Who likes physics?

[mikeyg]

Solve these and I'll give you a cookie.

 A loaded penguin sled weighing 70 N rests on a plane inclined at 20° to the horizontal. Between the sled and the plane the coefficient of static friction is 0.23, and the coefficient of kinetic friction is 0.17. 

(a) What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane?

(b) What is the minimum magnitude F that will start the sled moving up the plane?

(c) What value of F is required to move the block up the plane at constant velocity?




  A slide loving pig slides down a certain 28° slide in twice the time it would take to slide down a frictionless 28° slide. What is the coefficient of kinetic friction between the pig and the slide?



 A slab of mass 40 kg rests on a frictionless floor, and a block of mass 10 kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.30. The block is pulled by a horizontal force with a magnitude of 100 N

a) What is the resulting acceleration of the block?

(b) What is the resulting acceleration of the slab?



A bolt is threaded onto one end of a thin horizontal rod, and the rod is then rotated horizontally about its other end. An engineer monitors the motion by flashing a strobe lamp onto the rod and bolt, adjusting the strobe rate until the bolt appears to be in the same eight places during each full rotation of the rod. The strobe rate is 2000 flashes per second; the bolt has mass 45 g and is at radius 3.3 cm. What is the magnitude of the force on the bolt from the rod?



A banked circular highway curve is designed for traffic moving at 60 km/h. The radius of the curve is 205 m. Traffic is moving along the highway at 35 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding off the road?


I'll give you a cookie I swear.

[mikeyg]

You have until midnight tonite to win the cookie. 

If you get it wrong, I'll know because I have the answers in front of me.

[BoliverAllmon]

You have until midnight tonite to win the cookie. 

If you get it wrong, I'll know because I have the answers in front of me.

CHEATER!!!!!!!!!!

[mikeyg]

How so?  I already solved them in physics class this afternoon.  Well, actually my professor did. 

[BoliverAllmon]

How so?  I already solved them in physics class this afternoon.  Well, actually my professor did. 

See cheater. someone else did the work for you. remember cheaters go to Hell. LOL

[mikeyg]

In hell we party, in heaven you pray.  What's more fun?

[pianonut]

here's some wild guesses:

a) .23
b) .17
c) .40 squared

i guess that the pig went down a greased slide with a coefficient of kinetic friction about .23

i don't really know how i came to these conclusions.  don't ask me to explain.  they are guesses.

[BoliverAllmon]

In hell we party, in heaven you pray.  What's more fun?

you must not have gone to a charismatic church before.

[Torp]

Solve these and I'll give you a cookie.

 A loaded penguin sled weighing 70 N rests on a plane inclined at 20° to the horizontal. Between the sled and the plane the coefficient of static friction is 0.23, and the coefficient of kinetic friction is 0.17. 

(a) What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane?

(b) What is the minimum magnitude F that will start the sled moving up the plane?

(c) What value of F is required to move the block up the plane at constant velocity?

(a) 15.13 N - equation used was F=.23*Cos(20)*70N
(b) a little hazy on the concepts on this one, though I think it would either be the same number as (a) or slightly more than that.  As a definitive answer I'll say 15.13 N
(c) 11.17 N - same equation as (a) with kinetic friction coefficient.

Am I in the ball park?

[Torp]

A slide loving pig slides down a certain 28° slide in twice the time it would take to slide down a frictionless 28° slide. What is the coefficient of kinetic friction between the pig and the slide?

Don't we need velocity to figure this one out? i.e. how many ft/s or m/s the pig is traveling.

I was using the equation f=v/(g*t)  where:

f = coefficient of friction
v = velocity
g = gravity
t = time

Since gravity is constant and the time variance is given, velocity seems to be what is needed.  I'm probably just using the wrong equation though.

Man, it's been a long time since I've done physics...

[mikeyg]

Am I in the ball park?

No

[m1469]

... in heaven you pray.  What's more fun?

How do you know ? 

Man, it's been a long time since I've done physics...

And I never have, so I am going to watch this thread like a hawk and learn, learn, learn 

m1469

[musik_man]

(a) First you need to break the force of Gravity into it's x and y components

F_g,x=sin(20)*70=23.94
F_g,y=cos(20)*70=65.78

Then we can use the y part of gravity to calculate friction

F_f=.23*65.78=15.13

If we want the penguin to stay still, the sum of the forces on him must be 0.

