A box weighing 60 kg begins to be moved along a horizontal surface with an acceleration of 1 m / s ^ 2

A box weighing 60 kg begins to be moved along a horizontal surface with an acceleration of 1 m / s ^ 2, acting on it with a constant force directed at an angle of 30 degrees to the horizontal. Determine the force with which the box is pulled if the sliding friction coefficient is 0.2?

m = 60 kg.

g = 9.8 m / s ^ 2.

μ = 0.2.

a = 1 m / s ^ 2.

∠α = 30 °.

F -?

Let’s write 2 Newton’s law in vector form for a box: m * a = F + Ftr + N + m * g.

Let’s find expressions 2 of Newton’s law in projections on the axis.

ОХ: m * a = F * cosα – Ftr.

OU: 0 = F * sinα + N – m * g.

N = m * g – F * sinα.

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

m * a = F * cosα – μ * m * g + μ * F * sinα.

F * cosα + μ * F * sinα = m * a + μ * m * g.

F * (cosα + μ * sinα) = m * (a + μ * g).

F = m * (a + μ * g) / (cosα + μ * sinα).

F = 60 kg * (1 m / s ^ 2 + 0.2 * 9.8 m / s ^ 2) / (cos30 ° + 0.2 * sin30 °) = 185 N.

Answer: the box is pulled with a force of F = 185 N.