A light, inextensible thread is tied to a body weighing m = 10 kg lying on a horizontal surface.

A light, inextensible thread is tied to a body weighing m = 10 kg lying on a horizontal surface. if a force F = 39.2N is applied to the thread, directed under the 60 to the horizon, then the body moves uniformly. With what acceleration the body will move if this force is applied at an angle of 30 to the horizon.

m = 10 kg.

g = 9.8 m / s2.

F = 39.2 N.

∠α1 = 60 °.

∠α2 = 30 °.

but – ?

Since the body moves uniformly in a straight line, then according to 1 Newton’s law, the action of forces on it is compensated.

Ftr = F * cosα1.

m * g = N + F * sinα1.

N = m * g – F * sinα1.

Ftr = μ * N = μ * (m * g – F * sinα1).

μ * (m * g – F * sinα1) = F * cosα1.

μ = F * cosα1 / (m * g – F * sinα1).

μ = 39.2 N * cos60 ° / (10 kg * 9.8 m / s2 – 39.2 N * sin60 °) = 0.3.

m * a = F * cosα2 – Ftr.

m * g = N + F * sinα2.

N = m * g – F * sinα2.

Ftr = μ * N = μ * (m * g – F * sinα2).

a = (F * cosα2 – Ftr) / m = (F * cosα2 – μ * (m * g – F * sinα2)) / m.

a = (39.2 N * cos30 ° – 0.3 * (10 kg * 9.8 m / s2 – 39.2 N * sin30 °)) / 10 kg = 1 m / s2.

Answer: the body will move with acceleration a = 1 m / s2.