F_x=0=23.94-15.13-F

so...

F=8.81N

(b) In order to move the penguin up the hill we need to give enough force to overcome both gravity and friction.

F_x=0=F-23.94-15.13

F=39.07N

(c) You do this just like (b) except that you use .17 instead of .23 as your coefficient of friction

F_f=11.18

F_x=0=F-23.94-11.18

F=35.12N

I might do some of the other ones later   After the Magnetism/electricity Physics course I took last semester, all the kinetics stuff is kinda pleasant.

[mikeyg]

A cookie for you my good sir. 

[mikeyg]

3:44 minutes to deadline poeple.  Let's get on it.

[pokeythepenguin]

awwww...... I thought this would be fun physics; not highschool simple formula junk -.-

Ask something like..... hmmmm......... what's a good one........


ok.  In an infinite universe there are an infinite number of things that are blue, and an infinite number of things that are not, so the ratio of things that are blue to things that are not is 1:1

In an infinite universe there are an infinite number of things that are blue, an infinite number of things that are green and an infinite number of things that are neither blue nor green, making the ratio of things that are blue to green to neither 1:1:1, making the ratio of things that are blue to things that are not blue 1:2

In an inifinite universe there are an infinite number of colors, making the ratio of things that are blue to the ratio of things that are other colors 1:1:1:1:1:1:1:1:1:1:1:1 ad infinitum...... making the ratio of things that are blue to things that arent 1:(infinity), and 1/infinity is equal to zero.


how can all of these be correct?   THAT is fun physics ^^

[mikeyg]

you didn't answer the question.

I'm sure you want a cookie, and I'm sure you can solve these.

So earn that cookie.

I'll turn this into a more interesting thread after the questions are answered

[mikeyg]

2:50.  I thought you guys were smart?  :p

[allchopin]

What do you need these answered for if your professor already solved them?

I guess I'll pick up where everyone has left off.
Not sure what the point of the road's max turn velocity is, but at 35 km/h, you'd need .047 cof.

The force on the bolt is 3664.3 newtons.

Pokey:  What exactly is the problem?  The logic isn't exactly correct in setting up a ratio using infinity, because 2*inf is not more than inf - thus, a 2:1 ratio of infinite objects is erroneous.

[pianonut]

why is it that space is always explored and the common household appliance is left to flounder.  we women should be working less due to all the brains involved in solving problems.  why do washer and dryer still shake?  how come people still attach the dryer to a LONG exhaust hose and not a short one (saving energy and making it more efficient?).  all i care about is results.

cookie?

say, in my last airline travels i pulled the aviation magazine out of the front pocket and noticed a moped that runs on electricity (obviously saving fuel) and wondered if you need a license plate to drive one of these?  probably not allowed in regular freeway traffic, but since my son is turning 16 soon - could he get to work on one of these and plug it in while at work?  has anyone seen these new mopeds?  what do you think?  if you don't have a license plate, you don't need insurance, right?  just a helmet and some gumption.

[pianonut]

one other physics question that i have wondered for years.  say you are in a 757 boeing jetliner and it suddenly takes a nosedive.  you go for the exits and try to open the doors.  will there be so much suction that the air helps you open the door (and you fly out with it), or will it be impossible to open the door (as well as it weighing 50 lbs, haivng so many procedures to open, and finally having to actually heave the door out and quickly pull the string for the slide).  do people ever actually use the escape mechanisms - or is this a false sense of security.  and , how come they haven't changed much in 20 years? 

[xvimbi]

one other physics question that i have wondered for years.  say you are in a 757 boeing jetliner and it suddenly takes a nosedive.  you go for the exits and try to open the doors.  will there be so much suction that the air helps you open the door (and you fly out with it), or will it be impossible to open the door (as well as it weighing 50 lbs, haivng so many procedures to open, and finally having to actually heave the door out and quickly pull the string for the slide).  do people ever actually use the escape mechanisms - or is this a false sense of security.  and , how come they haven't changed much in 20 years? 

Do the following experiment: drive in a car at, say 40 miles per hour, then just open the door but don't push it outwards. Observe what is going to happen.

[pianonut]

but if you wait too long, the airliner will sink - leaving you to worry about getting the seat cushion off the seat and unwrapping yourself from the oxygen mask which you might hang from if you did that first - and didn't worry about getting a start on the door (and soon finding out the physics of opening the door under water).

maybe one doesn't have to worry about physics of all this since the jetliners usually break in half on impact.  if you survive, you have to trust your seat cushion?  how well do these seat cushions work?  has anyone tried to float on one?   

