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 $F_{L}$ acting downwards on the car is: (Assume forces on the four tyres are identical and g=acceleration due to gravity)

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$m (\frac{v^{2}}{μ_{S} R} + g)$

$m (\frac{v^{2}}{μ_{S} R} - g)$

$- m (g + \frac{v^{2}}{μ_{S} R})$

$m (g - \frac{v^{2}}{μ_{S} R})$

answer is B.

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Detailed Solution

$\begin{array}{l} F r i c t i o n p r o v i d e s c e n t r i p e t a l f o r c e . \\ \frac{m v^{2}}{R} = μ_{s} N \\ If we consider F_{L} is negative lift then N = mg + F_{L} \\ \Rightarrow μ_{s} (mg + F_{L}) = \frac{m v^{2}}{R} \\ \Rightarrow F_{L} = \frac{m v^{2}}{μ_{s} R} - m g \end{array}$

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