Physics – holding your hand out of a car window
January 23rd, 2020
This may seem like a silly question but I’ve just started college and I’m doing A level physics, so please bare with me.
We are studying Newtons third law and we have to complete a table, noting the action and reaction force for each example. One of the examples is holding your hand out of a car window.
Now I’m thinking that the action force is simply the force of the hand pushing against the air, and the reaction force is the force of the air pushing back. Is this correct? It’s probably beyond the physics we are doing right now, but all I am thinking is that if you stick your hand out the window in a moving car it’s hard to keep your hand in the same place sometimes. Are they really still equal forces as stated in the third law?
Many thanks in advance
Ok well when your hand is in the car, there are no forces acting on it in relation to the movement of the car. When you sick your hand out the window, it introduces resistance from the air. Now your hand isn’t an aerodynamic shape so the air doesn’t flow evenly around you hand. Depending on how tour hand is positioned, they could either be higher air resistance flowing over the hand then under, pushing your hand down. So the answer is, your hand isn’t an aerodynamic object and the air will flow over and under at different forces meaning it will be pushed around.
Thank you for your answer, but surely even if the car is not moving there are still balanced forces acting on it? Or would the balance be such that the forces are zero?
And just to clarify, in terms of my work and newton’s third law, the action force would be the hand pushing against the air, and the reaction force would be the air trying to push back?
Ok well when your hand is in the car, there are no forces acting on it in relation to the movement of the car. When you sick your hand out the window, it introduces resistance from the air. Now your hand isn't an aerodynamic shape so the air doesn't flow evenly around you hand. Depending on how tour hand is positioned, they could either be higher air resistance flowing over the hand then under, pushing your hand down. So the answer is, your hand isn't an aerodynamic object and the air will flow over and under at different forces meaning it will be pushed around.
You can demonstrate this to yourself by cupping your hand and angling it up and down.
A semi-aerofoil shape. Your hand will pull up and down on your arm, lifting and dropping like a plane (and for the same reasons – low/high pressures over surfaces)
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While your hand is inside the car it is travelling at the same speed as the air in the car.
When you put your hand out of the window it’s no longer in the car environment – but it is still attached to the body in the car. It’s now a semi-separate object travelling at the same speed as the car + or – external air/wind velocity.
If you were, hypothetically, to drive into a 100 Mph headwind (gale or cyclone?) at 100 Mph and put your hand out the window you could easily end up with a broken arm just from the speed/force/pressure variation.
While your hand is inside the car it is travelling at the same speed as the air in the car.Thank you, this makes sense – the explanation was helpful. I think I was partly getting confused with Newton’s first and second law too, so that wasn’t help.
When you put your hand out of the window it's no longer in the car environment - but it is still attached to the body in the car. It's now a semi-separate object travelling at the same speed as the car + or - external air/wind velocity.
If you were, hypothetically, to drive into a 100 Mph headwind (gale or cyclone?) at 100 Mph and put your hand out the window you could easily end up with a broken arm just from the speed/force/pressure variation.
Useful to know about the possibility of an arm being broken as well
Look up, make and play with, a vortex cannon and you can see the power air movement has.
Or have a look at the relevant QI episode – I forget which one, so you’d have to check IMDB.
Even gravity has that same effect – you can damage your eyes during a bungee jump because they are only “semi-connected” (meaning “not fully and solidly anchored”) to your head.
When you reach the bottom of the bungee jump and bounce back, they keep going for a micro-second or so. (Admittedly that’s more of an momentum/inertia thing)
For the purpose of science, look up “nude skydive” in google images – look at what happens to those boobies in the wind! (or someone’s face – Clarkson in Top Gear driving the super fast open top cars)
I’ve been following this discussion a bit. You make some interesting points ! Your examples about bungee jumping and such make sense, but I never considered it before…
Umm… isn’t the hand considered a plain point for the scope of the exercise?
I've been following this discussion a bit. You make some interesting points ! Your examples about bungee jumping and such make sense, but I never considered it before...
Powerful words from the guy who shut down the world with multiple satellite EMP bursts!!
