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 μₛ, 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)

Solution

Physics sir · 3 steps

Downforce pushes the car into the road, so it adds to the weight in N and buys extra friction.

1. Vertical balance. The car has no vertical acceleration. Weight mg and the negative lift F_L both push down, and the road pushes up with N, so N = mg + F_L.
2. Horizontal need. On a flat track only static friction points towards the centre. At the limiting speed it supplies the full centripetal force, so μₛ N = mv²/R.

   μₛ(mg + F_L) = mv²/R

3. So. Dividing by μₛ gives mg + F_L = mv²/(μₛ R), so F_L = m(v²/(μₛ R) - g). That is option (c).

The idea to keep. Friction is μ times the normal force, not μ times the weight. Find N from the vertical balance first, adding every extra push into the road, then let μₛ N supply mv²/R.

Answer from the NTA final answer key. Solution by PrepWiser.