Banking with Friction — Maximum and Minimum Speed

The situation

A banked road of angle $\theta$ has coefficient of static friction $\mu$ between tires and road. There is a range of safe speeds. Below $v_{min}$, the vehicle slides down the bank. Above $v_{max}$, it slides up the bank.

Finding $v_{max}$ — tendency to slide up

At maximum speed, the vehicle tends to slide up the bank (outward). Friction acts down the bank (inward, toward centre).

Forces on vehicle of mass $m$:

• Weight $mg$ downward
• Normal force $N$ perpendicular to bank
• Friction $\mu N$ down the bank (toward centre and down)

Resolving along and perpendicular to the bank is messy. Easier to resolve vertically and horizontally.

Vertical:

$N\cos\theta - \mu N\sin\theta = mg$

$N(\cos\theta - \mu\sin\theta) = mg \quad \text{...(1)}$

Horizontal (centripetal, inward):

$N\sin\theta + \mu N\cos\theta = \frac{mv_{max}^2}{r}$

$N(\sin\theta + \mu\cos\theta) = \frac{mv_{max}^2}{r} \quad \text{...(2)}$

Divide (2) by (1):

$\frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta} = \frac{v_{max}^2}{rg}$

Divide numerator and denominator by $\cos\theta$:

$\frac{\tan\theta + \mu}{1 - \mu\tan\theta} = \frac{v_{max}^2}{rg}$

$\boxed{v_{max} = \sqrt{rg \cdot \frac{\tan\theta + \mu}{1 - \mu\tan\theta}}}$

Finding $v_{min}$ — tendency to slide down

At minimum speed, the vehicle tends to slide down the bank (inward). Friction acts up the bank (outward, away from centre).

Vertical:

$N\cos\theta + \mu N\sin\theta = mg$

$N(\cos\theta + \mu\sin\theta) = mg \quad \text{...(3)}$

Horizontal (centripetal):

$N\sin\theta - \mu N\cos\theta = \frac{mv_{min}^2}{r}$

$N(\sin\theta - \mu\cos\theta) = \frac{mv_{min}^2}{r} \quad \text{...(4)}$

Divide (4) by (3):

$\frac{\sin\theta - \mu\cos\theta}{\cos\theta + \mu\sin\theta} = \frac{v_{min}^2}{rg}$

$\boxed{v_{min} = \sqrt{rg \cdot \frac{\tan\theta - \mu}{1 + \mu\tan\theta}}}$

The safe speed range

$v_{min} \leq v \leq v_{max}$

Within this range, static friction adjusts to whatever value is needed (between zero and $\mu N$) — no skidding.

Special cases

No friction ($\mu = 0$):

$v_{max} = v_{min} = \sqrt{rg\tan\theta} = v_{ideal}$

Only one safe speed — the ideal banking speed. Any deviation causes skidding.

$v_{min} = 0$: This happens when $\tan\theta = \mu$, i.e. the banking angle equals the angle of friction. The vehicle can be stationary on the bank without sliding down.

Condition for $v_{min}$ to be real

$\tan\theta \geq \mu$

If $\tan\theta < \mu$, the formula gives an imaginary $v_{min}$ — meaning the vehicle can remain stationary on the bank (friction is sufficient to prevent downward sliding even at $v = 0$).