Friction on Car

[joburnet]

I have a friend who thinks that a car that weighs more will get better gas mileage going 60mph on flat ground because the extra weight will help the car break through the wind. Please explain what will happen and why. Thanks.

 

[iwantanewcomputer]

if it is going constant speed and same aerodynamics/gearing, the force of air pushing against the front is the same...so the only thing we have to worry about is which has more slip between the wheels and ground.

more weight gives it a higher normal force which makes the force of friction between the tire and raod greater. higher friction means less slip means more of the work done by the engine is used against the wind resistance and less to spin the wheels.

think of an infinately heavy car... it would have no slip because there is too much friction

 

[f95toli]

The weight of the car is irrelevant in this case, weight won't help you against the wind.

Think of it in terms of energy, basically the car will loose kinetic energy (speed) due to the wind unless more (potential) energy is supplied by the engine (which in turn gets its energy from the fuel).

Hence, the only relevant factor in this case are the aerodynamic properties of the car.

 

[Armitage]

Originally posted by: iwantanewcomputer
if it is going constant speed and same aerodynamics/gearing, the force of air pushing against the front is the same...so the only thing we have to worry about is which has more slip between the wheels and ground.

more weight gives it a higher normal force which makes the force of friction between the tire and raod greater. higher friction means less slip means more of the work done by the engine is used against the wind resistance and less to spin the wheels.

think of an infinately heavy car... it would have no slip because there is too much friction

In general (steady state, dry roads, etc.) you won't have any wheel slip regardless of weight.

Weight is irrelevant in the steady state on level ground. Mass is only a factor if you're accelerating.

 

[DrPizza]

As Armitage stated, you're really not losing any energy due to friction with the surface of the road... there may be a very slight amount of slippage, but I highly doubt it's even close to significant.

HOWEVER, do lose energy to "friction", caused by the constant deformation of the tires so that the "patch" of the tire is flat against the surface of the road. The lower the pressure of the tires, the larger the patch is going to be. Also, them more the car weighs, the larger the patch is going to be. Thus, if the tire pressure remains constant, adding mass to the car is going to increase the rolling resistance, and thus decrease fuel economy.

Perhaps what your friend is thinking about is that with a given shape of vehicle, the force of air resistance is going to be the same, regardless of the mass of the car. However, using F=ma, the higher the mass, the lower the acceleration. Thus, a heavier vehicle won't lose speed as quickly due to air resistance. To understand this better, drop two identically sized balls - one lead, and one styrofoam. Notice that the styrofoam ball takes longer to get to the ground. However, if both were dropped in the absence of air, they'd both hit the ground at the exact same time.

So, this seems like heavier would be better.
However, because KE = 1/2 m v^2, and conservation of energy,
having a heavier car is going to require more energy to get up to a certain speed.

Since the air resistance is a force acting in the opposite direction of the motion of the car, the engine needs to provide a force (via the road/tire interface) equal in magnitude to the wind resistance to maintain a constant speed (net force = 0)
Both act on the car over the same distance, so using work = force * distance,
the amount of work the engine does has to be equal to the amount of work done to the car by the air resistance (= force * distance), else, the car will slow down. Therefore, regardless of the mass of the car, if the shape is identical, the work done against air resistance is the same.

So, the heavier car will use more energy because of increased tire deformation.
This also points to why it is so important to watch your tire inflation pressures on a regular basis... properly inflated tires can make a tremendous impact on miles per gallon; far more of a difference than typical weight differences may cause.

Want to check this hypothesis? It's pretty simple to do. Road trip! Travel 200 miles to a distant Home Depot or Lowes or some such place. Carefully measure mpg on the trip. Purchase 10-15 80 pound bags of concrete. Drive back home, avoid pot holes! Carefully measure mpg on the return trip. Even if you have a head wind on the way out, and a tail wind on the way back, I don't think it'll make that much of a difference compared to the difference caused by the weight difference. But, then again, you can even try to account for this in your experiment by returning the concrete to HD, then driving back home again. 🙂

 

[CycloWizard]

The other guys have explained it very well, but here's another way to look at it that might suit you better. Consider car in a wind tunnel. In one experiment, the car is empty (just four seats and a steering wheel). In the next, fill the car with concrete. As long as the windows are up in both experiments, the results should be exactly the same.

