Bicycle Pedal
A wire is tied to a cycle pedal that is stationary at the bottom of its arc. If someone pulls this wire backwards (while another person lightly holds the seat to keep the bike balanced), will the cycle move forward, backward or not at all?
Give reasons !
Give reasons !
Labels: funphysics, thinktank
posted by Rajesh Lal at
9:22 AM
37 Comments:
The cycle would move forward..
I think it will go forward, as it will be continuing its cycle.
Looking from the side the cycle is a clockwise circle and pulling backwards at the bottom of the arc is just continuing that cycle, hence forwards.
I fell of my bike as a child and never got back on, so I'm not sure about this....
@Karl Sharman
Looking from one side the pedal would be moving clockwise and looking from the other it would be moving counter-clockwise, this clockwise and counter-clockwise motion of the pedals would create a torque moving the cycle forward.It doesn't matter which pedal you pull back from the bottom of the arc, the cycle would move forward.
Seems like an obvious FORWARD motion ... but, I'm not so sure. Physics is not my cup of tea ... but it may have something to do with whether the force moving the pedal is produced by someone sitting on the bike, or if the force is applied externally, such as by the wire. Maybe some of you physics folks can prove or disprove my theory. Any thougthts?
If it does move forward ... which I have my doubts ... it would do so only until the pedal has moved 90 degrees ... then the wire restricts forward motion and the bike then is pulled backwards.
things to consider: does the bike have wheels if so is there a chain connecting the wheel to the pedals
if so are both wheels touching the ground? again if so is there anything else to stop the cycle from moving...
also interms of pulling the wire you are applying a backward force on the bike presumably equal to the force created by the chains movement, so even though the bike would try to move forward the backward force caused by you would be equal to the forward push of the rear wheel therefore the bike would remain stationary.
also without enough pressure pushing the bike down for it to get a good grip of a survace the wheel make just even skid.
i know this is a bad explanation but i will try to write better when i get home...
Griever.
My last thought ... the I will watch responses for a bit. When sitting on the bike, the pedaling motion obviously moves the bike forward. But when the force is applied by the wire, doesn't the bike then free-wheel backwards?
It wont move at all if the pedal is @ the bottom of its motion.
It moves backwards - assuming that the pedal moves more slowly (tangential speed) than the speed of the tyre when the bike is operated normally. That is what I'd expect for a typical bicycle.
Let the speed ratio be R (R > 1). Let F be the force applied to the pedal, then the force applied by the tyre will be F/R. So the net force acting on the bike will be F-F/R and is pulling the bike backwards. I'm ignoring gear losses and assuming only a small motion.
The pedal will move forward (with respect to the bike - i.e. in the opposite direction to normal cycling use). As this happens the backwards speed will increase. Just after the pedal is horizontal, it will quickly move to the opposite rear position (via the freewhell mechanism), and the active rearward aceleration will become a maximum. So the bike will keep on going backwrs as long as the force is applied.
If R was < 1 (e.g for very steep uphill use) then the bike would go forwards. But this time then pedal will be moing towards the rear (i.e. in same direction as with normal cycling). Once the pedal has moved through an angle A from downwards, the force applied by the rear wheel will be reduced to (F/R)*cos(A). The bike will come to an neutral force position when cos(A) = R (crudely speaking). Beyond that position, the net force will be tending to make the bike go backwards. It will ultimately come to rest when cos(A) = 1/R.
It wont really move at all. Think about it mechanically, there is a reason you stand atop a pedal with one of the most powerful muscles in your body and push down-with gravity-to go. if you could pull push your way around on a bike, they would have been made that way.
Further to actually trying it, the bike will not move. The force exerted on the pedal moving the bike forward is cancelled by the force on the pedal pulling it backwards.
SBA - dont follow your thinking, the pedal on the other side would be at the top of its arc, and would be going in a clockwise cycle when viewed from the other side of the bike
Anonymous, that's a great example of how not to look at the problem!
The force on the pedal is pushing the bike backwards, the force on the wheel is pushing the bike forwards, but is smaller than the force on the pedal. So the bike goes backwards.
