Maximum Lateral Wind Force on a Car – Detailed Solution

Maximum Lateral Wind Force on a Car

Question: A car of mass \(m\) is moving at a constant speed \(v\) around a circular track of radius \(R\). The maximum coefficient of static friction between the tires and the road is \(\mu\). In the absence of any additional force, the frictional force provides the necessary centripetal force for circular motion: \[ F_{\text{centripetal}} = \frac{mv^2}{R}. \] Suddenly, a lateral wind applies a sideways force \(F\) on the car (perpendicular to the original centripetal direction). The static friction now has to supply the vector sum of the centripetal force and the wind force. The car will begin to skid if this sum exceeds the maximum friction force, which is \(\mu mg\). Determine the maximum magnitude of the wind force \(F_{\text{max}}\) that can be applied without causing the car to skid.

Select the correct option:

• (1) \(F_{\text{max}} = \sqrt{(\mu mg)^2 - \left(\frac{mv^2}{R}\right)^2}\)
• (2) \(F_{\text{max}} = \mu mg - \frac{mv^2}{R}\)
• (3) \(F_{\text{max}} = \mu mg + \frac{mv^2}{R}\)
• (4) \(F_{\text{max}} = \sqrt{(\mu mg)^2 + \left(\frac{mv^2}{R}\right)^2}\)

Select the Correct Option

(1) \(F_{\text{max}} = \sqrt{(\mu mg)^2 - \left(\frac{mv^2}{R}\right)^2}\)

(2) \(F_{\text{max}} = \mu mg - \frac{mv^2}{R}\)

(3) \(F_{\text{max}} = \mu mg + \frac{mv^2}{R}\)

(4) \(F_{\text{max}} = \sqrt{(\mu mg)^2 + \left(\frac{mv^2}{R}\right)^2}\)

Step‑by‑Step Explanation

1. Friction as the Limiting Force:
   The maximum static friction available is: \[ f_{\text{max}} = \mu mg. \]
2. Effect of Wind Force:
   When no lateral force is applied, friction provides the centripetal force: \[ f_{\text{centripetal}} = \frac{mv^2}{R}. \]
3. Combined Requirement:
   When the lateral wind force \(F\) acts (perpendicular to the centripetal force), the net force that must be provided by friction is the vector sum: \[ F_{\text{required}} = \sqrt{\left(\frac{mv^2}{R}\right)^2 + F^2}. \] For equilibrium (no skidding), we need: \[ \sqrt{\left(\frac{mv^2}{R}\right)^2 + F^2} \le \mu mg. \]
4. Solving for \(F_{\text{max}}\):
   Setting the equality (at the threshold of skidding): \[ \sqrt{\left(\frac{mv^2}{R}\right)^2 + F_{\text{max}}^2} = \mu mg. \] Squaring both sides, \[ \left(\frac{mv^2}{R}\right)^2 + F_{\text{max}}^2 = (\mu mg)^2. \] Solving for \(F_{\text{max}}\): \[ F_{\text{max}}^2 = (\mu mg)^2 - \left(\frac{mv^2}{R}\right)^2, \] and thus, \[ F_{\text{max}} = \sqrt{(\mu mg)^2 - \left(\frac{mv^2}{R}\right)^2}. \]
5. Final Answer:
   The maximum lateral wind force that can be applied without skidding is: \[ F_{\text{max}} = \sqrt{(\mu mg)^2 - \left(\frac{mv^2}{R}\right)^2}, \] which corresponds to option (1).