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Physics · LAWS OF MOTION

हिन्दी

Refer to the given figure showing a car of mass \( m \) navigating a circular track of radius \( R \) banked at an angle \( \theta \). Under the maximum speed condition where static friction \( f \) acts down the incline, which pair of equilibrium equations correctly describes the force components along the vertical direction and horizontal centripetal direction?

  1. Vertical: \( N = mg \cos\theta \); Horizontal: \( f = \frac{m v^2}{R} \)
  2. Vertical: \( N \sin\theta = mg + f \cos\theta \); Horizontal: \( N \cos\theta + f \sin\theta = \frac{m v^2}{R} \)
  3. Vertical: \( N \cos\theta + f \sin\theta = mg \); Horizontal: \( N \sin\theta - f \cos\theta = \frac{m v^2}{R} \)
  4. Vertical: \( N \cos\theta = mg + f \sin\theta \); Horizontal: \( N \sin\theta + f \cos\theta = \frac{m v^2}{R} \)

Correct answer · Option 4

Vertical: \( N \cos\theta = mg + f \sin\theta \); Horizontal: \( N \sin\theta + f \cos\theta = \frac{m v^2}{R} \)

Explanation

Resolving forces for a banked road: Vertically, upwards component \( N \cos\theta \) balances downwards gravity \( mg \) plus friction component \( f \sin\theta \), giving \( N \cos\theta = mg + f \sin\theta \). Horizontally, \( N \sin\theta + f \cos\theta \) provides the centripetal force \( \frac{m v^2}{R} \).
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