Oh.. wait…
That was SNAKE Plissken.. not SNEAK Plisken (younger brother???)
Just teasing!
Umm… isn’t the hand considered a plain point for the scope of the exercise?It’s a variable surface – by keeping the fingers together you can then ‘cup’ it a little and turn it into a semi-aerofoil surface.
Cupping and flattening the hand will now vary that surface and the resulting pressure differentials.
Even a flat hand pointed upwards will do the same, the eddying on the top surface creates a low pressure region that gives lift.
Opening the fingers destroys the surface and each finger becomes it’s own separate entity attached to the palm just as the hand is attached the the arm.
In essence, if we have 50 foot hands we could glide (probably still to heavy to actually fly).
Disclaimer:
If you try this, please be aware of other cars and trees, etc!! Don’t come back with a law suit for me!
And also note: Wearing a Superman costume does not actually give the ability to fly!
Look up, make and play with, a vortex cannon and you can see the power air movement has.
Or have a look at the relevant QI episode - I forget which one, so you'd have to check IMDB.I think I’ve seen this episode… You have to love QI.It is quite scary to think how much power can be generated just by air – I think I’m going to take your word for it and not try flying and sticking my hand out of moving vehicles!
I handed my work in and it was correct, but I’m still a little confused. Sorry if I’m getting annoying… So I get that when you stick your hand out the window your hand tries to push the air, and the air pushes back. In theory, that’s a pair (as in Netwon’s third law), but that means there must be some other force acting because the two forces cannot be equal magnitude, otherwise they wouldn’t be a pair. So, what is this other force that causes your arm to get ripped off?… Or is it just that the two forces are not the same, in which case, why was my work marked right?
Many thanks
In theory, that's a pair (as in Netwon's third law), but that means there must be some other force acting because the two forces cannot be equal magnitude, otherwise they wouldn't be a pairWhat? Why they can’t be equal? Then your hand will move with steady speed (w/o acceleration).
Look up, make and play with, a vortex cannon and you can see the power air movement has.
Or have a look at the relevant QI episode - I forget which one, so you'd have to check IMDB.I think I've seen this episode... You have to love QI.
It is quite scary to think how much power can be generated just by air - I think I'm going to take your word for it and not try flying and sticking my hand out of moving vehicles!
I handed my work in and it was correct, but I'm still a little confused. Sorry if I'm getting annoying... So I get that when you stick your hand out the window your hand tries to push the air, and the air pushes back. In theory, that's a pair (as in Netwon's third law), but that means there must be some other force acting because the two forces cannot be equal magnitude, otherwise they wouldn't be a pair. So, what is this other force that causes your arm to get ripped off?... Or is it just that the two forces are not the same, in which case, why was my work marked right?
Many thanksPaired forces , but not necessarily equal forces. 20 –> and 40 <– for example.
Aerodynamics also plays a part – a streamlined car is not the same as a flat fronted big rig (cab-over-engine type).
(your arm will be pretty safe at 40 mph – it’s kinda childish fun)
♦
You’re right, but not in enough detail over the variance of forces.
100 Mph west into a 150 mph wind. Presume you suddenly stick you hand out the window, fast.
It goes from a (kind of) inert system (inside the car) where your hand AND the air around it are going west at 100 mph suddenly into an environment with a eastward force of 150 mph.
Total force on your arm-car boundary is 250 mph (that’s not a “force”, but this is simplifying it for my typing)
(I’m ignoring the effects of that headwind on the car itself, slowing it down, see below)
SNAP!!
♦
Lets ignore hand and arms – think of a car only.
If you do 59 mph in a 60 mph zone and suddenly a tailwind arrives of about 80 mph, you will go faster and break the speed limit – the differential is 20 mph, wind force to the car. You have to ease up on the accelerator to stay at 59 mph.
Same in reverse – 59 in a 60 zone and you hit a head wind:
a] you either slow down or
b] you press harder on the accelerator to overcome the ‘negative’ force and stay at the 59 mph (and use more fuel doing so).
