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Wind Force is Equal?

by 909kidd
Tags: force, friction, headwind, wind
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909kidd
#1
Mar12-12, 04:31 AM
P: 3
Against the Wind

One windless day you ride your bicycle at 25 km/h for one hour on a flat road. The next
day, you ride 5 km/h into a 20-km/h headwind for one hour on a flat road. Assuming that
the force of the wind (the primary "friction" in this case) scales with the square of the
relative velocity of the bike and wind, the wind's force is the same on both days. Which
ride seemed to take more effort on your part?
(a) The ride on the windy day was harder.
(b) The ride on the windless day was harder.
(c) Both rides were about the same.
Explain your answer.

This not Homework or anything. My class was discussing this Question, and must of use came up with the C), but not a lot of us came with the same Why?

I assumed that Force in both cases are Zero, but the work done on day one would be greater, because I am going at a 5x the speed on the first day. I know my answer is wrong. But can anyone explain why the answer is C?
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rcgldr
#2
Mar12-12, 04:51 AM
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russ watters has posted the correct answer below.
909kidd
#3
Mar12-12, 04:56 AM
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But if the wind is blowing directly at you at 20KM/hr, then wouldn't the 5km/hr day be harder, since the first day is windless?

russ_watters
#4
Mar12-12, 06:45 AM
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Wind Force is Equal?

The answer is b, even if we ignore friction in the bike. Power is force times speed, so while the force is the same, the speed in b is much higher, so the power is much higher.
rcgldr
#5
Mar12-12, 07:10 AM
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It might be easier to understand this if you imagine that you're riding a bike on the surface of a very long treadmill, and moving at 25 kph with zero wind. Situation (b) (windless day) day would correspond to the treadmill not moving, so the rider is supplying all the power in order to go 25 kph. Situation (a) (windy day) would correspond to the treadmill moving at 20 kph, supplying most of the power, with the rider only having to supply the power to move an additional 5 kph relative to the treadmill.
russ_watters
#6
Mar12-12, 08:10 AM
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Er, no. If there is no wind and the bike is moving at 25kph, then thats like the treadmill moving at 25kph, with a fan blowing in your face at 25kph. In the other situation, you're riding 5kph with a fan blowing at 25kph.
sophiecentaur
#7
Mar12-12, 08:39 AM
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It's a pity the word "effort" was used in the question because it is a bit ambiguous in this context. Effort is a term used for the Force applied to a Machine and is an 'instantaneous' quantity,
If the question is referring to the total Energy involved then pedalling against the wind with the same force as in still air will involve going slower - hence taking longer over the journey. This will involve more energy because you are doing the same amount of work against the wind per second - but for longer. Muscles are impossible to analyse accurately but you could imaging that you might drop a gear to go against the wind so your 'effort' force was the same in both cases. Hence, the Energy input would be proportional to the time taken. I'd go for 'a' - with my reading of the question.
rcgldr
#8
Mar12-12, 09:04 AM
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Quote Quote by rcgldr View Post
It might be easier to understand this if you imagine that you're riding a bike on the surface of a very long treadmill, and moving at 25 kph with zero wind. Situation (b) (windless day) day would correspond to the treadmill not moving, so the rider is supplying all the power in order to go 25 kph. Situation (a) (windy day) would correspond to the treadmill moving at 20 kph, supplying most of the power, with the rider only having to supply the power to move an additional 5 kph relative to the treadmill.
Quote Quote by russ_watters View Post
If there is no wind and the bike is moving at 25kph, then thats like the treadmill moving at 25kph, with a fan blowing in your face at 25kph. In the other situation, you're riding 5kph with a fan blowing at 25kph.
Apparently I worded that badly, so a redo:

Situation (windless day) analogy - Wind speed wrt ground is zero. Treadmill speed wrt ground is zero. Rider is pedaling the bike at 25 kph wrt treadmill.

Situation (headwind day) analogy - Wind speed wrt ground is zero. Treadmill speed wrt ground is 20kph. Rider is pedaling the bike at 5 kph wrt treadmill.
D H
#9
Mar12-12, 09:07 AM
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Quote Quote by russ_watters View Post
The answer is b, even if we ignore friction in the bike. Power is force times speed, so while the force is the same, the speed in b is much higher, so the power is much higher.
The windless day (answer b) is harder? That doesn't make sense. Ride a bike the same distance against a considerable headwind versus a calm day. You'll feel a much more tired on reaching the destination on the windy day compared to the calm day. However, this is not a problem of riding equal distances with respect to the ground. It is a problem of riding for equal amounts of time.

The correct answer is c.

The problem here is one of mixing reference frames. Energy (and thus work and power) are frame-dependent quantities. Ground speed is irrelevant since we're ignoring rolling friction, so using ground speed as the basis for reasoning is a red herring. Much better is to use a frame moving with the flowing air. This is the same as the ground frame in the case of a windless day, but not on the windy day.

In both cases (windy versus windless), the bicyclist is moving at 25 km/hr relative to the wind frame, making the aerodynamic force against the bicyclist is the same in both cases. The force exerted by the bicyclist is also the same in both cases by Newton's third law. In both cases, the distance traveled in this wind frame is the same, 25 kilometers. Same force, same distance, same amount of time means the work against the non-conservative aerodynamic drag is the same in both cases.

