Subtopic: Thermodynamic principles of I.C. engines
Question Statement:
A two-wheel drive tractor has a total weight of $24\text{ kN}$ with a wheelbase of $2.0\text{ m}$. In static condition on level ground, the center of gravity is located $0.7\text{ m}$ ahead of the rear axle centerline and $0.8\text{ m}$ above ground. If the tractor develops a horizontal drawbar pull of $8.0\text{ kN}$ at a hitch height of $0.4\text{ m}$, calculate the dynamic reaction on the rear axle in $\text{kN}$.
Answer Key:
17.2
Step-by-Step Solution & Explanation
Taking moments about the front wheel ground contact point to find dynamic rear reaction $R_r$:
$R_r \times x_w = W \times (x_w - x_r) + P \times y_h$
Where:
- Total weight $W = 24\text{ kN}$
- Wheelbase $x_w = 2.0\text{ m}$
- Distance of CG from rear axle $x_r = 0.7\text{ m}$, so distance from front axle is $(2.0 - 0.7) = 1.3\text{ m}$
- Drawbar pull $P = 8.0\text{ kN}$
- Hitch height $y_h = 0.4\text{ m}$
Static rear reaction:
$R_{rs} = W \times \frac{x_w - x_r}{x_w} = 24 \times \frac{1.3}{2.0} = 15.6\text{ kN}$
Dynamic weight transfer to rear:
$\Delta W = \frac{P \times y_h}{x_w} = \frac{8.0 \times 0.4}{2.0} = 1.6\text{ kN}$
Dynamic rear axle reaction:
$R_r = R_{rs} + \Delta W = 15.6 + 1.6 = 17.2\text{ kN}$