A car with a mass of 1 ton rises along a highway with an angle of 30 degrees under the action of a traction force of 7 kN.

A car with a mass of 1 ton rises along a highway with an angle of 30 degrees under the action of a traction force of 7 kN. The coefficient of friction between the tires of the car and the surface of the highway is 0.1. Find car acceleration.

m = 1 t = 1000 kg.
g = 10 m / s2.
Ft = 7 kN = 7000 N.
μ = 0.1.
∠α = 30 °.
a – ?
Let us write for the car 2 Newton’s law in vector form: m * a = Ft + Ftr + m * g + N, where m is the mass of the car, a is the acceleration of the car, Ft is the traction force, Ftr is the friction force, m * g is the force gravity, N is the reaction force of the road.
ОХ: m * a = Fт – Fтр – m * g * sinα.
OU: 0 = – m * g * cosα + N.
a = (Ft – Ftr – m * g * sinα) / m.
N = m * g * cosα.
The friction force Ffr is expressed by the formula: Ffr = μ * N = μ * m * g * cosα.
a = (Fт – μ * m * g * cosα – m * g * sinα) / m.
a = (7000 N – 0.1 * 1000 kg * 10 m / s2 * cos30 ° – 1000 kg * 10 m / s2 * sin30 °) / 1000 kg = 1.1 m / s2.
Answer: the car is moving with acceleration a = 1.1 m / s2.