A modern grand-prix racing car of mass m is t | vedclass

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Question

(A) For a car moving in a circular path on a flat track,the centripetal force is provided by the static friction force $f_{s}$.The maximum static fric

Vedclass · Accepted Answer

$m \left(\frac{v^{2}}{\mu_{s} R} - g\right)$

Vedclass · Answer

(A) For a car moving in a circular path on a flat track,the centripetal force is provided by the static friction force $f_{s}$.The maximum static friction force is given by $f_{s,max} = \mu_{s} N$,where $N$ is the normal force.The centripetal force required is $F_{c} = \frac{mv^{2}}{R}$.Thus,$\mu_{s} N = \frac{mv^{2}}{R}$,which gives $N = \frac{mv^{2}}{\mu_{s} R}$.The normal force $N$ on the car is the sum of its weight $mg$ and the downward negative lift force $F_{L}$ (aerodynamic downforce).Therefore,$N = mg + F_{L}$.Substituting the expression for $N$,we get $\frac{mv^{2}}{\mu_{s} R} = mg + F_{L}$.Solving for $F_{L}$,we get $F_{L} = \frac{mv^{2}}{\mu_{s} R} - mg = m \left(\frac{v^{2}}{\mu_{s} R} - g\right)$.

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