The inclined plane is located at an angle α = 30 ° to the horizontal. What force must be applied along

The inclined plane is located at an angle α = 30 ° to the horizontal. What force must be applied along the plane to pull a 50kg load upwards with an acceleration of 2m / s²? Friction coefficient 0.2

∠α = 30 °.

m = 50 kg.

g = 9.8 m / s ^ 2.

a = 2 m / s ^ 2.

μ = 0.2.

F -?

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

ОХ: m * a = F – Ftr – m * g * sinα.

OU: 0 = – m * g * cosα + N.

F = m * a + Ftr + m * g * sinα.

N = m * g * cosα.

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

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

F = 50 kg * (2 m / s ^ 2 + 0.2 * 9.8 m / s ^ 2 * cos30 ° + 9.8 m / s ^ 2 * sin30 °) = 430 N.

Answer: to pull the body in, you need to apply a force F = 430 N.