Physics question

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Hi there,

Just wondering could anyone help me out with this physics MCQ. The question is; A car and a truck are both being driven at 60km/h on a uniform level surface. Each driver applies a maximum breaking force, and both car and truck stop in the same distance. Why is this the case?

A. The frictional force between tyres and road is proportional to the weight of the vehicle, so the heavier truck has a greater stopping force.

Could anyone explain the reason for this answer as simply as possible? I chose "The bigger tyres on the truck provide a greater friction force than the narrow tires on the car" as my answer, which was incorrect.

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@lad123,

lad123 wrote:

Hi there,

Just wondering could anyone help me out with this physics MCQ. The question is; A car and a truck are both being driven at 60km/h on a uniform level surface. Each driver applies a maximum breaking force, and both car and truck stop in the same distance. Why is this the case?

A. The frictional force between tyres and road is proportional to the weight of the vehicle, so the heavier truck has a greater stopping force.

Could anyone explain the reason for this answer as simply as possible? I chose "The bigger tyres on the truck provide a greater friction force than the narrow tires on the car" as my answer, which was incorrect.

I'm no physicist, but the answer given seems to be self-explanatory. Because the truck is much heavier, its tires "push down" on the road with more force. In essence it seems like the equivalence of inertial and gravitational forces. Heavier objects do not "fall slower" because "gravity" has to exert more force to move them. All objects fall at the same rate.

Apparently all vehicles stop at the same rate, too, since the greater or lesser weight of any vehicle is offset by the greater or lesser pressure the tires exert on the road.

Apparently the "level surface" is important here. I don't think a heavy truck can stop nearly as fast as car if both are going down a steep incline. In that case the downhill momentum of the truck is much greater.

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@lad123,

As a former Physics teacher, I must say the question is bogus. There are several other possible answers to this question, and without setting the other parameters (tire size, air resistance etc,) the question can't be answered definitively. If I could, I would say this to your teacher ... Wink

.

That being said, let's answer the question.

1) The assumption is that if the stopping distance is the same, than the acceleration is the same. (This, of course, assumes constant acceleration which is another problem with this bogus question).

2) Newton's second law says F = ma. This can be written as A = F/m (the acceleration is equal to the Force (in this case friction) divided by the mass of the vehicle.

3) The Frictional force (Ff) is equal to the coefficient of friction times the Weight of the Vehicle. (The weight is the mass * gravity...)

So this stupid question wants you to assume that the coefficient of friction is the same in both cases.

In this case when you increase the Mass of the vehicle you also increase the Weight of the vehicle.

So for the car, The mass is lower, the frictional force is lower.
So for the truck. The mass is highter, the fictional force is Higher.

But in both cases the acceleration is the same sinc with A = F/m. You increase both F and m, and the cancel each other out making the Acceleration the same in both cases.

This is still a very stupid Physics question.

2 Replies

@lad123,

lad123 wrote:

Hi there,

Just wondering could anyone help me out with this physics MCQ. The question is; A car and a truck are both being driven at 60km/h on a uniform level surface. Each driver applies a maximum breaking force, and both car and truck stop in the same distance. Why is this the case?

A. The frictional force between tyres and road is proportional to the weight of the vehicle, so the heavier truck has a greater stopping force.

Could anyone explain the reason for this answer as simply as possible? I chose "The bigger tyres on the truck provide a greater friction force than the narrow tires on the car" as my answer, which was incorrect.

It has to do with inertia.

The greater the mass the greater the inertia. A heavier object has more inertia however the heavier object exhurts more friction on the surface.

In this case their velocity is equal but their mass and inertia are not. However; the truck exhurts more friction than the car. Therefore they are equal in stopping force. So they would decelerate within an equal distance.

Its inertia that slightly overcomes the friction otherwise the truck would actually stop in a shorter distance than the car. But with greater inertia than the car the friction cancels out the increased inertia so they stop within the same distance.

0 Replies

@maxdancona,

maxdancona wrote:

But in both cases the acceleration is the same sinc with A = F/m. You increase both F and m, and the cancel each other out making the Acceleration the same in both cases.

Ya kinda lost me there, Max. Who said either vehicle was accelerating at all? What's wrong with the answer I gave?

Edit: I guess you're talking about the "acceleration" of slowing down, eh? But your answer is not the "simple" explanation that the OP asked for.

1 Reply

@layman,

An acceleration means a change in velocity. If a vehicle is slowing down (i.e. coming to a stop) there is an acceleration. I don't know if there is anything wrong with your answer layman.... I was just trying to answer in a way that discussed Physics.

I thought of a simpler way to explain this mathmatically

(1) Ff = mA (Newton's second law)
(1a) A = Ff/m (solve for A)

(2) Ff = mu * Fn (the friction equation)

(3) Fn = m * g (Normal force is mass times the acceleration of gravity)

(4) Ff = mu * g * m (plug (3) in to (2))

(5) A = [mu * g * m] / m (plug (4) into (2) )
(5) A = mu * g

So we get the result that the acceleration that the vehicle experiences is unrelated to the mass of the vehicle. The reason is seen in step (5). The mass cancels out of the equation.

This is what your Physics teacher wants you to see. I still think the question is bogus.

1 Reply

@maxdancona,

maxdancona wrote:

An acceleration means a change in velocity. If a vehicle is slowing down (i.e. coming to a stop) there is an acceleration.

Yeah, I edited my response to acknowledge that before seeing your answer.

This is an area where you and I have disagreed before. Math is not an "answer," in the sense that it explains concepts. To me the concept is important here, and it is better "explained" in non-mathematical terms.

1 Reply

I'm with maxancona. The question is bogus.

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@layman,

Quote:

Math is not an "answer," in the sense that it explains concepts.

