Question 18: Car Slope Climbing Capability
Solution Analysis
1. Derivation of Normal Reactions ($N_r, N_f$):
Consider the torques about the contact points. Let $l$ be the wheelbase and $h$ be the height of the Center of Mass (CM). The CM is at distance $l/2$ from both wheels.
Calculating $N_f$ (Torque about Rear Wheel Contact):
$$ \Sigma \tau_{Rear} = 0 $$
$$ N_f \cdot l – (mg \cos \alpha) \cdot \frac{l}{2} – (mg \sin \alpha) \cdot h = 0 $$
(Note: $mg \sin \alpha$ acts at height $h$, creating a clockwise torque that reduces load on front).
$$ N_f = \frac{mg}{l} \left( \frac{l}{2} \cos \alpha – h \sin \alpha \right) $$
Calculating $N_r$ (Torque about Front Wheel Contact):
$$ \Sigma \tau_{Front} = 0 $$
$$ -N_r \cdot l + (mg \cos \alpha) \cdot \frac{l}{2} + (mg \sin \alpha) \cdot h = 0 $$
$$ N_r = \frac{mg}{l} \left( \frac{l}{2} \cos \alpha + h \sin \alpha \right) $$
2. Friction Limits (Climbing Condition):
The force required to climb is $F_{drive} = mg \sin \alpha$.
This force is limited by friction: $F_{drive} \le \mu N_{drive}$.
Case A: Rear Wheel Drive (RWD)
$$ mg \sin \alpha \le \mu N_r $$
$$ \sin \alpha \le \mu \left( \frac{1}{2} \cos \alpha + \frac{h}{l} \sin \alpha \right) $$
Dividing by $\cos \alpha$:
$$ \tan \alpha \le \frac{\mu}{2} + \frac{\mu h}{l} \tan \alpha $$
$$ \tan \alpha \left( 1 – \frac{\mu h}{l} \right) \le \frac{\mu}{2} \implies \tan \alpha \le \frac{\mu l}{2(l – \mu h)} $$
Case B: Front Wheel Drive (FWD)
$$ mg \sin \alpha \le \mu N_f $$
$$ \sin \alpha \le \mu \left( \frac{1}{2} \cos \alpha – \frac{h}{l} \sin \alpha \right) $$
$$ \tan \alpha \left( 1 + \frac{\mu h}{l} \right) \le \frac{\mu}{2} \implies \tan \alpha \le \frac{\mu l}{2(l + \mu h)} $$
3. Conclusion:
Comparing the denominators, $l – \mu h < l + \mu h$. A smaller denominator yields a larger maximum $\tan \alpha$. Thus, Rear-Wheel Drive is superior for climbing slopes.
Using values: $\mu=0.8, l=3, h=0.75$.
$$ \tan \alpha_{max} = \frac{0.8 \times 3}{2(3 – 0.8 \times 0.75)} = \frac{2.4}{2(2.4)} = 0.5 $$