A vehicle weighing 5 tons moves evenly uphill. Determine the traction force developed by the car engine

A vehicle weighing 5 tons moves evenly uphill. Determine the traction force developed by the car engine if the coefficient of friction is 0.7 and the lift angle is 30 °.

Given: m (mass of a car moving evenly uphill) = 5 t (in SI m = 5000 kg); μ (coefficient of friction) = 0.7; α (ascent angle) = 30º; g (acceleration due to gravity) ≈ 10 m / s2.

For uniform movement, equality is true (the projection of forces on the X-axis (the direction of the selected X-axis coincides with the direction of movement)): Ft = Ftr + Ftyl * sinα = μ * N + m * g * sinα = μ * m * g * cosα + m * g * sinα = m * g * (μ * cosα + sinα).

Calculation: Ft = 5000 * 10 * (0.7 * cos 30º + sin 30º) = 55 311 N ≈ 55.3 kN.