[i_m_robot]

and , how come they haven't changed much in 20 years? 

just the commercial ones

who knows what the heck we have nowadays

and would we trust it as much as the common image of a plane

[pianonut]

anyone see sean connery's version of the soviet sub that wasn't rescued in time.  can you imagine the frustration of those guys.  they must have been really miffed.

[asyncopated]

Which way does the door in a 757 face?  I haven't been on a 757.  Are the hinges on the left or the right? 

I don't think that the doors are supposed to be opened during a nose dive.  They are supposed to be after the plane has crashed.  Presumably, if the plane does crash on water, it has some buoyancy if the hull is not compromised and will float for a while.  If it does crash on land, hopefully the pilot has the sense to find a reasonably clear spot  (open fields etc.) and try to land the plane softly.

Boeings and other passenger planes are built for stability, not for aerobatics.  It's actually not that easy to make the plane nose dive if you don't want to.  In theory, the plane can even glide to a certain extent without any of its engines -- not that I would want to try.

As for your question, I can only give you an approximate answer.  Lets assume that you don't push the door open but a machine disengages all the hinges and throws the door open.  A very rough estimate is this

Force needed to open door = (Pressure outside - Pressure in cabin) * Area + ½*density * speed^2 *Area.

The density is the density of air at that altitude.  By pressure outside, I mean static pressure.  Speed^2 is the square of the relative speed of the plane, that is speed = ground speed-wind speed.

The last term comes from Benoulli's equation, which presumes an incompressible fluid with no eddies.  So it is very much a first order approximation, but is roughly correct.

I suspect that the term that will dominate will be the Benoulli term with speed^2.  So it will be very difficult to open the door.   

My advice -- just pray that you plane does not crash.

al.

[Torp]

No

Given Musik_Man's solution, I'd say I was.  I had at least properly calculated the coefficient of friction.  I knew I was missing something.

Perhaps since your teacher had solved the problems for you, you just weren't aware that I was on the right track.

[asyncopated]

I would like to start the second question.

First assume that the rectagular (cuboidal) pig. 

[asyncopated]

Ok, here is the proper answer.  Assuming the pigs don't fly,

Weight of pig = m g
Normal force = m g  cos(theta)

Force along the frictionless slide is given by
m g sin(theta)

With friction it is
m g [ sin(theta) – mu cos(theta) ]

Assuming the pig starts with velocity 0 and acceleration is constant with a=f/m
when there is no friction, acceleration is
a_{nf} = g sin(theta)
with friction, it is
a_{f} = g (sin(theta) – mu cos(theta)),
where mu is the coefficient of friction

Formula for constant acceleration, by integrating
v =  a t
and distance
x =  ½ a t^2

The slides have the same length
2 t_{nf}= t_{f} 

Equating the lengths
½ a_{nf} t_{nf}^2 = ½ a_{f} t_{f}^2

substituting the relavant formulae we get
¼ a_{nf} t_{f}^2 = a_{f} t_{f}^2

Substituting the accelerations,
¼ sin(theta)=sin(theta)-mu cos(theta).

The coefficient of friction is therefore
mu = ¾ tan(theta)

Voila. 

Alternatively you can write :=

The Hamiltonian for such a system is not time independent, and thus a temporal varying Largrangian must be introduced.  In this case, a Routhian is likely to be the best choice.   Moreover, as in this case where the Hamiltonian does not vary in an adiabetic manner, it is difficult if not impossible to give an ad hoc solution or apply simple perturbation to this problem.

You should get full marks for that.

al.

[xvimbi]

The Hamiltonian for such a system is not time independent, and thus a temporal varying Largrangian must be introduced.  In this case, a Routhian is likely to be the best choice.   Moreover, as in this case where the Hamiltonian does not vary in an adiabetic manner, it is difficult if not impossible to give an ad hoc solution or apply simple perturbation to this problem.

Not if the teacher knows anything about higher math. Then you'd be sent home as a bull***ter

[popndekl]

awwww...... I thought this would be fun physics; not highschool simple formula junk -.-

yep, that's right - we had such exercises in our test papers...
but pokey, soOo high physics begin to become philosophy, and that's what i hate. period.
the famous popndekl has spoken. lol.

