Physics question...

[jh_on_the_cape]

I am sitting here grading online physics homework and here is the question and thought I would post:

"Why can you exert greater force on the pedals of a bicycle if you pull up on the handlebars?"

This is part of Newton's laws of motion in a conceptual physics class, aka Physics for Poets.

How would you answer?

[BumbleBeeDave]

"For every action there is an equal and opposite reaction."

Pulling up on the bars exerts an equal and opposite reaction downward, which is added to whatever force your legs are already exerting downward.

Do I get a gold star? . . .

BBD

[54ny77]

Because theyre carbon bars of course!

[rwsaunders]

Deep drop, ergo or moustache?

[Ken Robb]

depending on how much force you can exert with your arms you can negate the equal/opposite force generated by pressure on the pedals or add to that because the bars will have little to no flex so all pulling will be applied as downforce on the pedals.

OTOH I was a history major in The School of Business (now Kellogg) and I never took Physics so I'm only guessing. :beer:

[Louis]

Because if you don't pull down on the bars all you got pushing you down onto the pedal is the force due to gravitational acceleration, which is limited by your mass. To put it another way, F = m x g.

To boost that you can add to F by pulling up on the bars, which increases the force pushing down in the pedals.

Happy riding! :bike:

Edit: recumbent riders can do this more easily by pushing back against the seat.

[fogrider]

Because if you don't pull down on the bars all you got pushing you down onto the pedal is the force due to gravitational acceleration, which is limited by your mass. To put it another way, F = m x g.

To boost that you can add to F by pulling up on the bars, which increases the force pushing down in the pedals.

Happy riding! :bike:

Edit: recumbent riders can do this more easily by pushing back against the seat.
agreed, kind of like a spring. without something to press against, the spring can only exert the force equal to the weight of the spring. with a handlebar to "push" against, the spring can exert more force than its weight.

[Don49]

"For every action there is an equal and opposite reaction."

Pulling up on the bars exerts an equal and opposite reaction downward, which is added to whatever force your legs are already exerting downward.

+1

Newton's Third Law of Motion.

[Louis]

The ISS has at least one stationary bike (see pic below, sorry it's so big) but if you look carefully, it seems to me that there are hand-holds behind him, to help him push "down."

Also, even without any gravity, you can briefly accelerate your body upward, which allows you to push down on the pedal, even if you are in zero-G. Of course, eventually you have to stop accelerating, otherwise you would rip your leg or foot off, so at that point only gravity can help push down on the pedal.

http://www.nasa.gov/images/content/114309main_iss011e05137_hires.jpg

[prometheus20]

Well I know that Newton's second law is that force is the product of mass X acceleration, or F=ma. So the force of pulling on the handlebars is added to the force of pushing on the pedals, thereby increasing the total net force exerted on the pedals, making the chain ring turn faster.

[fourflys]

expected answer...

"who rides a bike, I have a car?"

[Ti Designs]

I could answer that a number of ways:

1) Simple machines - it's a lever. Your fulcrum is the saddle, your input force are the handlebars, the pedals are somewhere between those two points on a horizontal model.

2) Vector math. If the pedal in question is +/- 90 degrees from the normalized force, it adds to the force (a diagram would help a lot here)

3) Learn how to ride.

[William]

...now it's more "Epic" :p


http://i1108.photobucket.com/albums/h420/SodaFuel/Fixie%20wheels/IMG_0001.jpg






William ;) :D

[d_man16]

Torque,

You are able to use the handle bar and frame as a lever arm. Think of a lever from your hands to the tire's contact point.

This will "increase" the amount of force applied to the pedals.

~D

[Chance]

I am sitting here grading online physics homework and here is the question and thought I would post:

"Why can you exert greater force on the pedals of a bicycle if you pull up on the handlebars?"

This is part of Newton's laws of motion in a conceptual physics class, aka Physics for Poets.

How would you answer?
You can’t. It’s a trick question.

In theory it’s not necessary to pull up on the bars at all for a rider to exert all the force his legs can muster.
When pulling up on one pedal while pushing down on the other the only limitation is rider strength.
In theory he could develop near infinite forces without even touching the bars.

In practice that’s a different subject ….