Inside the car, your arm is a part of the car.
Outside the car it’s a separate entity BUT still attached to the car – forces act upon it with relation to the arm-car join.
If you have a titanium arm it would be resistant to higher “force differentials” than a flesh and blood arm.
( I’m not using the correct terms – I’m paraphrasing to simplify the tale)
♦
It’s complicated by many other factors – engine power, tyre grip, etc.
Imagine this:
A hover car, jet engine, no road contact at all, engine is low power.
It’s doing 40 mph north and a 50 mph head wind arrives – if the engine is very low power, the hover car will end up going backwards.
If the engine is high power, it adds force to the equation and keeps the hover car moving forwards into the headwind.
–> 50 <–40 = –> 10
OR
–>50 <–40 (+40/engine) = <– 30
(simplified, again, yes, but understandable).
To speak more complkicated:
A car moving forwards has multiple sources of positive and negative force.
Tyre grip, road inclination/declination (gravity), headwind/tail wind, engine power, aerodynamic shape (which affects its reaction to said head/tail wind), and so on.
Tyre grip also includes how much energy can be transmitted to the road as well as the actual friction on the road from the tyres.
Even your hand out the window, doing “air-planes” will introduce another variable.
But, most are so small as to be insignificant to the average person – tyre grip friction, your hand, etc.
Head wind, car momentum (current speed), engine power (acceleration or “staus quo” maintaining speed) are what really matter.
♦
The other issue is the speed of introduction of the arm to the outside environment: A skydiver reaches quite high speed during a fall but his arms don’t snap – becase he gets to that high speed slowly and evenly.
If he jumped out and suddenly went from zero-relative to 200-relative he’d be in some trouble. (just go look for those youtube naked dive videos hehehe)
Plus, a human is reasonable aerodynamic (smooth surfaced; some smoother than others ).
http://lvmarciano.com/wp-content/uploads/2009/07/naked_skydive.jpgSemi NSFW, but look at that inversion on that woman’s breasts!!
In theory, that's a pair (as in Netwon's third law), but that means there must be some other force acting because the two forces cannot be equal magnitude, otherwise they wouldn't be a pairWhat? Why they can’t be equal? Then your hand will move with steady speed (w/o acceleration).I was thinking that the force of the air hitting your hand cannot be the same as the force of your hand pushing against the air, otherwise you could stick you hand out of a car moving at 100 mph and not be affected.
As has pointed out, it might be ok if you gained that speed at a steady rate, but even so its likely that your arm will be bent backwards (depending on how strong you are).
Paired forces , but not necessarily equal forces. 20 --> and 40 <-- for example. But we have been taught that according to Netwon’s third law, the opposing forces but be the same magnitude but in opposite directions.So basically, there is a paired force between the hand and the air. However because of the car’s acceleration and velocity, and wind speed, the forces become unbalanced around the pair and there are extra forces pushing against your hand. Is this correct?
[quote=””]
In theory, that's a pair (as in Netwon's third law), but that means there must be some other force acting because the two forces cannot be equal magnitude, otherwise they wouldn't be a pairWhat? Why they can’t be equal? Then your hand will move with steady speed (w/o acceleration).
I was thinking that the force of the air hitting your hand cannot be the same as the force of your hand pushing against the air, otherwise you could stick you hand out of a car moving at 100 mph and not be affected.
Not quite - you still need something capable of sustaining the forces on it.
If you have a ton of force on one side and a ton on the other they cancel out - but the thing in between still need to endure two tons of force.
Force on force cancels out evenly.
Force on object with counter force is still force; even if cancelled out, it still affects the object.
As has pointed out, it might be ok if you gained that speed at a steady rate, but even so its likely that your arm will be bent backwards (depending on how strong you are).
Paired forces , but not necessarily equal forces. 20 --> and 40 <-- for example. But we have been taught that according to Netwon's third law, the opposing forces but be the same magnitude but in opposite directions.Ahh – I’m seeing you point now, and your question properly – you are right, but it’s something that is updated by micro-seconds.
As you accelerate, you increase the force forwards and also increase the wind pressure backwards.
