I saw this puzzle in a local newspaper:

Consider a normal bicycle set to stand in its upright position, and its pedal is set to the position as shown in this figure. One man slightly hold the seat to keep the bicycle from falling. (Actually this is not important, there can be many ways to keep the bicycle from falling without affecting the experiment result). A non-elastic string is tied to the end of the pedal crank arm as in here. Another man then pulls the string backward.

The question is: In which direction will the bicycle move? Forward, backward, or standstill? Discuss possible cases.

enter image description here

Hint: The answer is non trivial. The readers are encouraged to try this experiment before coming to a hasty conclusion, since it's quite simple to carry out.

Despite the "Hint" part, and the "counter-intuitive" in the title, which is a quite clear suggestion that the bicycle will move in a counter-intuitive direction (backward instead of forward), many people still submitted answer such as "Forward, why not!!", or "Standstill" or "Forward, then as the crank move to 9 o'clock position, it will move backward". Of course they are not the correct answer.

Some people answered that it will move backward, but couldn't give a decent explanation.

My best guess is that this must have something to do with the size (radius) of the wheel, the radius of the crank arm, and the ratio of the crank gear and the wheel gear as well. If the bicycle moves forward, the crank bolt/shoulder will move as well, then the displacement of the crank does not simply equal to the displacement cause by the wheel rotating...

So what can be a simple, short explanation? And what can be a detail explanation with math and equations involved?

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    $\begingroup$ @AccidentalFourierTransform I think the gearing makes this question a little bit more complicated. $\endgroup$ – fibonatic Jan 17 '16 at 1:09
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    $\begingroup$ It depends on whether the man pulling the string retains a fixed position relative to the ground or relative to the bike. In the former case, the bike will move backward (and the pedals will rotate counterclockwise). In the latter case (similar to him being on the bike), the bike will move forward (and the pedals will rotate clockwise). NB: The former case assumes that the bike does not have a backpedal braking mechanism, and that gearing is 'normal'. It is the lack of appreciation of the distinction between being fixed to the bike or to the ground that may make things look cntrintuitive. $\endgroup$ – Keep these mind Jan 17 '16 at 6:37
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    $\begingroup$ I was planning on writing an answer. But since this question got closed as duplicate, I will just post the animation demonstrating both outcomes here (since my answer would really apply to the "duplicated" question, because of the lack of a gearing ratio). $\endgroup$ – fibonatic Jan 17 '16 at 20:06
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    $\begingroup$ @LarsH I used the free software Algodoo. The "pedal" on the left one is on the gear (red thing). $\endgroup$ – fibonatic Mar 8 '16 at 12:00
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    $\begingroup$ Meta discussion here. Please weight in! $\endgroup$ – AccidentalFourierTransform Sep 20 '18 at 20:05

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.


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

  • $\begingroup$ "I think that the answer is clearly that it all depends on what bicycle gear you are in." no, it does not: independently of the size of the gears, the bicycle will move backwards; otherwise, the exerted work $W=\boldsymbol F\cdot \boldsymbol r$ would be positive (and the first law sais this cannot be...) $\endgroup$ – AccidentalFourierTransform Jan 16 '16 at 21:30
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    $\begingroup$ @AccidentalFourierTransform - No, that's not correct. I think that the point you're trying to make is that the pedal at the point where it is attached to the rope cannot move forward. I believe that that's true. But it's not true that the bicycle as a whole cannot move forward. $\endgroup$ – Samuel Weir Jan 16 '16 at 21:37
  • $\begingroup$ the question is actually a duplicate (see my comment in the OP). In the other question, they explain why it moves backards. $\endgroup$ – AccidentalFourierTransform Jan 16 '16 at 21:57
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    $\begingroup$ I agree with Samuel Weir. What some people are missing is that the ground is exerting a forward force on the rear wheel of the bike when torque is applied to the pedal. So the force balance has to include both the backward pulling force by the string and the forward pushing force by the ground. Samuel's analysis is flawless. It is possible to analyze this in greater detail quantitatively, taking into account the torque on the pedal and the gear ratio, to determine which force wins out. $\endgroup$ – Chet Miller Jan 17 '16 at 2:12
  • $\begingroup$ @SamuelWeir You could put your solution in an answer! $\endgroup$ – Volker Siegel Jan 3 at 21:54

In the reference frame fixed to the bike, a backward movement of the pedal by $dx$ moves the ground backward by $\alpha dx$. So the pedal moved backward $(1-\alpha)dx$ with respect to the ground.