[pianonut]

dear asyncopated,

thanks for taking the time to seriously answer my level 1 physics questions.   being quite reassurred especially that 757's are not built for acrobatics and usually stay level for enough time to restart engine or glide  - and are capable of soft landings, etc.  i really thought that they would soon nose dive if the engines quit.  thanks!

impressed that you know so much about math and physics.  i took a class years ago and promptly forgot the formulas.  you have to use it all the time, i think.  sort of like foreign languages.  my dad is an engineer.  we never got along because whenever i'd ask for math help, he'd explain it once and i would listen and then start asking questions.  then he'd explain it again - this time talking louder.  then, the last time he'd be yelling.  and, of course, to end the whole thing - he'd say something like - 'you wear your feelings on your sleeve.'

ok.  i try to notice details, but you know - i don't think there were hinges on this door.  to properly do the experiment at 40+ mph, maybe i should take up gliding.  i believe this door to be hingeless.  therefore making it possible to heave as one complete door.  it would be very frustrating to me if one were 'in charge' of the exit, only to find that the pilot was the only one that could open the door.  now what kind of reassurance is that? especially if the pilot is dazed, confused, or dead.  maybe they're built to unlock upon a major impact? or there's a red button somewhere that the stewardesses know about.  why do they make you worry so much if you sit near the exits when you couldn't open it if you wanted to.  oh well.

i always do pray for a safe take-off and landing.  somehow, i find lightening interesting.  we had a bolt strike while flying and it was fairly nearby.  having done my utmost to not perterb the pilot (turning my cell phone off) - i suddenly wondered what a cell phone would do compared to a strike of lightening?  oh, well. 

 

[mikeyg]

Ok, here is the proper answer.  Assuming the pigs don't fly,

Weight of pig = m g
Normal force = m g  cos(theta)

Force along the frictionless slide is given by
m g sin(theta)

With friction it is
m g [ sin(theta) – mu cos(theta) ]

Assuming the pig starts with velocity 0 and acceleration is constant with a=f/m
when there is no friction, acceleration is
a_{nf} = g sin(theta)
with friction, it is
a_{f} = g (sin(theta) – mu cos(theta)),
where mu is the coefficient of friction

Formula for constant acceleration, by integrating
v =  a t
and distance
x =  ½ a t^2

The slides have the same length
2 t_{nf}= t_{f} 

Equating the lengths
½ a_{nf} t_{nf}^2 = ½ a_{f} t_{f}^2

substituting the relavant formulae we get
¼ a_{nf} t_{f}^2 = a_{f} t_{f}^2

Substituting the accelerations,
¼ sin(theta)=sin(theta)-mu cos(theta).

The coefficient of friction is therefore
mu = ¾ tan(theta)

Voila. 

Alternatively you can write :=

The Hamiltonian for such a system is not time independent, and thus a temporal varying Largrangian must be introduced.  In this case, a Routhian is likely to be the best choice.   Moreover, as in this case where the Hamiltonian does not vary in an adiabetic manner, it is difficult if not impossible to give an ad hoc solution or apply simple perturbation to this problem.

You should get full marks for that.

al.

you get a cookie as well.  And another for being sexy.

[musik_man]

you get a cookie as well.  And another for being sexy.

Why don't I get a sexy cookie???

[mikeyg]

Why don't I get a sexy cookie???

Because sexy cookies are oly given out today.  So answer another one and I'll give you a sexy cookie.

[xvimbi]

I suspect that the term that will dominate will be the Benoulli term with speed^2.  So it will be very difficult to open the door. 

I somehow doubt this. In my car example above, the door will actually open on its own. The reason is that there is a pressure difference between the outside and the inside of the car. The pressure difference is zero when the velocity of the car relative to the surrounding air is zero, but as soon as there is wind outside the car, the pressure will be lower on the outside. This is the reason why people get sucked out of a plane when there is a hole. In actual fact, they get blown out of the plane. In addition, the air pressure at high altitude is already lower than inside the plane, even without the plane moving through the air. So, I would say, the doors will open, and man they will just fly off the hinges!

[allchopin]

I somehow doubt this. In my car example above, the door will actually open on its own. The reason is that there is a pressure difference between the outside and the inside of the car. The pressure difference is zero when the velocity of the car relative to the surrounding air is zero, but as soon as there is wind outside the car, the pressure will be lower on the outside. This is the reason why people get sucked out of a plane when there is a hole. In actual fact, they get blown out of the plane. In addition, the air pressure at high altitude is already lower than inside the plane, even without the plane moving through the air. So, I would say, the doors will open, and man they will just fly off the hinges!

