A load weighing 100 kg is moved uniformly across a horizontal surface, applying a force of 200 N directed at an angle

A load weighing 100 kg is moved uniformly across a horizontal surface, applying a force of 200 N directed at an angle of 30 ° to the horizontal. With what acceleration does the body move if the coefficient of friction is 0.1?

m = 100 kg.

g = 10 m / s2.

F = 200 N.

∠α = 30 °.

μ = 0.1.

a -?

Let us write Newton’s 2 law in vector form: m * a = F + m * g + N + Ftr, where F is the force with which the load is pulled, m * g is the force of gravity, N is the surface reaction force, Ftr is the friction force.

ОХ: m * a = F * cosα – Ftr.

OU: 0 = F * sinα – m * g + N.

a = (F * cosα – Ftr) / m.

N = m * g – F * sinα.

The friction force Ffr is determined by the formula: Ffr = μ * N = μ * (m * g – F * sinα).

The acceleration of the body a will be determined by the formula: a = (F * cosα – μ * (m * g – F * sinα)) / m.

a = (200 N * cos30 ° – 0.1 * (100 kg * 10 m / s2 – 200 N * sin30 °)) / 100 kg = 0.9 m / s2.

Answer: the load will move with acceleration a = 0.91 m / s2.