What force must be applied to lift a trolley weighing 600 kg along an overpass with an inclination

What force must be applied to lift a trolley weighing 600 kg along an overpass with an inclination angle of 20 degrees, if the coefficient of resistance to movement is 0.05.

To find out the force applied in the specified trolley (the movement is considered uniform), we project the acting forces onto the overpass surface: m * a = 0 = F – Ftr – Ft * sinα, from which we can express: F = Ftr + Ft * sinα = μ * mw * g * cosα + mw * g * sinα = mw * g * (μ * cosα + sinα).

Const: q – acceleration due to gravity (q ≈ 9.8 m / s2).

Data: mw – trolley weight (mw = 600 kg); α – tilt angle of the overpass (α = 20º); μ – coefficient of resistance to movement (μ = 0.05).

Calculation: F = mw * g * (μ * cosα + sinα) = 600 * 9.8 * (0.05 * cos 20º + sin 20º) ≈ 2287.35 N ≈ 2.29 kN.

Answer: To lift the specified trolley, a force of 2.29 kN is required.