It’s an equation (that also fitted mine above)
<– 40 = –>40 but you also have to add the original wind speed.
If the wind was still then you have a forwards speed of 40 which also equals a wind speed of 40 (you are in essence hitting stationary wind at 40 which equals the opposing force of 40).
If the wind is doing 20 towards you the resultant forces are 20 backwards (so you have to increase power to maintain the same speed), ignoring aerodynamic efficiency.
If your hand moves backward relative to you (or the car, essentially the same thing) then there is more force from the front, and visa versa.
So basically, there is a paired force between the hand and the air. However because of the car's acceleration and velocity, and wind speed, the forces become unbalanced around the pair and there are extra forces pushing against your hand. Is this correct?The “equal and opposite reaction” does apply, you are pushing the wind with the same force as the wind is pushing you. But, because you are in a moving system, they don’t cancel out.
But that’s the thing – those “equal and opposite” forces are the forces that can do the damage.
At the surface of your hand there is the forward force of the car’s momentum and acceleration AND the force of the wind pushing back.
…
Study question:
If we breathe out at X force and there is an “equal and opposite” forcev(pushing back), how can we actually breathe?
If we breathe out at X force and there is an "equal and opposite" forcev(pushing back), how can we actually breathe? And then you wonder why I would bombard earth with multiple satellite EMP bursts? Dunno, it may have something to do with the equal and opposite forces going on inside your thorax creating a difference in pressure between inside your lungs and the outside?
So basically, there is a paired force between the hand and the air. However because of the car's acceleration and velocity, and wind speed, the forces become unbalanced around the pair and there are extra forces pushing against your hand. Is this correct?Essentially, yes.
A hand at rest has an internal pressure of appox 1 atmosphere (has to be, higher and we’d balloon out, lower and we’d feel compressed) to match air pressure. (This is why feel feel strange at altitude, until we re-balance).
In essence, the wind is increasing the air pressure in the localised area of your hand to some amount calculable from the wind speed. As you push into it it does push back but it also compressed on the leading side away from your hand.
——–> <—————————- | ————————-> (hand)
wind ….. pressure caused by hand | equal/opposite force .. your hand
It’s complicated because air is not a single entity.
Try moving your hand around in the bath and you’ll see the effects in a more structurally linked medium.
This is how we swim, after all – we kick, pressure backwards = we move forwards. Equal and opposite (taking into account losses due to friction, etc)
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If you stop breathing with your mouth open, the pressure in our lungs matches the pressure outside – we compress our intercostal muscles (rib muscles) to create a region of higher pressure inside our lungs, forcing the air out.
The we expand the muscles and create a low pressure region – external 1 atmosphere of pressure forces air in.
Air is not a finite article, like a car, it’s made up of minute separate-but-reactive ‘articles’ – as we breathe out they react to the forces on them and compress ahead of our breath (which is why we can create a small wind and blow on things)
Think of this:
You are pushing a small cabinet filled with books (40kg total) on a wooden surface.
You start pushing the cabinet but it doesn’t move so you push a little harder. You keep increasing pressure until it starts to slide along the floor.
You could equate this to 40 kg + floor friction/intertia = 45 kg. you have to equal that 45 kg pressure.
As you increase your oush, the cabinet ‘resist’ with equal force back to you.
When you break the “friction barrier”, the cabinet now only ‘resists’ with its own 40 kg plus the reduced friction/inertia quotient (as it’s now moving).
So you push with 45 kg, it resists with, say, 41 kg = the cabinet moves at the speed that 4kg of force will give it.
If someone reached in without you knowing and pulled out 30 kg of books, the cabinet would shoot off and you’d fall over (to much laughter) – because you are pushing with 45, resist is now 11 (the -30 of the books), so the force is now 34 kg and off it shoots (hopefully hitting the muppet who pulled the books out!).
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Have you ever looked at something and thought it was heaver that it was?
When you lift it, it shoots up – because you applied a force level you thought matched it’s combined mass/inetia/friction.
