: the angle of the inclined plane is ∠α = 53 °.