In my 2:27 pm post I made a typo. The last statement should have been "It will ultimately come to rest when cos(A) = R"
I really mucked up my 2:27 PM post. I had started writing it assuming infinitesimal motion, but hastily tacked on covering macroscopic motion. In consequence I didn't include the cos(A) in the first half of my response. So here's a corrected and neater version:
The force produced by the rear wheel is (F/R)*cos(A), where F is the force applied by the wire, R is the speed ratio and A is the angle of the pedal; A = 0 when the pedal points downwards. The net force acting on the bike is therefore, F - (F/R)*cos(A) = F(1 - cos(A)/R). If R > 1, then the force will be pushing the bike backwards for all values of A. If R < 1, then the force can change direction as A varies, then the force = 0 when cos(A) = R.
For the pedants, R is actually the mechanical advantage of the system, and will be a fraction (e.g. 0.9) of the speed ratio.
it may also depend on the bike. i have 2 bikes and one the bike goes nowhere if u pedal backwards its a bmx bike so theres no gears that propel it backwards only forwards and the other will engage the brakes. there are
The cycle cannot move forwards. To pull on the pedal the attached wire must be under tension; this tension will be pulling backwards on the pedal and therefore also pulling backwards on the cycle, preventing it from being able to move forwards. Under these circumstances forward motion could only be achieved if the wire could stretch, but if it could stretch as the cycle moved forward then it would also stretch as it was pulled taught. If the wire was stretchy to this extent then it would not exert enough force to move the pedal until it had reach full extent of its “stretchability” anyway at which point it cannot stretch further to allow forward motion of the cycle.
So the only options remaining are for the cycle to remain static, or move backwards. The cycle will remain static if the force pulling on the wire is less than the force of friction acting of the tyres preventing them from sliding backwards, as they cannot roll backwards due to the wire acting on the pedal prevent anti-clockwise rotation.
If the force acting on the pedal overcomes the force of friction acting on the tyres then the cycle will begin to move backwards. At this point, depending on the smoothness of the surface over which it is sliding, one of two things will happen.
If the cycle is being pulled over a smooth frozen lake, for example, then it is likely to slide very easily without losing contact with the surface it is on, in which case the cycle will slide backwards and the pedal will not move.
If he cycle is being pulled over an uneven, gravely surface for example, then as it “bounces” backwards the wheels will momentarily lose grip on the surface as it moves and the force of friction acting on the tyres will be removed. Each time this happens the pedal will be able to rotate backwards towards the 9 o’clock position, from which point it will rotate no further.
The wire is being pulled by "someone". It is obvious that the pull is meant to be a constant force (as far as that is practical). The wire introduces no constraints, it's just mind candy for there is an external force applied to the bike at the pedal and it is pointing backwards. The salient difference between the situation being described and the normal way a bike is used is that in this case the force is applied to the bike externally rather than internally (by a rider).
Introducing icy surfaces is crazy. Why not consider the whole problem in free fall, or after spraying glue everywhere, or at the heart of the sun? Or just for a hoot, why not try an ordinary bike on an typical flat high friction surface so that the tyre doesn't slip?
Chris
I don't think it's crazy to introduce the notion of icy surfaces. DualAspect is just showing what would happen in the two extreme cases i.e. a very low friction environment and a very high friction environment. Any 'normal' situation will be between those two. Since in both cases the bike moves backwards, in the 'normal' case the bike will also move backwards.
I prefer to visualise it. If you imagine the wire attached to lorry which is behind the bike and facing away from it, and that the wire is taught and parallel to the ground. The lorry then drives off away from the bike. I think it's absurd now to imagine that the bike goes forward. It would clearly be dragged backwards behind the lorry. There may initially be a slight turning moment which would cause the bike to bounce.
If the wire were parallel when the pedal had rotated to 9o'clock / 3o'clock the bike would simply be dragged along the ground.
It is obvious that the bike isn't being pulled with an irresistable force so that it is dragged along the ground. A lorry is far more powerful than the "someone" that is specified in the problem and will overcome any friction between the bike's driving tyre and the ground. Lorries, icy surfaces, stretching and non-stretching wires etc., are not about answering the posted problem - their introduction is inappropriate.
