A car weighing 4 tons moves uphill with an acceleration of 0.2 m / s square. Find the pulling force if the slope

A car weighing 4 tons moves uphill with an acceleration of 0.2 m / s square. Find the pulling force if the slope is 0.02 and the drag coefficient is 0.04. sine alpha = 0.02, cosine alpha = 1. slope-ratio n / s = 0.02

m = 4 t = 4000 kg.

g = 10 m / s2.

a = 0.2 m / s2.

h / S = sinα = 0.02.

cos = 1.

μ = 0.02.

Fт -?

For the movement of a car on an inclined plane Let us write 2 Newton’s law in vector form: m * a = Ft + m * g + N + Ftr, where Ft is the thrust force of the engine, m * g is the force of gravity, N is the reaction force of the surface of the inclined plane, Ftr – friction force.

ОХ: m * a = Ft – Ftr – m * g * sinα.

OU: 0 = – m * g * cosα + N.

Ft = m * a + Ftr + m * g * sinα.

N = m * g * cosα.

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

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

Ft = 4000 kg * (0.2 m / s2 + 0.02 * 10 m / s2 * 1 + 10 m / s2 * 0.02) = 2400 N.

Answer: when driving on an inclined plane, the traction force of the car engine is Ft = 2400 N.