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I was doing my car license's questions, and it came up this question, why would the stoppage distance (the distance which the car stops after breaking) increase if you add more weight on top of the motorcycle?

I thought it would decrease, since the friction force depends on the normal force which increases after adding weight to it, isn't it?

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  • $\begingroup$ If you think of it as weight, it will be mysterious. But if you realised that adding weight is actually accomplished by adding mass, then the increase in inertia should be a good hint. $\endgroup$ Sep 6, 2023 at 6:19
  • $\begingroup$ @naturallyInconsistent can you expand on that inertia part please?, The only inertia I know of is the moment of inertia I which is used in rotation $\endgroup$
    – Ulshy
    Sep 6, 2023 at 11:32
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    $\begingroup$ The mass in weight = m g uses gravitational mass, whereas the mass in N2L, F = m a, is the inertial mass. They are equal: the more massive something is, the more difficult it is to change its state of motion. In particular, the more massive, the more difficult to stop. $\endgroup$ Sep 6, 2023 at 12:00

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A heavier vehicle contains more kinetic energy than a lighter one, and needs more braking power to stop it. If both vehicles have the same brake arrangement that generates the same braking force, the heavier vehicle will require more distance (over which that braking force acts) to stop.

This assumes that the wheels roll on the pavement without skidding/sliding in both the light and heavy vehicle cases.

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  • $\begingroup$ You are definitely correct, but the question was also correctly pointing out that the greater weight means greater frictional force possible, i.e. not the same braking force. $\endgroup$ Sep 6, 2023 at 6:47
  • $\begingroup$ @naturallyInconsistent, will edit. -NN $\endgroup$ Sep 6, 2023 at 6:53
  • $\begingroup$ It is usually the contact forces between the tyres and the road rather than the contact forces within the braking system which limit the braking of a car. Without ABS etc putting ones foot hard on brake pedal will result in a skid. $\endgroup$
    – Farcher
    Sep 6, 2023 at 8:31
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The situation is more complex then you might think in that for pneumatic tyres the maximum value of the coefficient of static friction depends on the vertical load and actually decreases as the load increases. This phenomena is called load sensitivity.

Thus the maximum braking force does not increase linearly with the mass of the car which results in the breaking distance increasing as the mass of the car is increased.

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  • $\begingroup$ I am not disputing the correctness of your answer, and the exam references real-world measurement data too, but the real-world measured dataset happens to correspond well with the linear assumption of the loading situation. i.e. The essence of the question can be answered without the complication you are referring to. $\endgroup$ Sep 6, 2023 at 8:10
  • $\begingroup$ @naturallyInconsistent I think that what the OP is suggesting is that if the normal force is $N = mg$ then the frictional force is $F= \mu N= \mu \,mg$ and as $F=ma$ then $ma = \mu\,mg \Rightarrow a=\mu g$, ie the acceleration is independent of the mass. What I have written is that $\mu$ is not constant but decreases with load, $mg$. $\endgroup$
    – Farcher
    Sep 6, 2023 at 8:27
  • $\begingroup$ So because of load sensitivity, adding mass always increases stopping distance? In other words, the increased KE is always the dominating factor over increased friction? $\endgroup$
    – AVS
    Sep 6, 2023 at 8:34
  • $\begingroup$ @Farcher again, you must be correct in the general sense, but I am saying that the real world data, the regime that appears in the driving exam and in the textbook for it, happens to exhibit only very little deviation from linearity, i.e. deviates only slightly from $a=\mu g$, such that the deviation due to load sensitivity is not the main phenomenon to consider. $\endgroup$ Sep 6, 2023 at 8:42
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For a top-heavy vehicle, the limit is not friction but stability. Braking doesn't simply decelerate the vehicle: it also applies a torque. The braking torque is opposed by torque due to the upward force of the ground on the front wheel(s). This is normally greater in magnitude than the braking torque: it is also opposed by the torque due to the upward force of the ground on the real wheel(s), which stabilizes the system.

But if the torque is sufficient to unload the rear wheel(s), the vehicle flips over. For a two wheeled vehicle breaking torque may unload the rear wheel enough that it locks and skids, leading to a loss of control even if the rear wheel isn't completely unloaded.

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