I have also given the condition under which the bike will move forward (although I don't expect that many bikes have such a low gear ratio for that to happen) - a lorry would drag the bike backwards regardless of that condition being met.
What totally stuns me is people's inability to recognise the most obvious and simple interpretation of the posted problem(s) especially if physics is involved.
People have different ways of looking at things. In fact the introduction of all those various different ideas you mention in your first paragraph are all ways of looking at the problem that arguably aim to get to the essence of the problem.
So saying "it is obvious" is simply incorrect. You should be saying "it is obvious to me" without presuming to speak for anyone else. What one person considers obvious another person may think completely idiotic. If that weren't the case the world would be a lot more homogenous than it in fact is.
I may be wrong but it would seem that you think your way of looking at things is the right one and other people's are wrong. Have you considered that maybe it's you that's missing the essence of the problem?
Chris,
Strangely enough, even though you have slated my post if you read what I have written it seems as if we actually agree with each other on most aspects. You have given your answer in the form of mathematical formulae, which is perfectly valid to do, and mine is written in the form of a real world visualisation, which is also a valid response.
We both agree that the cycle would move backwards
I would question one thing about your answer though: in the introduction to your 2:27 post you stated “It moves backwards - assuming that the pedal moves more slowly (tangential speed) than the speed of the tyre when the bike is operated normally.” I would agree with this entirely except for if the cycle was on a very slippery surface (the question doesn’t specify the type of surface), whereby the lack of friction against the tyres would not allow the work done on the pedal by the wire to overcome the friction inherent within the cycle’s transmission mechanism before overcoming the minimal friction working between the tyres and the surface resulting in the cycle sliding without the pedal moving at all. This is partly where I believe that my real world example becomes relevant, and perhaps a mathematical solution does not cover all variables.
I'll admit to being unusually scathing in my comments. I am very good at interpreting problems. I don't mind people exploring extensions to problems - in fact the extensions can be very interesting. However, that isn't the case here.
I agree that I should have said "it is obvious to me" rather than "it is obvious", but that's nit picking.
I am quite sure that my interpretation of the posted problem is the only appropriate one. I see it as a simple physics problem. I seem to be the only person that can actually properly deal with the problem. Almost everyone else is waffling.
Actually, thinking on it a little more, if the bike was geared to the point where the wheels turned more quickly than the pedal then the amount of force required to turn the pedal would be very large (imagine trying to start pedalling a bike from stationery when starting out in 15th gear, it's very difficult). In this case the surface would not have to be so slippery for the bike to start sliding backwards before sufficient force could be applied to turn the pedal.
I think the idea that the cycle could be geared to instigate forward motion before backward motion took over is very unlikely, because the amount of friction preventing the tyres from sliding would need to be unfeasibly massive.
So my responses which deal with kinetics, work, forces, friction, gearing mechanisms etc are not related to the "simple physics" involved????
I don't see that a discussion regarding surface friction is introducing a variable, it's an inherent part of the "simple physics" that you are such an expert in.
While we seem to agree on the result although disagree about the method it got me to thinking. I have a bike with shoes which clip onto the pedals to aid transmission of force. It would be possible to attempt to pedal in the same way as described in the original problem i.e. if the left pedal is at the bottom putting your force though you left leg, pulling backwards and upwards while not bothering to do anything with your right leg. I haven't tried this but surely in this case the bike should go forwards? Or at worst stay still? It certainly would not go backwards. That being the case, what is the relevant distinction between this instance and the one with the wire?
Also, another interesting aside. I have actually seen this done: it is possible to construct a bike with square wheels that will provide a perfectly smooth ride. What conditions are necessary for this to happen?
Mr F, in your scenario you are on the bike moving forward with it and not pulling against the bike, whereas the person pulling the wire is providing an external force. The 2 are not comparable.
Does the bike with square wheels work on water?
Suppose it might work on water but that's not what I was thinking of.
If the surface of your world - or your cycle path - is not flat but consists of what look like a series of connected semicircles (actually hyperbolic cosines with an arc length the same as that of the square tyre) then as long as you only go in that direction of these you'll travel perfectly smoothly. You can see a demonstration of this at the Exploratorium Museum in San Francisco (and elsewhere too I'm sure)
http://www.youtube.com/watch?v=LgbWu8zJubo
Hi DualAspect. I posted my last post before your one appeared. I think your and my posts have almost no convergence.