I am a Physics teacher who has a Physics degree. This is a thread about Physics from a Physics student trying to understand a Physics problem.

In that context, math is not only "an answer", it is the answer. I don't believe that I have ever said that Physics is the only way to have an understanding of the world, or that math is the only way to understand the world.

But Physics is about understanding the world through math. If you take a physics class, you will be doing math.

0 Replies

@lad123,

lad123 wrote:

Could anyone explain the reason for this answer as simply as possible? I chose "The bigger tyres on the truck provide a greater friction force than the narrow tires on the car" as my answer, which was incorrect.

If I recall my physics correctly, the surface area does not matter. The equation for friction force references the perpendicular (normal) force from the object, and does not reference surface area at all.

Why doesn't friction depend on surface area?

Quote:

Question
Why doesn't friction depend on surface area?

Answer
Although a larger area of contact between two surfaces would create a larger source of frictional forces, it also reduces the pressure between the two surfaces for a given force holding them together. Since pressure equals force divided by the area of contact, it works out that the increase in friction generating area is exactly offset by the reduction in pressure; the resulting frictional forces, then, are dependent only on the frictional coefficient of the materials and the FORCE holding them together.

lad123 wrote:

A. The frictional force between tyres and road is proportional to the weight of the vehicle, so the heavier truck has a greater stopping force.

Force (friction) = Coefficient of friction * Normal force

For a level plane, Normal force = mass * gravity

So friction force = Coefficient of friction * mass * gravity

Also, Force = mass * acceleration

acceleration = Force/mass

acceleration = (coefficient of friction * mass * gravity) / mass

acceleration = (coefficient of friction * gravity)

So if the coefficient of friction is the same for both vehicles, the acceleration will be the same. If they have the same starting speed, they will decelerate equally and stop at the same time.

3 Replies

@DrewDad,

Yes, DrewDad is correct. I had forgotten about surface area (although this is what the math says in the last post). Maybe the question isn't as bogus as I first thought. Although rather than remembering results as "facts", I would rather a student learn the mathematical reasoning behind it.

So yes, tire size has nothing to do with it. I believe that air resistance does Wink

. I don't know what this question buys us.

1 Reply

@DrewDad,

DrewDad wrote:

Quote:
Question
Why doesn't friction depend on surface area?

Answer
Although a larger area of contact between two surfaces would create a larger source of frictional forces, it also reduces the pressure between the two surfaces for a given force holding them together. Since pressure equals force divided by the area of contact, it works out that the increase in friction generating area is exactly offset by the reduction in pressure; the resulting frictional forces, then, are dependent only on the frictional coefficient of the materials and the FORCE holding them together.

See Max, this is what I'm talking about. This answer explains the situation conceptually, without a single digit of "math."

You say:

Quote:

I would rather a student learn the mathematical reasoning behind it.

I'm the opposite. I would rather that a student learn the physical (not mathemantical) reasoning behind the issue. That can always later be translated into a mathematical form, but understanding the "formula" does not, in itself, give any understanding of the physical principles in play.

1 Reply

@layman,

DrewDad wrote:

r. Since pressure equals force divided by the area of contact, it works out that the increase in friction generating area is exactly offset by the reduction in pressure; the resulting frictional forces, then, are dependent only on the frictional coefficient of the materials and the FORCE holding them together.

Max wrote:

A = mu * g

DrewDad and I said the same thing. (I also wrote the steps to derive this fact from basic principles, but DrewDad and I both end up in the same place).

I don't know why you think what DrewDad says is simpler than what I said. I do know that you can't understand or study Physics without a strong level of fluency in mathematics.

1 Reply

@maxdancona,

maxdancona wrote:

I don't know why you think what DrewDad says is simpler than what I said.

See what I added to my last post, eh?

1 Reply

@layman,

Most of us who study Physics see Mathematics as a language for expressing ideas. You need Mathematics because non-mathematical language is imprecise. Once you get to a certain level of Physics, it is simply impossible to discuss without mathematics.

We start students off with mathematics, because that is what Physicists do. What you are calling the "physical reasoning" is the exact same thing as the "mathematical reasoning". Non-mathematical language is simply not precise enough to explain Physics. It is important in Physics to understand the experiments and observations on which we base our understanding of physical phenomina. This is best (and perhaps only) communicated using the language of mathematics.

There are some cases where teachers of students in middle or high schools try to teach Physics without mathematics. In my opinion this is a big mistake. The students get the wrong impression because they are learning concepts in a non-mathematical languages about concepts that are implicitly mathematical. They always end up learning the wrong thing that must then be untaught (I have seen this happen again and again).

A perfect example of this is electron orbitals. It is simply impossible to understand electron orbitals without understanding PDEs. People are taught a bunch of rules and "concepts", but without the mathematics... they always end up with misconceptions.

It is a fact that if you undertake any serious study of Physics, you will be doing mathematics.

1 Reply

@maxdancona,

Thanks maxdancona!

0 Replies

@maxdancona,

maxdancona wrote:

There are some cases where teachers of students in middle or high schools try to teach Physics without mathematics. In my opinion this is a big mistake...

It is a fact that if you undertake any serious study of Physics, you will be doing mathematics.

You seem to be missing my point completely. To me this is not some false dichotomy which says you must either learn math OR concepts. The two go together, but it is much more important to understand the physical concepts (first) before diving into math, which is essentially meaningless if you don't understand the concepts the math is being applied to.

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@DrewDad,

Thanks! I was confused about the surface area, that clears it up, appreciate your response!

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@maxdancona,

maxdancona wrote:

So yes, tire size has nothing to do with it. I believe that air resistance does

https://imgs.xkcd.com/comics/experiment.png

0 Replies

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