A body weighing 100 kg is acted upon by a force of 1500 N, angle alpha = 40 °, mu = 0.1

A body weighing 100 kg is acted upon by a force of 1500 N, angle alpha = 40 °, mu = 0.1. find mg, friction force Ffr and acceleration a

m = 100 kg.

g = 9.8 m / s ^ 2.

F = 150 N.

∠α = 40 °.

μ = 0.1.

Ftr -?

a – ?

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

Let us express Newton’s 2 law in projections.

ОХ: m * a = F * sinα – Ftr.

OU: F * cosα + N = m * g.

Ftr = μ * N.

N = m * g – F * cosα.

Ftr = μ * (m * g – F * cosα).

Ftr = 0.1 * (100 kg * 9.8 m / s ^ 2 – 150 N * cos40 °) = 83.1 N.

a = (F * sinα – Ftr) / m.

a = (150 N * sin40 ° – 83.1 N) / 100 kg = 0.011 m / s ^ 2.

Answer: the body is affected by the friction force Ffr = 83.1 N, the acceleration of the body is a = 0.011 m / s ^ 2.