The bar rolls off an inclined surface with an acceleration of 4 m / s 2.

The bar rolls off an inclined surface with an acceleration of 4 m / s 2. If the angle of inclination of this surface is 30 °, then what is the coefficient of sliding friction between this surface and the bar?

g = 10 m / s2.

a = 4 m / s2.

∠α = 30 °.

μ -?

The bar moves along an inclined plane under the action of the projection of the force of gravity.

2 Newton’s law when sliding a bar along an inclined plane in vector form will have the 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 = m * g * sinα – Ftr.

OU: 0 = – m * g * cosα + N.

Ftr = m * g * sinα – m * a = m * (g * sinα – a).

N = m * g * cosα.

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

μ * m * g * cosα = m * (g * sinα – a).

μ = (g * sinα – a) / g * cosα.

μ = (10 m / s2 * sin30 ° – 4 m / s2) / 10 m / s2 * cos30 ° = 0.116.

Answer: the coefficient of sliding friction between this surface and the bar is μ = 0.116.