Gear ratios of a four-speed car

Take $g = 9.81\ \text{m/s}^2$, $W = mg = 14715$ N.

(a) First-gear ratio

Slope angle $\theta = \tan^{-1}(0.3) = 16.70^\circ$; $\sin\theta = 0.2873$, $\cos\theta = 0.9578$.

Tractive force needed (gradient plus rolling resistance):

$$\begin{aligned} F &= W(\sin\theta + f\cos\theta) \\ &= 14715\,(0.2873 + 0.015\times0.9578) = 4440\ \text{N} \end{aligned}$$

Torque at the wheels: $T_w = F r = 4440 \times 0.32 = 1421\ \text{N m}$.

Overall ratio from engine to wheels:

$$G_o = \frac{T_w}{T_e\,\eta} = \frac{1421}{120\times0.88} = 13.45$$

First-gear ratio of the gearbox:

$$G_1 = \frac{G_o}{G_f} = \frac{13.45}{4.1} = 3.28$$

(b) Intermediate ratios (GP)

For four speeds, $G_1, G_2, G_3, G_4 = G_1,\ G_1/k,\ G_1/k^2,\ G_1/k^3$ with $G_4 = 1$:

$$k = (3.28)^{1/3} = 1.486$$

(c) Maximum speed in top gear

Engine speed $\omega_e = 5500\times2\pi/60 = 575.96$ rad/s.

$$v = \frac{\omega_e\, r}{G_4 G_f} = \frac{575.96\times0.32}{1\times4.1} = 44.95\ \text{m/s} = 161.8\ \text{km/h}$$

Traction check

Maximum tractive force available $= \mu\,(0.55\,W\cos\theta) = 0.7\times0.55\times14715\times0.9578 = 5426$ N.

Since $5426 > 4440$ N, the wheels do not slip on the 30% slope.

Answer: $G_1 = 3.28$; $G_2 = 2.21$, $G_3 = 1.49$; top speed about 162 km/h; traction is sufficient (5426 N available against 4440 N required).