A modern grand - prix racing car of mass m is travelling on a flat track in a circular arc of radius R with a speed v. If the coefficient of static friction between the tyres and the track is μ s, then the magnitude of negative lift FL acting downwards on the car is: (Assume forces on the four tyres are identical and g = acceleration due to gravity) (a) m(g - (v²)/(μs R)) (b) -m(g + (v²)/(μs R)) (c) m((v²)/(μs R) - g) (d) m((v²)/(μs R) + g)

Answer: (c). Normal force on the car: N = mg + FL (downforce adds to weight). Friction provides the centripetal force: μs N = mv²/R. μs(mg + FL) = mv²/R ⇒ FL = (mv²)/(μs R) - mg So the answer is (c) m((v²)/(μs R) - g).

A modern grand - prix racing car of mass m is travelling on a flat… | JEE Main 2021 (17 Mar, Shift 1)

Q12JEE Main 2021 · 17 MAR · SHIFT 1Difficult

Level Road & Rotating Platform

A modern grand - prix racing car of mass m is travelling on a flat track in a circular arc of radius R with a speed v. If the coefficient of static friction between the tyres and the track is $\mu$ s, then the magnitude of negative lift FL acting downwards on the car is: (Assume forces on the four tyres are identical and g = acceleration due to gravity)

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Answer: (c)

Normal force on the car: $N = mg + F_L$ (downforce adds to weight).
Friction provides the centripetal force: $\mu_s N = \frac{mv^2}{R}$ .
$\mu_s(mg + F_L) = \frac{mv^2}{R} \Rightarrow F_L = \frac{mv^2}{\mu_s R} - mg$
So the answer is (c) $m\left(\frac{v^2}{\mu_s R} - g\right)$ .

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