The locomotive develops a constant traction force of 147 kN on a horizontal track length of 600 m

| January 31, 2021 | education

The locomotive develops a constant traction force of 147 kN on a horizontal track length of 600 m. The train speed increases from 36 km / h to 54 km / h. determine the force of resistance to movement, considering it constant. train weight 1000 tons.

To find the value of the resistance force to the movement of the locomotive, we project all the acting forces onto the horizontal axis: m * a = Ftyag – Fcopr, whence Fcopr = Ftyag – m * a = Ftyag – m * (Vk ^ 2 – Vn ^ 2) / 2S.

Values ​​of variables: Ftyag – constant traction force (Ftyag = 147 kN = 147 * 10 ^ 3 N); m is the mass of the locomotive (m = 1000 t = 10 ^ 6 kg); Vк – final speed (Vк = 54 km / h = 15 m / s); Vн – the initial speed of the locomotive (Vн = 36 km / h = 10 m / s); S is the length of the horizontal section (S = 600 m).

Let’s perform the calculation: Fcopr = Ftyag – m * (Vk ^ 2 – Vn ^ 2) / 2S = 147 * 10 ^ 3 – 10 ^ 6 * (152 – 10 ^ 2) / (2 * 600) = 42.83 * 10 ^ 3 H = 42.83 kN.

Answer: The force of resistance to the movement of the locomotive is 42.83 kN.



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