Now, to confuse the issue: the only things that have an effect on wind resistance are the car aerodynamics and velocity. In this case, the car is stationary and the air velocity varies. However, fluid dynamics tells us that this type of experiment is exactly the same, only the frame of reference has changed (Eulerian vs Lagrangian perspective).

 

[TankGuys]

Originally posted by: CycloWizard
The other guys have explained it very well, but here's another way to look at it that might suit you better. Consider car in a wind tunnel. In one experiment, the car is empty (just four seats and a steering wheel). In the next, fill the car with concrete. As long as the windows are up in both experiments, the results should be exactly the same.

Now, to confuse the issue: the only things that have an effect on wind resistance are the car aerodynamics and velocity. In this case, the car is stationary and the air velocity varies. However, fluid dynamics tells us that this type of experiment is exactly the same, only the frame of reference has changed (Eulerian vs Lagrangian perspective).

Please don't talk about Lagrangian mechanics, it brings back too many painful memories from my advanced mechanics class 🙂

 

[Calin]

You should also mention that there are other ways to loose energy because of increased mass: increased friction in wheels bearings, mostly (there might be other things to take into consideration)

 

[Armitage]

Originally posted by: DrPizza
As Armitage stated, you're really not losing any energy due to friction with the surface of the road... there may be a very slight amount of slippage, but I highly doubt it's even close to significant.

HOWEVER, do lose energy to "friction", caused by the constant deformation of the tires so that the "patch" of the tire is flat against the surface of the road. The lower the pressure of the tires, the larger the patch is going to be. Also, them more the car weighs, the larger the patch is going to be. Thus, if the tire pressure remains constant, adding mass to the car is going to increase the rolling resistance, and thus decrease fuel economy.

Perhaps what your friend is thinking about is that with a given shape of vehicle, the force of air resistance is going to be the same, regardless of the mass of the car. However, using F=ma, the higher the mass, the lower the acceleration. Thus, a heavier vehicle won't lose speed as quickly due to air resistance. To understand this better, drop two identically sized balls - one lead, and one styrofoam. Notice that the styrofoam ball takes longer to get to the ground. However, if both were dropped in the absence of air, they'd both hit the ground at the exact same time.

So, this seems like heavier would be better.
However, because KE = 1/2 m v^2, and conservation of energy,
having a heavier car is going to require more energy to get up to a certain speed.

Since the air resistance is a force acting in the opposite direction of the motion of the car, the engine needs to provide a force (via the road/tire interface) equal in magnitude to the wind resistance to maintain a constant speed (net force = 0)
Both act on the car over the same distance, so using work = force * distance,
the amount of work the engine does has to be equal to the amount of work done to the car by the air resistance (= force * distance), else, the car will slow down. Therefore, regardless of the mass of the car, if the shape is identical, the work done against air resistance is the same.

So, the heavier car will use more energy because of increased tire deformation.
This also points to why it is so important to watch your tire inflation pressures on a regular basis... properly inflated tires can make a tremendous impact on miles per gallon; far more of a difference than typical weight differences may cause.

Want to check this hypothesis? It's pretty simple to do. Road trip! Travel 200 miles to a distant Home Depot or Lowes or some such place. Carefully measure mpg on the trip. Purchase 10-15 80 pound bags of concrete. Drive back home, avoid pot holes! Carefully measure mpg on the return trip. Even if you have a head wind on the way out, and a tail wind on the way back, I don't think it'll make that much of a difference compared to the difference caused by the weight difference. But, then again, you can even try to account for this in your experiment by returning the concrete to HD, then driving back home again. 🙂

You're right - the increased deformation of the tires would likely impact the mileage somewhat. I think you'd have to increase the weight quite a bit to see it, but it would be there. Similarly for the wheel bearings, as Calin pointed out.

You're road trip won't work very well to test the original hypothesis though, unless it's a very flat route with very few stops. Increased weight will definitely impact your mileage on hills, and accelerating. Maybe if you had one of those instantaneous MPG displays on your dash you could watch it on a particular flat stretch of road loaded & unloaded - but I doubt that has the precision needed.