# when braking a car, why do the brakes heat up?

My understanding is that if the car isn't slipping, that static friction at the point of contact between the tyre and the car is what decelerates the car, counteracting the torque which is being applied by the brakes.

This torque is due to kinetic friction (as the rotor has rotational velocity, and the brake pads are held in place) but since this kinetic friction is an internal force / not the net force, do the rotor/brake-pads still heat up? It appears as if they do in the extreme cases (F1 racing), even though the car doesn't appear to be slipping. I'm clearly missing something here.

• "...contact between the tyre and the car is what decelerates the car, counteracting the torque which is being applied by the brakes" doesn't make much sense. Could you expand that a bit with a quick diagram or something? Alternatively, that system could be the cause for your misunderstanding.
– user191954
Aug 15, 2018 at 16:11
• absolutely. I suspect this is where my picture is incomplete/incorrect, but this is correctly what I imagine the system to be: imgur.com/wVAuexU. The torque applied by the brakes is due to kinetic friction (brake pads/rotor at different velocities). The counteracting force (and thus, net force on the car) on the bottom in green is due to static friction, as the velocity at the point of contact between tyre/road is zero, so the tyre and the road are instantaneously at the same speed. Aug 15, 2018 at 16:25
• Kinetic energy always causes heat loss. Regardless if it is internal or external. This is due to kinetic energy by its very nature always doing work $W=\int F\;dx$. This work must go somewhere - so it turns into heat. Aug 15, 2018 at 18:30