d = v2/2ug,
d = minimum stopping distance,
v = velocity when brakes were applied
u = coeff. of friction between wheels and road
g = gravitational constant,
The above equation shows that braking distance is independent of mass of vehicle. I know that depending on the value of slip, the value of 'u' will change, but lets assume that 'u' is constant. So does it mean that two different vehicles, one a normal car and other a sports car would have a same braking distance for a given same 'v'. I am confused because this does not seem to be the case. What am I missing?
Hi,
you could try to use D'Alemberts principle or the Energy balance to solve this dynamical problem.
If you use the energy balance then you compare to states. In the first state the car has only kinetic energy E1 = 1/2*m*v^2. Since the car breaks until standstill it has no kinetic energy in the second state. Due to the friction, however, the system is not conservative. The kinetic energy of the first state is consumed by the work of the friction force E2 = umgd. The energy balance now gives you
E1 = E2
And you get d = v2/2ug.
Best regards,
The weight does play a role. u = F_friction/F_normal. F_normal is the force that presses the road and tyres together, i.e. the weight of the car, F_normal = mg.
If we are comparing two cars, in your case a sedan car and a sports car, assuming that the mass is different for both cars (m_sports > m_sedan), say:
g = 10N/kg
m_sports = 2kg, F_sports = 2kg * g = 20N
m_sedan = 1kg, F_sedan = 1kg * g = 10N
say u = 1 for both cars,
then the only way u can be constant is by F_friction of both cars to change accordingly,
u_sedan = F_friction/F_sedan = 10N/10N = 1
u_sports = F_friction/F_sports = 20N/20N = 1
i.e. if the sports car has bigger mass, then the tyres need to be bigger as well. So if we say that u is constant, we are actually saying we are compensating for the mass of the cars with frictional force.
In reality though, for the same v, two cars (sports and sedan) do not have the same u, that is why you will see one go off a cliff.
For the sake of Argument we assume that u is equal in both cases and since the frictional force only depends on the normal force and u the mass does not Play a role theoretically.
Even in reality that should be the case if u is constant and acceleration is the same for both cars.
Best,
Theoretically it is but practically it is not. This is because of the varying operating variables such as normal loads (unladen and laden state) and load transfer effects during braking. As such, we can refer to EEC Directive 75/524 as guide for brake efficiency and adhesion utilisation analyses in order to incorporate suitable brake bias or proportioning valves to ensure good efficiency or adhesion utilisation trough out the possible vehicle operating variables.
You are right: theoretically, braking distance does not depend on mass. But in practice, yes. The explanation is that tire effective friction coefficient cannot be assumed constant: it slightly decreases with increasing normal force. So braking distances are longer when more mass is added. As a limit example, consider that on trucks, the tire coefficient usually is smaller than 0.65 even on dry roads>
