# A level Physics Help

Guys,in Physics May 2021 Unit 1 AS Edexcel,in the MCQs,there was this question:
A van travels at a straight,horizontal road at a constant velocity.Which of the following statements is correct?

A.The frictional force of the road on the tyres can be ignored.
B.The frictional force of the road of the tyres is equal to the resultant force on the van.
C.The frictional force of the road on the tyres is in the direction of motion of the van
D.The frictional force of the road on the tyres is in opposite direction of the van.

The correct answer is C.But I think it should have been D.I didn't understand why the answer should be option C.
Can anyone please explain it to me?
Well to answer the question as put and explain. When talking about tyres and vehicles your mind zooms into standard mechanics that friction always opposes motion. It does here but not in the way the typical candidate imagined. I do not think the answers from the first response are desperately helpful.

When considering forces that tend to slow the wheel, the questioner implies that road tyre friction is responsible. They are wrong. The thing slowing the wheel down is hysteresis from contracting and expanding the rubber.

IF it was friction the the friction would need to slow the tyre down. You need to apply a force opposing the rotation so at the top of the wheel that would be backwards but towards the bottom it would be forward and partly down or up depending on which side of the wheel we are talking about.

Whether the road can do this is a moot point but see my next post for that, meanwhile…

To focus on C. The van is going at constant velocity. If you Imagine running beside the van looking at the tyres from the kerb. The wheel is rotating anti clockwise and you want to change the velocity by applying a force to the bit of the tyre in contact with the road. This is static friction for the driven wheels but no friction for a non driven wheel. Not what is discussed in any detail in the official Edexcel physics text that I have and there is nothing in the question to say we are dealing with a driven wheel.

If you push backwards, this will tend to increase the anti-clockwise speed and hence speed up the van. If you want to slow down the van, you need to push forward and this will tend to oppose the anti-clockwise motion. Hence C.

This works regardless of any driving force, that could be a rocket or a linear induction motor for all we know. Hence the reason D is wrong is not what @o1618 tells us. The reason D is wrong is in my explanation above.

A is wrong because we can’t ignore friction. Nothing to do with what is driving the van.
B is wrong because constant velocity tells us that the resultant force is zero.

The driving force is not relevant to the question because it is not mentioned in the question. A real van will be slowed down by hysteresis and wind resistance. A driving force has to oppose this. If the driving force is an engine and transmission then the force is applied via static friction. If it were a linear induction motor or a rocket or something like that then the friction is zero.

This is a really tricky question IMHO I have further thoughts which I will add
(edited 7 months ago)
Ok, this question was asked in international Gce unit1 p66393a which I guess is a sample unit but I may be wrong.

In my view the question presents a misunderstanding of what rolling resistance is. The point of contact between the road and the tyre is stationary relative to the road and there is little or no friction. Tyres would not last up to 50000 miles otherwise. Friction occurs when one body is in sliding contact with another. I will add more to this later

In the case of a real van, there has to be a force opposing the air resistance and this comes from static friction. The van stationary in a high wind from the front is in the same situation. This is why C is correct but the question is grossly under specified.

Later: well “micro-friction” according to some tyre manufacturers contributes less than 5% to rolling resistance. The big contribution is from hysteresis losses. This gives us A as the answer. We end up with the conclusion that friction between road and tyre is as nothing compared to other losses and from high speeds we may as well forget it. At a level we should be talking about ideal situations anyway which makes A the only possible answer but not for the reasons the question setter anticipated.

The reason I have a bit of an issue with this question and similar ones is that the best students I have ever taught would be confused by them. They would think of the boundary condition of a rolling wheel.

Too many questions test thing like alertness to SI units and the definition of a thing. A correct answer to this question is not within the a level physics specification and the text books limit discussion of friction (correctly) to sliding contact forces. The a level physics specification has had so much removed from it that examiners seem to be forced to resort to trick questions to remove marks or to pretend that a-level physics questions have to relate to “real” situations which end up with questions like this one.
(edited 7 months ago)
Original post by nerak99
Well to answer the question as put and explain. When talking about tyres and vehicles your mind zooms into standard mechanics that friction always opposes motion. It does here but not in the way the typical candidate imagined. I do not think the answers from the first response are desperately helpful.

When considering forces that tend to slow the wheel, the questioner implies that road tyre friction is responsible. They are wrong. The thing slowing the wheel down is hysteresis from contracting and expanding the rubber.

IF it was friction the the friction would need to slow the tyre down. You need to apply a force opposing the rotation so at the top of the wheel that would be backwards but towards the bottom it would be forward and partly down or up depending on which side of the wheel we are talking about.

Whether the road can do this is a moot point but see my next post for that, meanwhile…

To focus on C. The van is going at constant velocity. If you Imagine running beside the van looking at the tyres from the kerb. The wheel is rotating anti clockwise and you want to change the velocity by applying a force to the bit of the tyre in contact with the road. This is static friction and not what is discussed in any detail in the official Edexcel physics text that I have.

If you push backwards, this will tend to increase the anti-clockwise speed and hence speed up the van. If you want to slow down the van, you need to push forward and this will tend to oppose the anti-clockwise motion. Hence C.

This works regardless of any driving force, that could be a rocket or a linear induction motor for all we know. Hence the reason D is wrong is not what @o1618 tells us. The reason D is wrong is in my explanation above.

A is wrong because we can’t ignore friction. Nothing to do with what is driving the van.
B is wrong because constant velocity tells us that the resultant force is zero.

The driving force is not relevant to the question because it is not mentioned in the question. A real van will be slowed down by hysteresis and wind resistance. A driving force has to oppose this. If the driving force is an engine and transmission then the force is applied via static friction. If it were a linear induction motor or a rocket or something like that then the friction is zero.

This is a really tricky question IMHO I have further thoughts which I will add

How do we know that the wheel is moving in anti -clockwise?
Well that’s because I said from the kerb. From the other side you just swap things round
Original post by nerak99
Well that’s because I said from the kerb. From the other side you just swap things round

Oh ok.Thank you!