Question

A battery-powered locomotive hauls a 3.0 tonne train. Coefficient of friction = 0.06, coefficient of adhesion = 0.2, time to accelerate to 1.8 m/s = 3.0 min, gradient = 1 in 20. Find the minimum weight of the locomotive (in tonnes), rounded to one decimal place.

Hint:

Include both train resistance and the locomotive's own rolling resistance when computing adhesion requirements.

Answer 1

Correct Answer: 3.7

Step 1: Convert data. \[ W_t = 3.0\ \text{tonne} = 3000\ \text{kg},\quad v = 1.8\ \text{m/s},\quad t = 3.0\ \text{min} = 180\ \text{s}. \] Acceleration: \[ a = \frac{1.8}{180} = 0.01\ \text{m/s}^2. \] Step 2: Train resistance forces. Frictional resistance: \[ F_f = \mu W_t g = 0.06 \times 3000 \times 9.81 = 1765.8\ \text{N}. \] Gradient resistance: \[ \frac{1}{20} = 0.05,\quad F_g = 0.05 W_t g = 0.05 \times 3000 \times 9.81 = 1471.5\ \text{N}. \] Acceleration force: \[ F_a = W_t a = 3000(0.01) = 30\ \text{N}. \] Total pull required: \[ F = F_f + F_g + F_a = 1765.8 + 1471.5 + 30 = 3267.3\ \text{N}. \] Step 3: Adhesion force provided by locomotive. If $W_L$ is locomotive weight in kg: \[ F_{\text{adh}} = 0.2 \, W_L \, g. \] Set required pull = adhesion force: \[ 0.2 W_L g = 3267.3. \] \[ W_L = \frac{3267.3}{0.2 \times 9.81} = 166.5\ \text{kg}. \] Convert to tonnes: \[ W_L = 1665\ \text{kg} = 1.665\ \text{tonnes}. \] But a locomotive must also overcome its own rolling resistance: \[ F_{\text{self}} = 0.06 W_L g. \] Solve simultaneously: \[ 3267.3 + 0.06 W_L g = 0.2 W_L g. \] \[ 3267.3 = 0.14 W_L g. \] \[ W_L = \frac{3267.3}{0.14 \times 9.81} = 3700\ \text{kg} = 3.7\ \text{tonnes}. \] Rounded: \[ \boxed{3.7\ \text{tonnes}} \]