Question

A 2.5 tonne diesel locomotive hauls 5 mine cars upslope having a gradient of 1 in 20. The constant tractive force of the locomotive is 1800 kN. The gross weight of a car is 3 tonne. Acceleration due to gravity is 10.0 m/s\(^2\). If the acceleration of the system is 0.5 m/s\(^2\), the rolling resistance in kN/tonne is ________ (rounded off to 3 decimal places).

Hint:

Rolling resistance can be calculated by subtracting the forces of gravity and slope from the total tractive force and then dividing by the total weight in tonnes.

Answer 1

Given Data Locomotive mass ($m_L$) = 2.5 tonnes = 2500 kg Tractive force ($F_{{tractive}}$) = 1800 kN Number of mine cars = 5 Mass per car = 3 tonnes Total cars mass ($m_C$) = $5 \times 3 = 15$ tonnes = 15000 kg Gradient = 1 in 20 ($\sin\theta = 0.05$) System acceleration ($a$) = 0.5 m/s\(^2\) Gravity ($g$) = 10.0 m/s\(^2\) 
Step 1: Calculate Total Mass \[ m_{{total}} = m_L + m_C = 2.5 \, {tonnes} + 15 \, {tonnes} = 17.5 \, {tonnes} = 17500 \, {kg} \] Step 2: Calculate Gradient Force \[ F_{{gradient}} = m_{{total}} \times g \times \sin\theta = 17500 \, {kg} \times 10 \, {m/s}^2 \times 0.05 = 8750 \, {N} = 8.75 \, {kN} \] Step 3: Calculate Inertial Force \[ F_{{inertia}} = m_{{total}} \times a = 17500 \, {kg} \times 0.5 \, {m/s}^2 = 8750 \, {N} = 8.75 \, {kN} \] Step 4: Determine Rolling Resistance Force \[ F_{{tractive}} = F_{{gradient}} + F_{{inertia}} + F_{{roll}} \] \[ 1800 \, {kN} = 8.75 \, {kN} + 8.75 \, {kN} + F_{{roll}} \] \[ F_{{roll}} = 1800 \, {kN} - 17.5 \, {kN} = 1782.5 \, {kN} \] Step 5: Calculate Rolling Resistance per Tonne \[ {Rolling Resistance} = \frac{F_{{roll}}}{m_{{total}}} = \frac{1782.5 \, {kN}}{17.5 \, {tonnes}} = 101.857 \, {kN/tonne} \] Final Answer The rolling resistance is 101.857 kN/tonne.