Balancing Pressure: The Impact of Wheel Size on Weight Distribution

[Atomic_Sheep]

If we have a car with no wheels an one of those sporty aero flat underside sitting on ice, then the pressure is just equally distributed throughout the whole underside and is relatively low, so a thin layer of ice can support it.

What happens when we put the car on wheels? The contact patch now becomes the size of a matchbox on each of the four corners, so if the car weights 1T, then I'm guessing each wheel's contact patch is holding 250kg per square matchbox?

 

[anorlunda]

I'm not sure about the matchbox size, but yes it is true that spreading the weight over more area makes it less likely to break through ice.
That seems obvious.

 

[Rive]

so a thin layer of ice can support it.

This part is a bit tricky, since ice would not break below the car, but around the car. If it is not thick enough then it won't be able to spread the stress, regardless of the pressure at the center.

 

[Dr.D]

The point being made by @Rive, as I understand him, is that it is not a matter simply of penetration directly below the contact point, but also a matter of the bending stress induced in the ice sheet. Some distance away from the car, the stress distribution will be exactly the same for both types of contact. It is the bending stress that will break the ice, not the contact stress.

 

[Atomic_Sheep]

I'm more trying to understand why the pressure under each wheel is a quarter of the whole car? This to me doesn't seem intuitive/obvious. I only brought the ice example because it's a classic way of talking about the problem. I don't understand why the car can't weigh a tonne and the contact point of each wheel having a tonne transferred through it? I don't have a second set of scales to weight myself standing one foot per scale, but intuitively, I find it hard to believe half my weight will be reflected on the respective scale.

 

[Lord Jestocost]

I don't have a second set of scales to weight myself standing one foot per scale, but intuitively, I find it hard to believe half my weight will be reflected on the respective scale.

Have a look at https://en.wikiversity.org/wiki/Force_equilibrium

 

[russ_watters]

I'm more trying to understand why the pressure under each wheel is a quarter of the whole car? This to me doesn't seem intuitive/obvious. I only brought the ice example because it's a classic way of talking about the problem. I don't understand why the car can't weigh a tonne and the contact point of each wheel having a tonne transferred through it? I don't have a second set of scales to weight myself standing one foot per scale, but intuitively, I find it hard to believe half my weight will be reflected on the respective scale.

Put your other foot on the floor.

I'm not sure why this isn't intuitive to you.

 

[Asymptotic]

I'm more trying to understand why the pressure under each wheel is a quarter of the whole car? This to me doesn't seem intuitive/obvious. I only brought the ice example because it's a classic way of talking about the problem. I don't understand why the car can't weigh a tonne and the contact point of each wheel having a tonne transferred through it? I don't have a second set of scales to weight myself standing one foot per scale, but intuitively, I find it hard to believe half my weight will be reflected on the respective scale.

Try this thought experiment.
Imagine a plywood platform with a rather unconventional chair situated on it that instead of four legs has 64 legs, and with a weigh scale under each chair leg. Let's say the combined weight of this chair and the person sitting on it is 640 kg.

If each scale read 640 kg then (64 legs x 640 kg) total weight would be 40960 kg, and the platform would almost certainly collapse.
Total mass doesn't change, so what happens is each leg supports a fraction of it (640 kg/64 legs = 10 kg per leg).

 

[Rive]

This to me doesn't seem intuitive/obvious.

Then instead of looking for explanation first you should look for examples and absurdities to challenge your 'intuition'.

For example, if one man can't lift a weight and calls for help, how would be able two lift the same weight if weight not shared but every contact point holds the same (original) weight?

If you push a weight with one hand or you push it with two hand, will you push twice of the weight with two hands?

If every contact point holds the same (original) weight, then to push a pole down the ground would just require a fluffy feather with maaaany contact points?

And so on.

 

[Chestermiller]

I'm more trying to understand why the pressure under each wheel is a quarter of the whole car? This to me doesn't seem intuitive/obvious. I only brought the ice example because it's a classic way of talking about the problem. I don't understand why the car can't weigh a tonne and the contact point of each wheel having a tonne transferred through it? I don't have a second set of scales to weight myself standing one foot per scale, but intuitively, I find it hard to believe half my weight will be reflected on the respective scale.

Have you tried drawing a free-body diagram of this rigid body and taking moment balances.

 

[Dr.D]

Have you tried drawing a free-body diagram of this rigid body and taking moment balances.

I appreciate the intent here, but this approach is apt to raise more questions than it answers. The difficulty lies in the fact that with four support points, the system is statically indeterminant. This is true with the classic problem of the four legged table or chair. The problem goes away if you consider a three legged stool.

 

[Chestermiller]

He could still answer the question regarding the rear wheels vs the front wheels, and the right wheels vs the left wheels. That should be enough to address his issue.