Newton's Laws of motion -- Bicyclist pedaling up a slope

[haruspex]

it has to be better than what I was thinking before.

No, it just adds irrelevant details. The question asks for one force; it can only mean net force. Breaking it into all the different forces / torques exerted on pedals, handlebars and saddle just confuses the picture.
Likewise, it is not that important whether the acceleration specified is instantaneous, constant or average; whichever adjective applies to that applies to the force.
As you originally wrote, net force on cyclist $=m\vec a$.
Force bicycle exerts on cyclist $= m(\vec a-\vec g)$.
Force cyclist exerts on bicycle $= m(\vec g-\vec a)$.
The flaw in the question is that to get the given answer it should ask for the component parallel to the plane.

 

[erobz]

No, it just adds irrelevant details. The question asks for one force; it can only mean net force. Breaking it into all the different forces / torques exerted on pedals, handlebars and saddle just confuses the picture.
Likewise, it is not that important whether the acceleration specified is instantaneous, constant or average; whichever adjective applies to that applies to the force.
As you originally wrote, net force on cyclist $=m\vec a$.
Force bicycle exerts on cyclist $= m(\vec a-\vec g)$.
Force cyclist exerts on bicycle $= m(\vec g-\vec a)$.
The flaw in the question is that to get the given answer it should ask for the component parallel to the plane.

I'm sorry for the confusion. Those were just general thoughts on the mechanics of pedaling a bike (exploration for the sake of amusement). They weren't supposed to be specifically applied to solve this problem.

 

[Orodruin]

Ok, I can see that as possible. Imagine a riding without hands in a sitting position and standing up during the downward pedal stoke. It's very difficult to do in practice...I've tried it.

I did not say it would be advisable. It was merely a theoretical statement regarding the possible size of the pedal force.

 

[Orodruin]

Absent toe clips or a firm grip on the handlebars, it is difficult to achieve an average downforce on the pedals in excess of the person's weight. Acceleration can help with peak force. Not so much with average.

Well, maybe if one is cycling in the globe of death.

Which is why I had carefully added the word ”instantaneous” …

Edit: When attending a conference in Beijing some 9 years ago or so I also visited an acrobat show. The globe of death act was great. 8 motorcycles in the globe at the same time.

 

[Lnewqban]

Force bicycle exerts on cyclist $= m(\vec a-\vec g)$.
Force cyclist exerts on bicycle $= m(\vec g-\vec a)$.
The flaw in the question is that to get the given answer it should ask for the component parallel to the plane.

But the problem in essence asks about the force on the rider — not rider+bike.

But the bicyclist can’t push the pedal downwards and the bike uphill without having the bike pushing his mass along, so he stays on the pedal during the slowdown process.

Want it or not, his energy or work is accelerating the mass of the bike and his own mass simultaneously, while gravity is doing the opposite work, ...and winning.

Where is our expert mountan biker @berkeman when we need him?

 

[PeroK]

No, it just adds irrelevant details. The question asks for one force; it can only mean net force. Breaking it into all the different forces / torques exerted on pedals, handlebars and saddle just confuses the picture.
Likewise, it is not that important whether the acceleration specified is instantaneous, constant or average; whichever adjective applies to that applies to the force.
As you originally wrote, net force on cyclist $=m\vec a$.
Force bicycle exerts on cyclist $= m(\vec a-\vec g)$.
Force cyclist exerts on bicycle $= m(\vec g-\vec a)$.
The flaw in the question is that to get the given answer it should ask for the component parallel to the plane.

In that case, it's not important that the cyclist is the one putting the energy into the system. The original question was:

Homework Statement:: A 65 kg cyclist on a 10 kg bicycle is moving uphill on a 9° slope. How much force does he provide
if the bicycle slows at a rate of 0.3 m/s2?

Just change bicycle/cyclist for car/driver:

A 65 kg driver in a 1000 kg car is moving uphill on a 9° slope. How much force does he provide if the car slows at a rate of 0.3 m/s2?

The answer is the same. But saying that the driver provides the force in this case is, I would say, simply wrong.

Or, a sailor in a boat being blown along by the wind.

 

[erobz]

Where is our expert mountan biker @berkeman when we need him?

Probably on his Mt Bike, right where I should be...having fun. Instead of tormenting myself!

 

[berkeman]

Where is our expert mountan biker @berkeman when we need him?

Out riding, where I belong.

 

[haruspex]

But saying that the driver provides the force

It is an awkward use of "provides", but I would not say it is wrong. "exerts" would be better.

 

[haruspex]

Want it or not, his energy or work is accelerating the mass of the bike and his own mass simultaneously

Don't confuse force with work. If I pull on both ends of a spring I do work without exerting a net force on it. That is very similar to what the cyclist is doing with the forces exerted on pedals etc.

 

[Orodruin]

In that case, it's not important that the cyclist is the one putting the energy into the system. The original question was:

Just change bicycle/cyclist for car/driver:

A 65 kg driver in a 1000 kg car is moving uphill on a 9° slope. How much force does he provide if the car slows at a rate of 0.3 m/s2?

The answer is the same. But saying that the driver provides the force in this case is, I would say, simply wrong.

Or, a sailor in a boat being blown along by the wind.

Force is not the same thing as energy.
A force is a transfer of momentum, not a transfer of energy. These need to be kept separated as they are separate concepts.

It has already been made clear early on in this thread that the question is badly posed. In particular the phrasing regarding ”providing” force.

 

[Orodruin]

But the bicyclist can’t push the pedal downwards and the bike uphill without having the bike pushing his mass along, so he stays on the pedal during the slowdown process.

Irrelevant to the resolution of the problem.

Want it or not, his energy or work is accelerating the mass of the bike and his own mass simultaneously, while gravity is doing the opposite work, ...and winning.

