What is the direction of rolling friction?

Click For Summary
The discussion centers on the direction of rolling friction when a forward force is applied to a wheel on a horizontal surface. When the wheel rolls without sliding, the frictional force acts in the opposite direction to the wheel's motion, which is backward. If a torque is applied while the wheel is already spinning, friction can act forward to maintain motion. The original poster questions the answer key, which incorrectly states that the friction force is upward. Ultimately, the consensus is that the correct answer is C, indicating that friction acts in the backward direction when the wheel rolls without sliding.
Raymond Zheng
Messages
1
Reaction score
0

Homework Statement


A forward force on the axle accelerates a rolling wheel on a horizontal surface. If the wheel
does not slide the frictional force of the surface on the wheel is:
A. zero
B. in the forward direction
C. in the backward direction
D. in the upward direction
E. in the downward direction

Homework Equations


None.

The Attempt at a Solution


I've already seen another post on the exact same question, but I'm still unclear as to how the solution is obtained.

If a forward force (force to the right, which we take to be positive) is applied at the axle (simply pushing), then, normally the wheel would just move forward without spinning... but because of friction from the surface, the wheel spins in a clockwise direction, meaning that the direction of friction is backwards (to the left). Answer should be C.

If a torque is applied to the wheel, then the wheel is already spinning. Contact with the surface pushes the wheel in the forward direction. Therefore, in this case, friction is pushing the wheel in the forward direction. Answer should be B.

However, the answer key states that the answer is D...?

I'm wondering if the answer key, or any of my answers, are wrong.
Thanks!
 
Physics news on Phys.org
The force that acts upwards is the "normal force" which, in this case, exactly opposes the weight of the wheel.
We would not normally think of this is a friction force - so the answer D is incorrect.

The way to think about it is what would happen without friction - the wheel would slide, and the addition of friction would be to oppose the relative sliding motion at the point of contact. So whichever way the rim would normally slide, the friction is the opposite direction. This is the same rule whether, as in this case, the wheel is being pulled by the axle or in the case often discussed where there is a driving torque applied to the wheel.

The former would happen (say) when your car is being towed and the latter when you step on the gas.

You've tried to work through the above process, but you should try again more slowly.
 
  • Like
Likes berkeman
First of all your answer key is definitely wrong, the friction force cannot be upwards (unless you posted the question wrong and it actually asks for the normal force).

Seems to me the correct answer is C, and your justification is almost correct. Because (as it given by the problem statement) the wheel is rolling but does not slide, this essentially means that the velocity of the c.o.m of the wheel is equal to the rotational velocity of the point of contact, for any time t. But the force on the axle (which as given by the statement, is accelerating the wheel, which essentially means it increases the velocity of the c.o.m) increases the velocity of the c.o.m, so something must increase the rotational velocity as well in order to remain equal. That increase in the rotational velocity can only come from the torque of friction (the torque of the force in the axle is zero).
 
Last edited:
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 30 ·
2
Replies
30
Views
1K
  • · Replies 30 ·
2
Replies
30
Views
4K
Replies
24
Views
3K
Replies
4
Views
3K
Replies
7
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
16K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K