Question

A highway curve with a radius of 750 m is banked properly for a car traveling 120 kph. If a 1590 kg car takes the turn at a speed of 230 kph, how much sideways force must the tires exert against the road if the car does not skid ?

• A. 6230 N
• B. 5800 N
• C. 7100 N
• D. 5150 N

Answer 1

Answer: (A)

6230 N

Answer 2

The banking angle $\theta$ is determined by the design speed $v_b$ using $\tan \theta = v_b^2 / (Rg)$. When the car travels at a higher speed $v$, it tends to slide outwards. The friction force $F_f$ acts down the incline. Resolving forces perpendicular and parallel to the banked surface leads to the equation $N = mg \cos \theta + F_f \sin \theta$ and $N \sin \theta + F_f \cos \theta - mg \sin \theta = mv^2/R$. Substituting $N$ and solving for $F_f$ yields $F_f = \frac{mv^2}{R} \cos \theta - mg \sin \theta$.

First, convert speeds to m/s: $v_b = 120 \text{ kph} = 120 \times \frac{1000}{3600} = \frac{100}{3} \text{ m/s}$ $v = 230 \text{ kph} = 230 \times \frac{1000}{3600} = \frac{575}{9} \text{ m/s}$

Calculate the banking angle using $g \approx 10 \text{ m/s}^2$: $\tan \theta = \frac{v_b^2}{Rg} = \frac{(100/3)^2}{750 \times 10} = \frac{10000/9}{7500} = \frac{10000}{67500} = \frac{4}{27}$

From $\tan \theta = \frac{4}{27}$, we construct a right triangle with opposite side 4 and adjacent side 27. The hypotenuse is $\sqrt{4^2 + 27^2} = \sqrt{16 + 729} = \sqrt{745}$. So, $\sin \theta = \frac{4}{\sqrt{745}}$ and $\cos \theta = \frac{27}{\sqrt{745}}$.

Now, calculate the friction force $F_f$ using the derived formula: $F_f = \frac{mv^2}{R} \cos \theta - mg \sin \theta$ $m = 1590 \text{ kg}$ $v = \frac{575}{9} \text{ m/s}$ $R = 750 \text{ m}$ $g = 10 \text{ m/s}^2$

$F_f = \frac{1590 \times (\frac{575}{9})^2}{750} \times \frac{27}{\sqrt{745}} - 1590 \times 10 \times \frac{4}{\sqrt{745}}$ $F_f = \frac{1590 \times \frac{330625}{81}}{750} \times \frac{27}{\sqrt{745}} - \frac{63600}{\sqrt{745}}$ $F_f = \frac{1590}{750} \times \frac{330625}{81} \times \frac{27}{\sqrt{745}} - \frac{63600}{\sqrt{745}}$ $F_f = \frac{53}{25} \times \frac{330625}{3} \times \frac{1}{\sqrt{745}} - \frac{63600}{\sqrt{745}}$ $F_f = \frac{699925}{3\sqrt{745}} - \frac{190800}{3\sqrt{745}} = \frac{509125}{3\sqrt{745}}$

Using decimal approximations for intermediate steps to match the answer's precision: $v \approx 63.889 \text{ m/s}$ $\tan \theta \approx 0.14815$, so $\theta \approx 8.426^\circ$ $\sin \theta \approx 0.14675$, $\cos \theta \approx 0.98925$

$\frac{mv^2}{R} = \frac{1590 \times (63.889)^2}{750} \approx 8653.39$ $mg \sin \theta = 1590 \times 10 \times 0.14675 \approx 2323.3$

$F_f = (8653.39 \times 0.98925) - 2323.3 \approx 8558.7 - 2323.3 \approx 6235.4 \text{ N}$

The slight difference from 6230 N is due to rounding in intermediate calculations or the assumed value of $g$. The calculated value is very close to the provided answer.