What speed can a tram develop on an uphill slope of 10 °, having four electric motors of 55
What speed can a tram develop on an uphill slope of 10 °, having four electric motors of 55 kW each, if the weight of the tram with passengers is 30 tons, and the friction coefficient is 0.05?
∠α = 10 °.
P1 = 55 kW = 55000 W.
m = 30 t = 30,000 kg.
g = 10 m / s2.
n = 4.
μ = 0.05.
V -?
Since the tram moves uniformly, it means, according to 1 Newton’s law, the effect of forces on it is compensated.
F + Ftr + N + m * g = 0 – vector.
ОХ: F – Ftr – m * g * sinα = 0
OU: N – m * g * cosα = 0.
N = m * g * cosα.
F = Ftr + m * g * sinα.
Ftr = μ * N = μ * m * g * cosα.
F = μ * m * g * cosα + m * g * sinα = (μ * cosα + sinα) * m * g.
The power P of the tram with uniform movement is expressed by the formula: P = F * V = (μ * cosα + sinα) * m * g * V.
P = n * P1, where n is the number of electric motors, P1 is the power of one motor.
(μ * cosα + sinα) * m * g * V = n * Р1.
V = n * Р1 / (μ * cosα + sinα) * m * g.
V = 4 * 55000 W / (0.05 * cos10 ° + sin10 °) * 30000 kg * 10 m / s2 = 3.3 m / s.
Answer: the tram will move at a speed of V = 3.3 m / s.