If you stop breathing with your mouth open, the pressure in our lungs matches the pressure outside - we compress our intercostal muscles (rib muscles) to create a region of higher pressure inside our lungs, forcing the air out.
The we expand the muscles and create a low pressure region - external 1 atmosphere of pressure forces air in.
Air is not a finite article, like a car, it's made up of minute separate-but-reactive 'articles' - as we breathe out they react to the forces on them and compress ahead of our breath (which is why we can create a small wind and blow on things) When you exhale or inhale? Or it doesn’t matter? What about the thingy (don’t know the English term) in your throat that shuts off your euhm tubes? Maybe I’m going on too much about this specific example, but when you put an end to the inhaling or exhaling process, you feel like there’s a certain pressure in your throat. So I’m not sure about the equality in pressure.When you exhale or inhale? Or it doesn't matter? What about the thingy (don't know the English term) in your throat that shuts off your euhm tubes? Maybe I'm going on too much about this specific example, but when you put an end to the inhaling or exhaling process, you feel like there's a certain pressure in your throat. So I'm not sure about the equality in pressure.We breathe by creating a pressure differential between our lungs and the external environment.
Yes, at the ends of the specific cycle there’s a pressure felt in our muscles to continue the cycle
I meant that if you just breathe out gently with your mouth open and just stop, the pressure in the lungs is equal to that outside, at that moment of stopping, with your mouth open.
You will soon feel a need to breathe in, this will be expressed as a feeling of emptiness in your chest, but it’s a biological process, mental and muscular combined.
Breathing is semi-autonomous. We can over-ride it but eventually the autonomous side will take over.
All of the ‘extras’ just help us control our breathing and divert one pipe into two and stop us choking when we drink.
I understand the process, but even if I tried it gently I get the feeling something locks in my throat when I stop. Maybe I should have my throat checked… Anyways, it’s been a while since I dabbled with physics, but I still think it’s a fun subject!
I understand the process, but even if I tried it gently I get the feeling something locks in my throat when I stop. Maybe I should have my throat checked... Anyways, it's been a while since I dabbled with physics, but I still think it's a fun subject!There is a mechanism for holding your breath – so you can go underwater.
You can over-ride it, sort-of, by relaxation.
You might be overly tense?
Try breathing in and our like this:
In slowly, count to 5 as you breathe in.
At five, hold it for a count of one.
Breathe out for a count of 5, hold for a one count and repeat.
in 1-2-3-4-5 – hold – out 1-2-3-4-5 – hold – in 1-2-3-4-5
Do this for a little while (stop the sequence if you feel dizzy (oxygen overload, it happens if you’re not used to the procedure) it’s not a dangerous thing to do).
(this is also an excellent relaxation technique in general)
Now, breathe at your normal resting rate; breathe out and just let it taper off and stop (rather than holding the breathe, just let it stop).
It’s like the end of a deep sigh – you just stop and at that moment you are equalised with external pressure.
After a short time your autonomous breathing will kick in and you’ll breathe in again.
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It’s interesting – our breathing is carbon dioxide driven rather than oxygen driven.
We breathe in to capture low concentration-of-CO2 air for osmotic transfer from our blood, we breathe out to expel than air when it’s ‘pumped’ with CO2.
It’s the CO2 concentration in the blood that defines the need to exchange the air in our lungs with fresh air. (not JUST that, there are O2 sensors too).
CO2 concentrations in the brain and arterial blood.
For example, when a healthy person takes several deep and fast breaths, CO2 in the lungs and blood falls. The breathing centre detects this drop and stops or reduces stimulation and work of the respiratory muscles. The person naturally holds their breath until the CO2 level reaches the initially pre-set value. Conversely, breath holding accumulates more carbon dioxide. The breathing centre senses this increase and intensifies breathing. This overbreathing is going to continue until extra CO2 is removed and the pre-set value is reached again.
We breathe more heavily during physical exercise, when our bodies produce more carbon dioxide. However, the rate of CO2 production matches the rate of CO2 removal in such a fashion that CO2 and O2 values in the arterial blood changes during exercise only slightly.