Calculate the specific tractive effort required to overcome a 1.5% uphill gradient for a 4500-ton freight train, assuming a total resistance of 35 kN for rolling and air resistance.

To calculate the specific tractive effort required, we must first determine the total force opposing the train's movement, which comprises gradient resistance and rolling and air resistance. Tractive effort is the total force a locomotive must exert to move a train, while specific tractive effort normalizes this force by the train's mass, typically expressed in Newtons per ton (N/ton). This standardized value is useful for comparing the performance demands on trains of different weights or on different gradients.

First, let's calculate the gradient resistance. Gradient resistance is the component of the train's weight that acts parallel to the incline, opposing upward motion. A gradient of 1.5% means that for every 100 units of horizontal distance, the track rises 1.5 units vertically. For railway engineering calculations involving small gradients, the resistance caused by gravity can be approximated as 98.1 Newtons per ton for every 1% of gradient. This value is derived from considering the force of gravity (mass × acceleration due to gravity) acting on a one-ton mass on a 1% slope (1000 kg × 9.81 m/s² × 0.01).

Given the train's mass is 4500 tons and the uphill gradient is 1.5%, the gradient resistance is calculated as follows:
Gradient Resistance = Train Mass × Gradient Percentage × 98.1 N/(ton · %)
Gradient Resistance = 4500 tons × 1.5 % × 98.1 N/(ton · %) = 662,175 Newtons (N).
To align with the unit of the other given resistance, we convert this to kilonewtons (kN): 662,175 N = 662.175 kN.

Next, we incorporate the total resistance due to rolling and air resistance. Rolling resistance is the force that opposes the motion of a train due to friction between the wheels and the rails, and deformation of the materials. Air resistance is the force opposing motion caused by the air particles hitting the train, increasing significantly with speed. The problem states that this combined total resistance is 35 kN.

The total tractive effort required by the locomotive is the sum of all forces it must overcome to move the train at a constant speed up the gradient. This includes both the gradient resistance and the total resistance from rolling and air.
Total Tractive Effort = Gradient Resistance + Total Resistance (Rolling and Air)
Total Tractive Effort = 662.175 kN + 35 kN = 697.175 kN.

Finally, to find the specific tractive effort, we divide the total tractive effort by the train's mass. This provides the force required per ton of train weight.
Specific Tractive Effort = Total Tractive Effort / Train Mass
To express this in N/ton, we convert the total tractive effort from kilonewtons back to Newtons: 697.175 kN = 697,175 N.
Specific Tractive Effort = 697,175 N / 4500 tons = 154.9277... N/ton.
Rounding to two decimal places, the specific tractive effort required is 154.93 N/ton.