A bar weighing 0.2 kg is evenly pulled with a dynamometer on the horizontal

A bar weighing 0.2 kg is evenly pulled with a dynamometer on the horizontal surface of the table. Dynamometer reading 0.5 H. What is the coefficient of sliding friction?

m = 0.2 kg.

g = 10 m / s2.

F = 0.5 N.

μ -?

We will assume that the bar is pulled evenly across the table. The condition for uniform rectilinear movement is the balancing of all forces that act on the bar. Let us write 2 Newton’s law in vector form: 0 = F + m * g + N + Ftr, where F is the force with which the load is pulled, which is shown by the dynamometer, m * g is the force of gravity, N is the reaction force of the table surface, Ftr – friction force between the bar and the table.

ОХ: 0 = F – Ftr.

OU: 0 = – m * g + N.

Ftr = F.

N = m * g.

The friction force Ffr is determined by the formula: Ffr = μ * N = μ * m * g.

F = μ * m * g.

μ = F / m * g.

μ = 0.5 N / 0.2 kg * 10 m / s2 = 0.25.

Answer: the coefficient of sliding friction of the bar on the table surface is μ = 0.25.