Register to reply 
Wind Force is Equal? 
Share this thread: 
#1
Mar1212, 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 20km/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? 


#2
Mar1212, 04:51 AM

HW Helper
P: 7,109

russ watters has posted the correct answer below.



#3
Mar1212, 04:56 AM

P: 3

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?



#4
Mar1212, 06:45 AM

Mentor
P: 22,286

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.



#5
Mar1212, 07:10 AM

HW Helper
P: 7,109

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.



#6
Mar1212, 08:10 AM

Mentor
P: 22,286

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.



#7
Mar1212, 08:39 AM

Sci Advisor
Thanks
PF Gold
P: 12,132

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. 


#8
Mar1212, 09:04 AM

HW Helper
P: 7,109

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. 


#9
Mar1212, 09:07 AM

Mentor
P: 15,153

The correct answer is c. The problem here is one of mixing reference frames. Energy (and thus work and power) are framedependent 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 nonconservative aerodynamic drag is the same in both cases. Another way to get an applestoapples 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. 


#10
Mar1212, 09:12 AM

HW Helper
P: 7,109




#11
Mar1212, 09:46 AM

Mentor
P: 15,153

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. 


#12
Mar1212, 09:48 AM

Sci Advisor
Thanks
PF Gold
P: 12,132

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. 


#13
Mar1212, 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. 


#14
Mar1212, 09:55 AM

Mentor
P: 15,153




#15
Mar1212, 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.



#16
Mar1212, 10:21 AM

Sci Advisor
Thanks
PF Gold
P: 12,132

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. 


#17
Mar1212, 10:35 AM

Mentor
P: 15,153

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?" 


#18
Mar1212, 10:40 AM

Sci Advisor
Thanks
PF Gold
P: 12,132

[QUOTE=D H;3811257]



Register to reply 
Related Discussions  
Wind scaling and increasing wind speed for wind tunnel problem  Mechanical Engineering  3  
Wind Power Vehicle Traveling Down Wind Faster Than The Wind  General Physics  305  
Equal and Opposite Force Help  General Physics  5  
Normal Force and Force of Gravity: When and why are they equal?  Introductory Physics Homework  5  
Centrifugal Force: Are these equal?  Introductory Physics Homework  2 