A locomotive is pulling a train with a mass of 180tonnes up an incline of 12° from rest to 80km/h for 2min. The rolling resistance of the train is 8500N. Calculate

Question

1) the acceleration
2 the output force of the train

Answer 1

To solve this problem, we first need to convert the given values and calculate the forces acting on the train.

Given data:

• Mass of the train (\( m \)): \( 180 , \text{tonnes} = 180,000 , \text{kg} \)
• Incline angle (\( \theta \)): \( 12° \)
• Final velocity (\( v_f \)): \( 80 , \text{km/h} = \frac{80 \times 1000}{3600} \frac{\text{m}}{\text{s}} = \frac{80000}{3600} \approx 22.22 , \text{m/s} \)
• Time (\( t \)): \( 2 , \text{min} = 2 \times 60 = 120 , \text{s} \)
• Rolling resistance force (\( F_r \)): \( 8500 , \text{N} \)

Part 1: Calculate the acceleration

To find the acceleration, we need to first determine the initial velocity (\( v_i \)), which is 0, and then use the equation of motion:

\[ v_f = v_i + a t \]

Rearranging, we get:

\[ a = \frac{v_f - v_i}{t} \]

Substituting the values:

\[ a = \frac{22.22 , \text{m/s} - 0}{120 , \text{s}} \approx \frac{22.22}{120} \approx 0.1852 , \text{m/s}^2 \]

Part 2: Calculate the output force of the train

The two main forces acting against the train as it moves uphill are gravity (due to the incline) and rolling resistance.

1. Gravitational force component acting down the incline:

\[ F_g = m \cdot g \cdot \sin(\theta) \] Where \( g \approx 9.81 , \text{m/s}^2 \)

Calculating the gravitational force component: \[ F_g = 180000 \cdot 9.81 \cdot \sin(12°) \]

Calculating \( \sin(12°) \): \[ \sin(12°) \approx 0.2079 \]

Now calculating \( F_g \): \[ F_g = 180000 \cdot 9.81 \cdot 0.2079 \approx 180000 \cdot 2.0421 \approx 36757.8 , \text{N} \]

2. Total force required to overcome rolling resistance: \[ F_{\text{total}} = F_g + F_r = 36757.8 , \text{N} + 8500 , \text{N} = 45257.8 , \text{N} \]

3. Net force required for acceleration: The net force required to accelerate the mass can be calculated using Newton's second law: \[ F_{\text{net}} = m \cdot a \] Substituting: \[ F_{\text{net}} = 180000 \cdot 0.1852 \approx 33336 , \text{N} \]

4. Total output force required by the locomotive: The total output force required by the locomotive to move the train uphill and overcome both the rolling resistance and gravitational force would be: \[ F_{\text{output}} = F_{\text{total}} + F_{\text{net}} = 45257.8 , \text{N} + 33336 , \text{N} = 78593.8 , \text{N} \]

Summary of Results:

1. The acceleration of the train is approximately \( 0.1852 , \text{m/s}^2 \).
2. The output force of the train is approximately \( 78593.8 , \text{N} \).