Irrelevant to the resolution of the problem. Work is not the same thing as force.

If you want to look at the flow of energy in the system this is certainly possible. It is however way more involved than the stated problem and therefore quite moot for the issue at hand.

 

[PeroK]

Force is not the same thing as energy.

I'm not saying it is.

 

[Orodruin]

I'm not saying it is.

Well, your argument was based on the energy flow so it certainly makes it seem so. The question regarding the net force is indeed not any different from the question with a car or a sailing boat. The difference appears in the study of internal forces and energy flow.

 

[Lnewqban]

Irrelevant to the resolution of the problem.

Not if the claimed solution of the problem excludes the mass of the cyclist.
Does it?

Irrelevant to the resolution of the problem. Work is not the same thing as force.

If you want to look at the flow of energy in the system this is certainly possible. It is however way more involved than the stated problem and therefore quite moot for the issue at hand.

As I see it, that instantaneous force is not able to keep the bicycle moving uphill.
Not considering the whole energy needed for that (or a constant input moment), seems to me not to have direct relation to the value of acceleration provided by the problem as a hint.

 

[haruspex]

that instantaneous force is not able to keep the bicycle moving uphill.

Remember we are discussing net force. As I showed in post #40, work can be done without any net force.

 

[Lnewqban]

Remember we are discussing net force. As I showed in post #40, work can be done without any net force.

Thank you much, @haruspex
I declare myself unable to understand your statements in both, this and post #10.
I am sure that you, as well as @Orodruin, are correct; but again, I simply can’t see the problem that way.
I will keep on following the discussion of this thread in the hope of finally understand.

 

[bob012345]

Remember we are discussing net force. As I showed in post #40, work can be done without any net force.

Like on a stationary bike.

 

[PeroK]

Well, your argument was based on the energy flow so it certainly makes it seem so.

And your argument is based on the idea that a cyclist propels the bike with his or her bottom!

 

[Orodruin]

And your argument is based on the idea that a cyclist propels the bike with his or her bottom!

No, it is not. You are completely missing the point. It has nothing to do with the propulsion mechanism. The question asks for the net force from the bike on the biker (in the travel direction). This has nothing to do with propulsion.

 

[PeroK]

The question asks for the net force from the bike on the biker (in the travel direction).

No it doesn't. That's not what the question asks at all.

That's the question that would yield the given answer.

 

[Orodruin]

No it doesn't. That's not what the question asks at all.

That's the question that would yield the given answer.

Well, as I pointed out in the beginning, the actual question quoted is badly worded and nonsense. If you are going to debate a nonsense question you will end up with nonsense and differing interpretation. I see no point in preferentially interpret ”providing force” as ”force used to propel the bike”. With that in mind, we have to attempt to interpret the question in the light of what the person posing the question wanted for an answer.

 

[bob012345]

Well, as I pointed out in the beginning, the actual question quoted is badly worded and nonsense. If you are going to debate a nonsense question you will end up with nonsense and differing interpretation. I see no point in preferentially interpret ”providing force” as ”force used to propel the bike”. With that in mind, we have to attempt to interpret the question in the light of what the person posing the question wanted for an answer.

The actual question made perfect sense to me. Just how would you have worded it to the student?

 

[Orodruin]

The actual question made perfect sense to me. Just how would you have worded it to the student?

”What is the component of the net force with which the biker acts on the bike in the direction of travel?”

As we have witnessed in this thread, the word ”provides” is not standard nomenclature and has caused a wide range of opiniated interpretations.

 

[jbriggs444]

The actual question made perfect sense to me. Just how would you have worded it to the student?

The question seemed badly strained to me.

As asked: "How much force does he provide"

As a native English speaker, the implication of "provide" is something the rider does beyond just sitting on the seat. As a reader, the obvious interpretation would then be that the rider "provides" motive power. He pedals. However that was not the writer's intent. [The astute reader realizes that the problem does not specify a gear ratio, and assumes that the writer intended the problem to be solvable. So the natural reading cannot be the intended one]

I would ask "What is the net force of rider on cycle in the direction of the upward slope?"

Or I might be trickier. Just "What is the net force of rider on cycle" and expect the correct answer to be in the form of a vector that might be expressed in a number of possible formats.

 

[Orodruin]

The answer is the same. But saying that the driver provides the force in this case is, I would say, simply wrong.

Disagree. Your beef is with the word ”provide” which does not have a standard meaning and even if you do interpret it as the force putting energy into the system it still makes the question unanswerable without further specification of gears etc. Making it pretty clear that this is not the intended interpretation.

 

[erobz]

[The astute reader realizes that the problem does not specify a gear ratio, and assumes that the writer intended the problem to be solvable. So the natural reading cannot be the intended one]

I'm having trouble seeing how the gear ratio comes into play. The force accelerating the rider at $-0.3 \rm{\frac{m}{s^2}}$ doesn't care what gear the bike is in. What force acting on the bike does?

 

[PeroK]

I'm having trouble seeing how the gear ratio comes into play. The force accelerating the rider at $-0.3 \rm{\frac{m}{s^2}}$ doesn't care what gear the bike is in. What force acting on the bike does?

The force applied to the pedals. Which is, of course, a critical factor for the cyclist.

 

[Orodruin]

I think we should also not rule out the possibility that the original problem was not posed in English and that ”provide” is simply a matter of a faulty translation and that the translation for the correct interpretation would be ”exert”.

(”Provide” could very well be the literal translation of the original word without being correct in the current technical setting)

 

[erobz]

The force applied to the pedals. Which is, of course, a critical factor for the cyclist.

Ok, I think I'm seeing it. That force applied to the pedals can grow in the direction normal to the slope due to the gear ratio.