# A bar weighing 1 kg moves uniformly across a horizontal surface under the action of a force F = 10n, the angle

**A bar weighing 1 kg moves uniformly across a horizontal surface under the action of a force F = 10n, the angle between the force F and the horizon is a = 30 degrees. The modulus of the friction force acting on the body is 1N. What is the coefficient of friction between the bar and the plane?**

m = 1 kg.

g = 10 m / s2.

F = 10 N.

∠α = 300.

Ftr = 1 N.

μ -?

Let’s write Newton’s 2 law in vector form: m * a = F + Ffr + m * g + N.

ОХ: m * a = F * cosα – Ftr.

OU: 0 = F * sinα – m * g + N.

N = m * g – F * sinα.

The friction force is determined by the formula: Ftr = μ * N, where μ is the coefficient of friction, N is the reaction force of the plane.

Ftr = μ * (m * g – F * sinα).

The formula for determining the coefficient of friction μ will have the form: μ = Ffr / (m * g – F * sinα).

μ = 1 N / (1 kg * 10 m / s2 – 10 N * sin300) = 0.2.

Answer: the coefficient of friction is μ = 0.2.