[jh_on_the_cape]

It's Newton's third law, action-reaction pair.
the upward force the rider puts on the bar has an equal and opposite force on the pedals. this is added to downward force provided by the legs, which has its equal and opposite force provided by gravity (wieght of the rider).

and if you are good, it is balanced by the upward pull on the other foot, if you are clipped in.

same reason you pull down on a rope to climb up it.

nothing to do with torque. only works with OS clamp carbon bars!

the space station pic is great. if he has a PERFECT spin, he does not have to hold on. you could work on only the pull up or pull down part of a spin by strapping up or down. neato.

[MadRocketSci]

Newton's third law will only tell you that if you pull up on the handlebars, the handlebars will pull down on your hand. All this other stuff about what happens at your feet or other parts are figured out using force/torque balances.

edit: nevermind, the question says "can," not "will"....carry on....

[jh_on_the_cape]

Newton's third law will only tell you that if you pull up on the handlebars, the handlebars will pull down on your hand. All this other stuff about what happens at your feet or other parts are figured out using force/torque balances.

edit: nevermind, the question says "can," not "will"....carry on....

This question was in the first week or two of class.

We do torques later. I actually include a lot of cycling in my class. Unfortunately that does not make it click, because most kids are not familiar with bikes.

Next semester I am going to measure the moment of inertia of a few bike wheels with a little gadget I made. Moment of inertia is the important measure of a bike wheel, not mass. That's tough to explain, though.

[Chance]

Newton's third law will only tell you that if you pull up on the handlebars, the handlebars will pull down on your hand. All this other stuff about what happens at your feet or other parts are figured out using force/torque balances.

edit: nevermind, the question says "can," not "will"....carry on....
You should have stuck to your previous point.

The question is framed based on the flawed assumption that pulling up on the bars is necessary to continue adding more force to pedals. And clearly that’s not the case. The question is therefore ambiguous in its nature and can have no one correct answer.

[Karin Kirk]

Good stuff JH!

My brother is a high school physics teacher and cyclist and he also uses a lot of cycling demos and analogies. My favorite is him sprinting a bike down the hallway and calculating how many horsepower he could make. Kids loved watching that and of course they all wanted to try it.

Here are some video clips that he uses to teach with. The bottom two involve finding the rotational inertia of bicycle wheels
http://serc.carleton.edu/sp/library/teachingwdata/video.html

He has oodles of bike-related ideas he's highly enthusiastic about innovative teaching. PM me if you want more info - sounds like you and he are on the same page.

[jh_on_the_cape]

You should have stuck to your previous point.

The question is framed based on the flawed assumption that pulling up on the bars is necessary to continue adding more force to pedals. And clearly that’s not the case. The question is therefore ambiguous in its nature and can have no one correct answer.

It's an essay question. I get some thoughtful answers, like how pulling up on the bars can make your a$$ hurt. which is true, I guess. If you did it sitting down with your feet off the pedals. So I guess the question is somewhat open.

[jh_on_the_cape]

Good stuff JH!

My brother is a high school physics teacher and cyclist and he also uses a lot of cycling demos and analogies. My favorite is him sprinting a bike down the hallway and calculating how many horsepower he could make. Kids loved watching that and of course they all wanted to try it.

Here are some video clips that he uses to teach with. The bottom two involve finding the rotational inertia of bicycle wheels
http://serc.carleton.edu/sp/library/teachingwdata/video.html

He has oodles of bike-related ideas he's highly enthusiastic about innovative teaching. PM me if you want more info - sounds like you and he are on the same page.

I will check that out and PM you AFTER I grade the finals this semester!

[Karin Kirk]

I will check that out and PM you AFTER I grade the finals this semester!
I hear you - good luck with the grading crunch.

[Chance]

Next semester I am going to measure the moment of inertia of a few bike wheels with a little gadget I made. Moment of inertia is the important measure of a bike wheel, not mass. That's tough to explain, though.
:confused: You can't start this and not follow up. :)

Please don’t keep us in suspense. Care to share why it’s “the” important measure.

To save time you don’t need to explain the differences if you don’t want to. That part is clear. Just want to know why you feel moment of inertia is so much more important than mass.

[MadRocketSci]

You should have stuck to your previous point.

