A bar weighing 0.4 kg is evenly moved along the horizontal surface of the table.

A bar weighing 0.4 kg is evenly moved along the horizontal surface of the table. The dependence of the modulus of the friction force Ffr on the modulus of the force of normal pressure N of the bar has the form: Ffr = 0.2 * N. Find: a) the coefficient of friction between the bar and the table. b) modulus of sliding friction force

m = 0.4 kg.

g = 10 m / s2.

Ftr = 0.2 * N.

μ -?

Ftr -?

F + Ftr + m * g + N = 0 – 2 Newton’s law in vector form, where Ftr is the friction force, F is the force with which the bar is pulled, m * g is the force of gravity, N is the normal pressure force of the bar.

ОХ: F – Ftr = 0.

OU: m * g – N = 0.

N = m * g.

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

Since in the condition of the problem Ftr = 0.2 * N, then μ = 0.2.

Ftr = 0.2 * 0.4 kg * 10 m / s2 = 0.8 N.

Answer: the sliding friction force is Ftr = 0.8 N, the friction coefficient μ = 0.2.