When a bicycle wheel is in motion, many forces are acting on it. The sum of the forces results in forward
motion. In which direction does friction from the ground act on the wheel?
[Physics] In which direction does friction from the ground act on the wheel
frictionnewtonian-mechanicsrotationrotational-dynamicsrotational-kinematics
Related Solutions
I think the way to look at the problem is this: The tricky thing about this problem is that the force $F$ is doing two things. First, it is pulling the entire bicycle backwards and, secondly, it is exerting a torque on the pedals. So let's try to separate those two effects. Consider, then, a person just sitting normally on the bicycle with his feet on the pedals and exerting the same force $F$ on the pedals (or, equivalently, a torque $rF$ on the pedal gear where r is the effective radius at which the pedals are located with respect to the center of the pedal gear). Also, assume that there is another man pulling backwards on the bicycle with a rope attached not to a pedal but to the frame of the bicycle. Now we have a situation which is equivalent to the problem originally posed. We have (1) the same torque as originally stated acting on the pedal gear and (2) we have the same force as originally stated acting to pull the bicycle backwards. Any problems with this picture thus far?
OK, so with this new but equivalent situation, what is the answer? Imagine that you are on the bicycling trying to pedal forward while exerting the torque stated above on the pedals while at the same time someone is trying to pull you backwards by pulling with a force $F$ on a rope attached to the frame of the bicycle. Remember that he is required to pull with exactly the same force $F$ that you are exerting on the pedals. No more and no less. Do you go forward or does he pull you backwards?
I think that the answer is clearly that it all depends on what bicycle gear you are in. If you are in low gear, you will be able to move forward. If you are in high gear, he will pull you backwards. In physics terms, it depends on whether the (clockwise) torque that you are exerting on the rear drive wheel by means of your feet on the pedals is greater than or less than the (counter-clockwise) torque that the other man is able to effectively exert on the rear drive wheel by pulling the bicycle in the backwards direction with a rope.
P.S.: With the single-speed bicycle shown in the picture, the gearing is such that the bicycle would most likely move backwards.
EDIT - SOLUTION FOUND
Found a discussion and solution for this puzzle on the Scientific American web site. It turns out that the bicycle can go either backwards or forwards, depending on the gearing ratio (which is the same result that I arrived at above). For normal gearing the bicycle will most probably go backwards, but for very low gearing the bicycle can also go forward. The argument they use is based on noting whether the point where the rope is attached to the pedal goes forward or backwards with respect to the ground. Here's the web link with an explanation video: Scientific American: Bicycle Puzzle
From definition of Friction:
Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.
Now consider your problem. Let the force be $F$, mass of spool be $m$
If Friction is absent
If Friction were absent, then due to torque of $F$, the spool will rotate clockwise. Now, the bottom edge of spool may have a velocity in forward direction, or backward direction, given by: $$\begin{align}\mathbb{v} &= v_{trans} -v_{rot} \\ \\ &=\frac{Ft}{m} - r\alpha t \\ \\ &= \frac{Ft}{m} - r\left(\frac{r\times F}{I}\right)t \\ \\ &= \frac{Ft}{m} - r\left(\frac{r F}{I}\right)t \\ \\ \end{align}$$ But we know that $\max(I) = mR^2$, so $\mathbb{v}$ above reduces to zero. $$\begin{align}&= \frac{Ft}{m} - r\left(\frac{r F}{mR^2}\right)t \\ \\ &= 0 \end{align}$$
But we also do know that for most rigid bodies, $I < mR^2$.
From this, we have : $$v_{rot} > \frac{Ft}{m}$$
Therefore we can say in general, for rigid body, that: $$\boxed{v_{rot} > v_{trans}}$$ Due to absence of friction, $\mathbb{v} \neq 0$ and also that the bottom point of wheel will have a net velocity in backward direction, and spool will slip.
If Friction is present
If Friction is present, it would make $\mathbb{v} = 0$, since it tends to remove relative motion between the bottom of spool and ground.
Now, the direction of friction will be decided by the velocity which was greater in the earlier case.
As we saw earlier, $v_{rot} > v_{trans}$ for a rigid body, $\mathbb{v}$ came out to be negative, hence friction will try to make $\mathbb{v}$ zero, or will be exerted in positive $x$ direction on spool.
Friction always opposes relative motion. So on ground, friction will act in -x direction (by third law of Newton).
Best Answer
Depends on the motion of the wheel, the direction of the friction force could be different. Remember the friction force from the ground is the only external force acted on the wheel (Ignore air resistance), 1) If the wheel is accelerating forward, the friction force will be pointing forward. 2) If the wheel is moving forward with a constant velocity, the friction force is zero (if air resistance is included, the friction force will be pointing forward, with the magnitude exactly canceling the air resistance). 3) If the wheel is slowing down,the friction force will be pointing backward.
Also, the external gravity and normal force are irrelevant in this problem.