The question is framed based on the flawed assumption that pulling up on the bars is necessary to continue adding more force to pedals. And clearly that’s not the case. The question is therefore ambiguous in its nature and can have no one correct answer.

all you can say is that if you are pulling up on the handlebars, and your cg is not moving down, then there will be an additional force somewhere pushing up on you. Depending on how you do it it "CAN" be at your foot. It could conceivably be at the wrong foot and thus hinder your pedaling. If you are sitting it could/would be at your butt. Or it could be your other hand. but at the foot pedaling downward is possible, which is what the question asks.

[Louis]

Re: Mass and Inertia

They are both important, just at different times. Mass more so when climbing, and inertia more so when accelerating on level ground.

Overall, given that the radius of the wheel limits how far out you can really put the material, and thus limits your inertia for a given mass, I'd say mass is a bigger deal. But it's really two sides of the same coin.

[jh_on_the_cape]

Re: Mass and Inertia

They are both important, just at different times. Mass more so when climbing, and inertia more so when accelerating on level ground.

Overall, given that the radius of the wheel limits how far out you can really put the material, and thus limits your inertia for a given mass, I'd say mass is a bigger deal. But it's really two sides of the same coin.

Mass is a measure of inertia. Moment of Inertia is something else.

[jh_on_the_cape]

:confused: You can't start this and not follow up. :)

Please don’t keep us in suspense. Care to share why it’s “the” important measure.

To save time you don’t need to explain the differences if you don’t want to. That part is clear. Just want to know why you feel moment of inertia is so much more important than mass.

You can say a bike wheel is 1000 g or whatever. But the location of the mass is also important.

Moment of Inertia is the integral of mass times radius squared. So if both wheels have the same mass, but one has the mass concentrated at the hub it will be easier to spin up or spin down. To change its rate of rotation. That is radial acceleration. Sprinting or braking.

So where should you try to drop mass on a wheel... the tire or the hub? THE TIRE! You can really notice this on a mtb if you go from heavy DH tires to skinny light XC tires.

I take a superlight front wheel and put on a heavy tire from a cheap hybrid bike. Then I take a bit heavier wheel and install a light skinny road tire with no tube. I get the masses almost the same. The first one is harder to spin up. I apply the same torque by letting a hanging mass unspool from the wheel, same for both.

Sorry, hard to explain right now. proctoring the final!

The total mass is important for getting the whole bike and rider up to speed or braking. And as we all know, climbing.

There are many other factors, of course... rolling resistance. RIDER STRENGTH!!!

[Chance]

You can say a bike wheel is 1000 g or whatever. But the location of the mass is also important.

Moment of Inertia is the integral of mass times radius squared. So if both wheels have the same mass, but one has the mass concentrated at the hub it will be easier to spin up or spin down. To change its rate of rotation. That is radial acceleration. Sprinting or braking.

So where should you try to drop mass on a wheel... the tire or the hub? THE TIRE! You can really notice this on a mtb if you go from heavy DH tires to skinny light XC tires.

I take a superlight front wheel and put on a heavy tire from a cheap hybrid bike. Then I take a bit heavier wheel and install a light skinny road tire with no tube. I get the masses almost the same. The first one is harder to spin up. I apply the same torque by letting a hanging mass unspool from the wheel, same for both.

Sorry, hard to explain right now. proctoring the final!

The total mass is important for getting the whole bike and rider up to speed or braking. And as we all know, climbing.

There are many other factors, of course... rolling resistance. RIDER STRENGTH!!!
As mentioned previously, you don’t have to explain what they are. And for what it's worth, the questions are not about being negative in any way, it’s just that it's inevitable to question general assumptions and or statements of fact to see if there are exceptions and therefore test for acceptable accuracy.

Louis’ point about mass versus (moment of) inertia is valid other than the fact that he left off “MOMENT OF”. If a rider is more concerned with climbing large mountain passes the wheels’ mass should be more important to him or her than what the moment of inertia is relative to other wheelsets. To suggest that one measure is better than another because it applies to acceleration more so than the other seems biased. In this case it comes across as the teacher valuing acceleration more than climbing. Do the students feel the same way or have similar needs?

Of much greater importance, how do you explain in the context of your general statement that moment of inertia is more important than mass when we can have two bikes of equal mass, with wheels of equal mass, and the one bike (of the two) with lower wheel moment of inertia actually accelerates slower?

