A bar weighing 2 kg lies on the table, the coefficient of friction between the bar and the table is 0.3.

A bar weighing 2 kg lies on the table, the coefficient of friction between the bar and the table is 0.3. What is the frictional force acting on the bar if a horizontal force of 4 N is applied to it?

m = 2 kg.

g = 9.8 m / s2.

μ = 0.3.

F = 4 N.

Ftr -?

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

With the horizontal direction of the force F with which the bar is pulled, the reaction force of the support N is equal to the force of gravity: N = m * g.

The sliding friction force Ftr will be determined by the formula: Ftr = μ * m * g.

Ftr = 0.3 * 2 kg * 9.8 m / s2 = 5.88 N.

Since a force F acts on the body, which is less than the sliding friction force Ffr, then the body will be at rest. At rest, the force of friction at rest acts on it, which is equal to the force that acts on the body: Ftr = Ft.

Ftr = 4 N.

Answer: the body is affected by the static friction force Ftr = 4 N.