If $\alpha>1$ then this means it either

  • moved forward, which would deliver energy to the string puller and thus makes no sense

  • didn't move at all ($dx=0$)

If $\alpha<1$ then the pedal moves backwards so the string puller is doing work on it, which makes sense.

Therefore the bike either stays still or goes forward depending on $\alpha$, which is a function of pedal length $L$, gear ratio $g$ and wheel radius $R$: $\alpha=g\frac RL$. The condition on $\alpha$ is the same as looking at the trajectory of the pedal: if it ever has backwards ground speed when you're riding, then pulling the string moves the bike forward.


Here is another way to look at this problem: we should consider the total torque relative to the two points where wheels touch the ground. Clearly in your picture you rotate the whole bicycle backwards regarding the back as well as front wheel (points where wheels touch the ground). Imagine that the wheels are nailed. If you would pull the rope forward the bicycle would go forward.

Me and my brother went to our bicycle to test this hypothesis. We put the bottom pedal slightly towards the front wheel and pulled it a lot down and slightly back so that the total projection of the force is backwards. But in this case the torque on the back wheel is such that it tends to rotate forward. And this torque is larger than the torque on the front wheel. So overall, even though the force projection was backwards the bicycle went forward.

Also this video confirms the idea: if the rope touches the pedal under the ground level, the torque on both wheels will make the bicycle move forward.

This explanation is good because it does not make you think about gears and wheels. What you need to know is the shoulder and the force for both wheels.

  • $\begingroup$ It is an appealing explanation, but it not correct, because the correct answer does depend on the gears. (See the video in the first answer.) $\endgroup$ – knzhou Sep 20 '18 at 20:43

It will definitely move backwards. The angular velocity of the wheel, multiplied by the radius of the wheel is much bigger then the angular velocity of the pedals, multiplied by their length (the ratio of angular velocities is constrained) so the backward motion of the bicycle is much more significant then the rotation of the pedals. When the horizontal pedal position is exceeded (string above pedal axis), the bicycle will unambiguously move backward (effects add instead of subtracting).

I assume the bicycle does not have any ratchet mechanism and no extremely low gears.

[sarcasm]Also, Physics SE is for questions and answers, not discussion or puzzles. By posting this question you are doing evil to the community! This post must be deleted immediately and the servers must be incinerated to stop the contamination![\sarcasm]

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    $\begingroup$ Reading your sarcasm part gives me a shiver :) I don't know how to interpret that part correctly. I thought about posting this on puzzling beta on SE but seems that people there are more involved with board game, missing patterns, riddles,... The original puzzle here does not ask for mathematical explanation. I asked for my curiosity. Definitely worth a nice physics question. $\endgroup$ – Jim Raynor Jan 16 '16 at 23:14
  • $\begingroup$ Discussion here means giving a full answer that cover different cases, not discussion in the sense of throwing out a question that would gather opinion-based answers.This site and other SE sub sites are exactly on the opposite end of the spectrum compared with Q...ra (another Q&A site). On one hand I don't like the way questions are spammed at Q...ra, yet on the other hand this site has been so ridiculously inflexible and discouraging to post anything to in recent years. $\endgroup$ – Jim Raynor Jan 16 '16 at 23:23
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    $\begingroup$ As you can see here, some people are obsessive about content being "on topic" and treat "off topic" content like radioactive fallout. $\endgroup$ – k-l Jan 16 '16 at 23:24
  • $\begingroup$ Ha, after posting my comment, I wondered why you only have 6 reputation, must have gone through some unpleasant downvote - so I landed exactly at your question you link above via your profile. I think now I understand your sarcasm part better. Been there too :) $\endgroup$ – Jim Raynor Jan 16 '16 at 23:29
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    $\begingroup$ And now they go and down-vote my answer and "protect" the question (protect it from answers?) I actually expected that. $\endgroup$ – k-l Jan 17 '16 at 0:55

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