I don't think you are taking into account the fact that the jet is moving at approximately 400mph.  Even on a windless day, the air moving past the plane and pressing on the door will keep the door from moving at all.  Unless the door is a sliding one or aided mechanically, theres no way a human would be able to open it.

[xvimbi]

I don't think you are taking into account the fact that the jet is moving at approximately 400mph.  Even on a windless day, the air moving past the plane and pressing on the door will keep the door from moving at all.  Unless the door is a sliding one or aided mechanically, theres no way a human would be able to open it.

Well, actually the faster the plane goes the faster the door will be flying off the hinges. The wind is not pressing against the door, it is sliding past it. It is the same principle that makes a plane fly in the first place: the air pressure is lower above the wings than below the wings, so the plane is moving upwards (Bernoulli's law). That's what happens with a car door, and I have envisioned a car-door-type door on that plane. Car doors open OUTWARD (see below for this important point). Before clarifying this, there is another experiment: hang a piece of curtain on an opening on the side of a plane. Which way will the curtain go? It will be pulled out of the plane and flap around on the outside.

Here is a little passage from some skydiving instructions:
"Wing Doors: These doors open out away from the fuselage and rest against the underside of the wing when open. They can be found on C-182s. This type of door is easy to unlock and open with the help of your pilot. Opening the door in straight and level flight can jerk the door from your hand, tend to pull you out the door or slam the door into the wing. This can easily be avoided by asking your pilot to put the plane in a slight left slip as you unlock the door. The slip relieves some of the air pressure on the door. The door should open about 6-10 inches. As the pilot returns to straight and level flight, the door will 'fly-up' under the wing by itself because of the aerodynamic forces. You don't have to push it up at all. The door won't slam into the wing as long as your pilot makes a smooth transition from the slip to straight and level flight."

And here is the important thing about modern planes:
"Obviously many people don't pay attention to the flight attendants or read the briefing cards, which explain in detail how to open the doors. You really should know how to do this. But in midflight, no, the doors will not open. That goes for both emergency exits and the main exits. They cannot open because of the outward-acting forces of the pressurized fuselage. The doors always open inward (usually inward, then outward or upward, but always inward first), and a person would not be capable of overcoming this force. So, in other words, the doors can't be opened until the aircraft is depressurized. You'll notice that on the sill of the main doors it often says DO NOT SIT. But in fact you could sit there all day, jiggling the handle to your heart's content; you aren't gonna get the door open. The reason they don't want you sitting there is to avoid messing with the inflatable escape slide that is contained inside the lower door structure. "

So, as long as the door open INWARD, one can't open it, because of the pressure differences that I alluded to (pressure is higher inside the plane, therefore pushing something into the plane will be difficult or impossible). If it opens OUTWARD, it'll be flying off the hinges!

[pianonut]

hmmm.  interesting.  sounds right. 

[allchopin]

Yes, but think of what happens once the door DOES fly open 'off its hinges' - a large metal rectangle perpendicular to the wind will not stay open like this.  Although I would assume there is a certain point (angle) where the suction of the low pressure is in balance with the pressure of the air pressing against it (due to the plane's movement).  Maybe this is the physics problem we should be solving instead,..

[xvimbi]

Yes, but think of what happens once the door DOES fly open 'off its hinges' - a large metal rectangle perpendicular to the wind will not stay open like this.  Although I would assume there is a certain point (angle) where the suction of the low pressure is in balance with the pressure of the air pressing against it (due to the plane's movement).  Maybe this is the physics problem we should be solving instead,..

Well, if it flies off the hinges, it is gone But even if it stays on and the hinges are on the side of the door that faces forward (i.e. in flight direction), the door will stay open. This is the same situation as with a car door. The reason is the same: the air pressure on the outside of the door is lower than behind the door. In fact the wider the door is open the greater the pressure difference, thus pulling the door open even more until it's about 90 degrees to the fuselage. Although you would think that air is pressing against the door and would close it, this is not the case. As you say, there will be an equilibrium. It's all about pressure differences. Objects often move in unexpected directions because of them. And we haven't even considered eddies yet that form behind the door once it is open.

