A block weighing 3 kg slides along an inclined plane with an inclination angle of 30 degrees.

A block weighing 3 kg slides along an inclined plane with an inclination angle of 30 degrees. The coefficient of friction between the bar and the plane is 0.3. What is the force of friction between the plane and the bar?

m = 3 kg.
g = 10 m / s2.
∠α = 30 °.
μ = 0.3.
Ftr -?
For the movement of a bar along an inclined plane, we write 2 Newton’s law in vector form: m * a = m * g + N + Ffr, where m * g is the force of gravity, N is the reaction force of the surface of the inclined plane, Ffr is the friction force.
ОХ: m * a = Ftr – m * g * sinα.
OU: 0 = – m * g * cosα + N.
Fт = m * a + m * g * sinα = m * (a + g) * sinα.
N = m * g * cosα.
The sliding friction force Ftr is expressed by the formula: Ftr = μ * N = μ * m * g * cosα.
Ftr = 0.3 * 3 kg * 10 m / s2 * cos30 ° = 7.8 N.
Answer: the friction force of the bar on the inclined plane is Ftr = 7.8 N.