# Braking distance vs. mass of the vehicle [closed]

We did an actual experiment where we checked braking distances of multiple trucks (rigid and semis), which had different GVW and consequently also axle loads. The results were very clear: for the same vehicle, the heavier it was, the longer was its braking distance (and it wasn't even close).

Now, we know that the mass of the vehicle has no effect on the braking distance, so I was looking for an explanation of the results. Is it true that the friction coefficient of a tire decreases with an increasing load (which would explain the result—see link below)?

https://en.wikipedia.org/wiki/Tire_load_sensitivity

• "we know that the mass of the vehicle has no effect on the braking distance" - how do you know that ? This chart from Wikipedia en.wikipedia.org/wiki/Braking_distance#/media/… says that the braking distances for a heavy truck (red bars) are roughly $40\%$ longer than for a passenger vehicle (orange bars), so it looks as if vehicle mass has some effect. Commented Jul 27 at 11:13
• @gandalf The longer braking distance of trucks vs cars can't be simply related to their larger mass. As discussed in the accepted answer to "Why do large trucks have a longer stopping distance than cars?", air brakes have an extra time lag and truck tires have lower friction coefficients than typical car tires. Commented Jul 27 at 16:37
• @Englishman Welcome to Physics Stack Exchange. I did not vote to close, but do think the question could be clarified (which might get it reopened). In your experiments, did the truck tires lock so the truck was skidding (in which tire friction is the primary factor) or did the tires continue to roll (in which case brakepad friction and ABS parameters are also relevant). You might also want to emphasize that your question is "Does a tire's friction coefficient decrease with increasing load, and if so why?", not "Why do otherwise identical but heavier vehicles take longer to stop?" Commented Jul 27 at 16:55
• @DavidBailey My point is that the assertion that "the mass of the vehicle has no effect on the braking distance" is not supported by the Wikipedia data. Yes, there may be several factors in play here. But the simple explanation of the experimental results is that the basic $d=\frac 1 2 \frac {v^2}{\mu g}$ model is not a good match to reality. Commented Jul 27 at 18:03

## 2 Answers

A really simple model which assumes no air resistance and using newton's laws and work energy theorem will give you a formula for braking distance of $$d = \frac{1}{2} v_{final\,velocity}^2/ \mu g$$.

When you do the math the mass term actually cancels. Your first assumption is a good one. So the only factor in that equation (or mathematical model if you feel fancy) that can change is the coefficient of friction which relates to the tyres. There would have to be a material explanation for it.

I did check the reference and wikipedia is correct in where you have referred to it. The relationship is true, but as the data suggests slip angle also seems to play a role. But this is related to lateral forces acting on the wheel which will probably be more related to cornering. That's my basic take on it as a high school physics teacher in China.

You may think of wheels as an extension of the road, on which you can grip more comfortably and apply the brakes indirectly.

On each of the wheels during braking, two forces are acting.

One between the road and the wheel, (the stopping force) and one between the braking pads and the wheel. Assuming no slip, during the time of braking, the torque provided by both these forces must match. Note that this is the max. critical value of braking force, above which the wheels would start to slip.

$$\tau_{on\:wheel\:due\:to\:road}=\tau_{on\:wheel\:due\:to\:brakes}$$

If the radial distance of the wheel axis to the braking pad is $$r_1$$ and that to the circumference of the wheel is $$r_2$$, and the normal force applied by the brakes on the wheel is B,

$$\mu_{between\:brake\:and\:wheel}\times B\times r_2=F\times r_1$$

Where F is the stopping force by the road on the wheel.

$$F=\mu_{between\:brake\:and\:wheel}\times B\times r_2/r_1$$

Notice that this stopping force is limited by the pressing force of the braking pads, and independent of the mass of the vehicle. Now the deceleration due to this stopping force is like 4 F/mass of truck (assuming there are 4 tires).

$$-4F=-Ma\implies a=4F/M$$

And the stopping distance would be:

$$S={u^2/2a}={M u^2/8F}$$

Stopping distance is directly proportional to Mass of the truck.