The question doesn't specify the nature of the surface because the author obviously (to me) was thinking of an ordinary frictional surface. If the surface is slippery, you may as well use a brick rather than a bike - hardly shedding light on the posted problem.
You claim to agree with my speed ratio argument, but demonstrate no understanding of it.
In your Dec 17, 4:16 pm post you first statement is wrong. The cycle can move forwards and I have given the condition (R < 1) when that is the case. You then say that the bike could only move forwards if the wire can stretch - that is wrong. The "someone" simply moves in the appropriate direction to keep a reasonable force applied to the pedal (and hence the bike). I have assumed that the wire does not stretch (for all practical purposes).
If the pedal was at the 9 o'clock (or 3 o'clock) position, the bike would move backwards - the force applied to the bike would be F, and the force applied by the wheel would be 0.
As a broad rule, all physics problems ultimately require the use of mathematics to adequately describe them. If I wanted, I could have explicitly included friction into my answers, but I would have done so using mathematics and without arm-waving and other vagueness. I hinted at that by noting that R is more accurately "mechanical advantage". If the tyre was slipping, then the mechanical advantage would become 0 and the bike would move backwards - using my equations with R = 0. Between my two main (earlier answers) I have covered everything, including the free wheel feature.
I'm being Mr. Intolerant today. Perhaps I'll be gentler tomorrow.
Hi DualAspect. I see you managed to get a few posts in while I was writing my last one - which was a response to your 7:20 pm post. In fact quite a few cross-posts have appeared, so it isn't clear what was a response to what.
I hadn't introduced friction as all that is required is that the wheels don't slip.
In view of your 8:18 pm post, I see no useful purpose in continuing the discussion.
Wow Chris?!
Bit stroppy. Is Aunt Flo visiting you this week?
I'm sorry everyone. I'm not usually such a grumpy old git.
I've got a dose of Merry Grimness.
First of all on any bike when you pull the pedal it locks it in place so the chain may not move in the opposite direction you are pedaling. 2nd The bike does not move at all.
You are pulling the pedal to move the bike toward you.. No matter how hard you pull back the bike will be trying to move forward at the same rate. Don't worry about gear sizes it doesn't matter. The force you apply will be the same force that is trying to move away. The only way to make the bike move forward is to start pulling the chain slightly upwards. The only way to move the bike backwards is to remove the chain that propels the bike.
Foregotten - none of what you say is correct.
... I'll elaborate. The only two ways of coupling the rear wheel that I've come across are fixed-wheel and free-wheel. WIth fixed wheel you can cycle backwards or brake using the pedals. With free-wheel, you can coast without the pedals turning - the free-wheel system presumably uses a ratchet to achieve that.
As to the second part - the gear ratio is crucially important. The most useful ratio to examine (for the posted problem) is the ratio of speeds of the pedals (with respect to the bike) and the speed of the tyre (ith respect to the bike). I have used the symbol "R" to denote this ratio. Whatever force F is applied to the pedal, a force (F/R)*cos(A) is applied by the rear wheel to the ground (and so by Newton's third law of motion), the ground pushes back with an equal and opposite force. The nett force acting on the bike (horizontally and backwards) is the sum of those two forces:
F-(F/R)*cos(A) = F(1-cos(A)/R).
As long as 1-cos(A)/R is positive, the nett force is rearwards. i.e if R > cos(A). This condition is met for all standard bikes. It is conceivable that a bike could have a very low gear ratio (for use on very steep hills), in which case the nett force on the bike will be forward as long as the pedal is pointing sufficiently downwards (so A is approx 0, i.e. near to 6 o'clock).
If the bike is the free-wheel type, then when the pedal is pointing in the top half, no force will be acting on the (whole) bike, but this will be for a short time as the pedal will quickly get to the 3 o'clock position (i.e. pointing backwards), then forces will be aplied to the (whole) bike again.
I'll also add that I've only been considering horizontal forces at all stages. I've also assumed that the bike is sufficiently heavy that the wheels stay in firm contact with the ground.
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