[TimD]

TiDesigns and MadRocketSci have it right, you need to sum the vector forces to understand this.

Imagine standing up on one pedal. Initially the pedal and crank arm are not moving and are at the bottom of the stroke, e.g., the "6 o'clock" position. Your one leg is resting on the pedal and holding your body up against the force of gravity:

/\
|
| F(n) - Normal force of pedal pushing up on foot holding you up.
|
O==
|
| F(g) - Force of gravity pulling you down.
|
\/

F(n) = F(g) = W, or your weight.

Now imagine the crank at the "3 o'clock" position and rotating. You are riding no hands and holding your leg rigid, meaning the only force acting on the pedal is your weight (no muscle extension applying force).

The pedal is pushing upwards on your one foot, but with less force than that required to hold up your body, because the crank is free to rotate.

Here's the diagram. F(g) is still pulling down, but the normal force is reduced (and your body is accelerating downward as a result).

/\
| k*F(n) - Reduced normal force of pedal pushing up on foot.
O==
|
| F(g) - Force of gravity pulling you down.
|
\/

k*F(n) < F(g). The difference in these forces is transmitted into the crank arm, applying a torque about the BB axle, rotating the chainring, and applying a pulling force on the chain. That, in turn, causes a torque on the cassette, which translates into a horizontal force applied to the road by the tire. F = m*a, so you and the bike will accelerate forward, assuming the horizontal force is large enough to overcome all drag, all losses, and any other horizontal forces acting on the bike (e.g., you are trying to ride uphill).

Notice that in this diagram the maximum downward force can only be F(g) - there's no other force acting on your body, nor on the pedal.

If you extend your leg using your quad then an additional downward force would be applied to the pedal, and you and the bike will accelerate more rapidly. However, there's a limit to how much force you can apply. Again, imagine the crank at the "6 o'clock" position, where it cannot move vertically. If you apply muscle force and cause F(N) to exceed F(g), your body will accelerate upwards (because the only thing holding it down is F(g)).

So, if we want to add force to the pedal and not fly off the bike, we need a source of opposing force.

Now add a diagram for the bars:


/\
|
| F(A) - Force of arms applied to bars
|
===\
----/
|
| F(B) - Force of bars applied to arms
|
\/

F(A) = F(B), approximately. "Approximately" in that there's some bending, normally your torso moves in response to the force applied to the bars (in the downward direction), and the bars carry a component of your weight (F(n)) .

But assume your torso is rigid and doesn't move and you are not using your arms to oppose F(n). In this case some other force has to balance F(B) or you are going to move, as there is no normal force pushing your arms back up in this diagram.

Since your arms are rigidly (sorta) linked to your torso, that upward force has to come from the pedal. If we assume the pedal is in the "3 o'clock" position, or leg / crank / pedal vector force diagram now looks like this:

/\
|
| k*F(n) + F(B)
|
|
o==
|
|
| F(g) + F(B)
|
\/

Woila, we have now increased to force on the pedal, the pulling force on the chain, the torque about the rear axle, and the horizontal force applied by the tire to the road. We accelerate and drop the peloton. Notice too that F=ma means less m means more a for a given F. :)

Hope that helps.

Tim

PS In my hypothetical country everyone is required to take introductory physics and introductory statics & dynamics.

[Mark McM]

"Why can you exert greater force on the pedals of a bicycle if you pull up on the handlebars?"

I think that a lot of the discussion and cross-discussion in this thread is because the wording of the question is a bit open. How is the rider positioned on the bike? Where are the handlebars with respect to the rider (and the pedals)?

For example, on this bike, pulling up on the bars has virtually no affect on how much force you can exert on the pedals:

http://www.recumbentblog.com/wp-content/uploads/2009/09/MariaOnSilvio.jpg

Maybe the question could be worded a little more specifically, like:

"Why can you exert greater downward force on the pedals of a bicycle if you pull up on the handlebars?"

[Louis]

Mass is a measure of inertia. Moment of Inertia is something else.

:confused: :confused: :confused: :confused: :confused:

Mass and inertia don't even have the same units.

Mass and the location of that mass relative to a reference point define mass moment of inertia.

For purposes of this conversation and wheels, "mass moment of inertia" is the same thing as "inertia."