Friction & Stopping: 18-Wheeler vs. Small Car On Thruway

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SUMMARY

In a discussion regarding the stopping distances of an 18-wheeler and a small car both traveling at 30 m/s, it was established that if the coefficient of friction is equal, both vehicles would theoretically stop over the same distance. However, real-world factors such as brake pressure, tire heating, and contact patch area significantly influence stopping distances. While the physics suggests equal acceleration due to friction being proportional to mass, practical considerations indicate that heavier vehicles like 18-wheelers require more brake pressure and experience faster tire degradation, leading to longer stopping distances.

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  • Understanding of Newton's laws of motion
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  • Basic knowledge of vehicle dynamics and braking systems
  • Awareness of tire performance under different conditions
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  • Explore the physics of vehicle dynamics during emergency braking scenarios
  • Learn about the impact of contact patch area on vehicle traction
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kreil
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My physics teacher told us today that if an 18-wheeler and a small car, both traveling at 30 m/s, and with equal brand tires (thus the same coefficients of friction) were to lock their brakes while on the thruway, they would both take the same distance to stop. He said that the although it takes more energy to stop the truck, the truck compensates for it by producing more friction from its massive weight.

His explanation makes sense to me, but it goes against my intuition. Is his statement really true?
 
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Assuming the coefficient of friction is the same, then both truck and car would have the same acceleration. The only horizontal force acting on each is friction, which is proportional to the mass (F = \mu m g), thus the acceleration of each is the same: a = F/m = \mu g, independent of the mass.
 
That is too generalized. Assuming that the weight per square inch of tire contact patch was equal then yes it works out. But if the car has (arbitrary numbers) 1 square foot of contact patch and weighs 3000lbs (thats 3K/Sqfoot) and the semi has 6 square feet but weighs 20,000 lbs (thats 3300 lbs/squarfoot) then it would stop slower and longer. and if the semi was unloaded and weighed only 8,ooo lbs it still has the same contact patch... but assuming the weight per sQ foot of tire contact was equal then it sounds about right to me, in a VERY dumbed down way.
 
Assuming an admittedly simplified model of friction, the acceleration does not depend on the weight per square inch.
 
Yes, I realize it is over-simplified, I did this on purpose because I just wanted a quick simple answer.

Thanks!
 
Doc Al said:
Assuming the coefficient of friction is the same, then both truck and car would have the same acceleration. The only horizontal force acting on each is friction, which is proportional to the mass (F = \mu m g), thus the acceleration of each is the same: a = F/m = \mu g, independent of the mass.
This is absolutely correct. There are however, other real world consideration. Locking up the wheels on a fully loaded 18 wheeler is going to take much greater brake pressure than say a bicycle, which means a big rig is going to roll farther than a bicycle before it locks up. Another problem is heat. The tires of a heavy truck heat up much faster than a light vehicle. When a tire gets too hot [which happens pretty quick in a high speed skid], the coefficient of friction starts dropping - especially when the tires start melting. So yeah, in the real world, the intuition that big truck = greater stopping distance is a good rule to live by.
 

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