Another way to get an apples-to-apples comparison is to look at things from the perspective of the bicyclist. The bicyclist is exerting the same amount of power in both cases. Work is average power times time, so once again, the answer is c.
rcgldr
#10
Mar12-12, 09:12 AM
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Quote Quote by D H View Post
In both cases, the distance traveled in this wind frame is the same, 25 kilometers.
Except the point of application of the force is at the ground. If the rider was driving a propeller to move the bicycle, then the power would be the same, but instead the rider power is output to the ground, in which case I think my treadmill analog may help explain the situation.
D H
#11
Mar12-12, 09:46 AM
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Quote Quote by rcgldr View Post
Except the point of application of the force is at the ground.
That's an irrelevant fact because we are ignoring ground friction here.

If the rider was driving a propeller to move the bicycle, then the power would be the same, but instead the rider power is output to the ground, in which case I think my treadmill analog may help explain the situation.
Your treadmill adds yet an unnecessary complication to the already unnecessary complication added by considering ground speed. Your propeller-driven bicycle is in fact a much better analogy.

Here's another analogous situation, using a canoeist rather than a bicyclist. On the first day, the canoeist paddles five kilometers across a lake in one hour. On the next day, the canoeist again paddles for one hour, but this time paddles a kilometer upstream in a river flowing at four km/hr. To an observer standing still on the lake shore / river shore, the paddler will have performed more work on the first day compared to the second. To the canoeist, the two days are of course identical.


The question asked "Which ride seemed to take more effort on your part?" It did not ask "Which ride seems to take more effort from the perspective an observer fixed with respect to the ground?" Energy is a frame dependent quantity. One has to be very careful that one is comparing apples to applies when looking at energy.
sophiecentaur
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Mar12-12, 09:48 AM
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Quote Quote by rcgldr View Post
Except the point of application of the force is at the ground. If the rider was driving a propeller to move the bicycle, then the power would be the same, but instead the rider power is output to the ground, in which case I think my treadmill analog may help explain the situation.
I think it adds to the confusion, actually (but I see where you're coming from).
If you assume that the cyclist can't tell what is causing the 'moving air' effects (energy loss) then he is constantly using the same Power in overcoming them. The journey into the wind will take him longer - hence more energy is needed for the whole journey.
anuragkanase
#13
Mar12-12, 09:52 AM
P: 27
W=F x s
F= m x a
here a=(a1 due to motion - a2 of wind)
So the answer goes b.
S=SPEED/T
So s is inversely proportional to T.
Answer goes a.
Taking over the overall situation, the whole Work done in Unit TIME, we are equating both thee equations. So the answer and the CORRECT answer is C.
The answer will totally change when you consider aerodynamics.
P. S. Correct me if I am wrong.
D H
#14
Mar12-12, 09:55 AM
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Quote Quote by sophiecentaur View Post
I think it adds to the confusion, actually (but I see where you're coming from).
If you assume that the cyclist can't tell what is causing the 'moving air' effects (energy loss) then he is constantly using the same Power in overcoming them. The journey into the wind will take him longer - hence more energy is needed for the whole journey.
The two journeys do not cover the same ground distance in this problem. They are instead of the same time duration. The same amount of power expended during the same amount of time means the same amount of energy will be expended in the two cases -- from the bicyclist's perspective. Other observers will have different thoughts because energy is a frame dependent quantity.
chingel
#15
Mar12-12, 10:10 AM
P: 261
Since the wind is at the same speed relative to him in both cases, the drag force is the same. But since he is pushing the ground, in one case he has to push the ground a longer distance, since the ground is moving faster relative to him. Work is force over a distance so he has to do more work in case b.
sophiecentaur
#16
Mar12-12, 10:21 AM
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A touch of the RTFQ, here. Woops!
If he's doing the same amount of work for the same time against 'air' then the other friction effects will be greater if he travels further (force times distance). That must mean that the longer journey over the ground (b) must use up more total energy.

a, b or c, depending on how carefully or pickily you read the question. Excellent conversations piece.
D H
#17
Mar12-12, 10:35 AM
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Quote Quote by sophiecentaur View Post
A touch of the RTFQ, here. Woops!
If he's doing the same amount of work for the same time against 'air' then the other friction effects will be greater if he travels further (force times distance). That must mean that the longer journey over the ground (b) must use up more total energy.
Read the question even more carefully. This answer, [i]b[/b], is an answer to the question "Which ride seems to take more effort from the perspective an observer fixed with respect to the ground?"

That is not the question that was asked. The question that was asked was "Which ride seemed to take more effort on [the bicyclist's] part?"
sophiecentaur
#18
Mar12-12, 10:40 AM
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[QUOTE=D H;3811257]
Quote Quote by chingel View Post
Read the question even more carefully. This answer, [i]b[/b], is an answer to the question "Which ride seems to take more effort from the perspective an observer fixed with respect to the ground?"

That is not the question that was asked. The question that was asked was "Which ride seemed to take more effort on [the bicyclist's] part?"
The sports commentator? It would all depend on what sums the ground-based observer was doing. I wouldn't have to be on the bike to get the answer right. (Assuming I'd bothered to RTFQ properly) I would be 'empathising' with that poor cyclist.


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