Here is another example along the same lines: A helium-filled ballon is tied to a car seat such that it doesn't touch anything inside the car. Now, start the engine and accelerate the car. Which direction (if any) does the ballon move?

[asyncopated]

Mikeyg --

Thanks for the cookies. At least i got that one right! 

Xvimbi --
Oh dear,  I suspect that you are right. I got my signs/direction wrong. 

Difference in for Force (out – in) = (Pressure outside - Pressure in cabin) * Area + ½*density * speed^2 *Area.

The direction of the force is pointing outwards not inwards, means that if the door were designed simply it would be easy, not difficult to open the door.  Opps. 

I'm not sure how a plane door is designed but i think xvimbi gives an accurate picture. 

my dad is an engineer. we never got along because whenever i'd ask for math help, he'd explain it once and i would listen and then start asking questions. then he'd explain it again - this time talking louder. then, the last time he'd be yelling. and, of course, to end the whole thing - he'd say something like - 'you wear your feelings on your sleeve.'
...

i always do pray for a safe take-off and landing. somehow, i find lightening interesting. we had a bolt strike while flying and it was fairly nearby. having done my utmost to not perterb the pilot (turning my cell phone off) -- i suddenly wondered what a cell phone would do compared to a strike of lightening? oh, well.

I don't think physics or maths is difficult.  Not anymore difficult than learning to play the piano or composing!  Sure the skills required are different.  I think the biggest stumbling block is the excuse, "but I don't have the talent for it".  Personally I think it's a load of bull.  The way i see it, there are three keys to learning (almost) anything and talent isn't one of them.  You just need to keep an open mind, understand positive/negative reinforcement and seek proper/good guidance.  I also think that the problems you have with your dad mostly about communication and certainly not a reflection on your abilities in science.

To answer your question about lightning, most airlines fly at an altitude of around 33000  feet, way above storm clouds.   The only time when there is opportunity for lighting to strike is really during take-off and landing, so the window is relatively small. 

If lightning does strike the plane, the consensus is that you will not be killed by electricity.   This is because of gauss's law -- a consequence of which is that any surplus change on a closed metal container will reside on the outside.  All the charges repel one another and move to the outside of the container.  So in a lightning storm, the safest place to stay, if you are on open terrain is in your car.  However, in a plane I think it will probably mess up the engines pretty badly.  I'm not sure what would happen.  I'm not sure if there are precautions taken to prevent this, but I suspect there are.  There are "lightning conductors" on tall buildings, especially here in singapore.  We have very bad tropical storms every other week.  This does not actually conduct away the electricity from the lightning after it has struck.  Instead, it works by neutralizing  charge build up, so that lightning does not strike were it is not suppose to.  The potential difference is  lower at places where you neutralize charges, and lightning will always follow the path with the highest potential difference (the easiest path).

What I can assure you is that planes are designed with ridiculous safety standards.  Usually the safety threshold for building or bridges, is about 3 to 4 times.  Engineers use materials/designs that can withstand about 3 to 4 times the amount of strain caused by regular usage or environmental effects like wind.  The typical safety threshold for an airplane is 7 times (I was told this by an aeronautical engineer).  Considering that all planes are custom built -- each component is build and fitted individually and that the designers of planes essentially try to cover every single eventuality for what is essentially a multi-million dollar object,  I would say you are in pretty good hands. 

If you still don't trust that, I can add that in most countries pilots are probably the most over trained people in the world and that actuaries cover a metallic flying object, loaded with highly volatile fuel very conservatively.


al.

P.s. Try xvimbi's last question. The answer is cool! 
   

[Torp]

Here is another example along the same lines: A helium-filled ballon is tied to a car seat such that it doesn't touch anything inside the car. Now, start the engine and accelerate the car. Which direction (if any) does the ballon move?

Total conjecture on my part (aka swag).  But, I think it would go forward.  The air pressure would be less at the front of the balloon thus making it move forward.  I wonder though, would this effect be offset by the magnitude of the airwaves coming out of my stereo at what are, reportedly, excessive volumes? 

[asyncopated]

Well, if it flies off the hinges, it is gone But even if it stays on and the hinges are on the side of the door that faces forward (i.e. in flight direction), the door will stay open. This is the same situation as with a car door. The reason is the same: the air pressure on the outside of the door is lower than behind the door. In fact the wider the door is open the greater the pressure difference, thus pulling the door open even more until it's about 90 degrees to the fuselage. Although you would think that air is pressing against the door and would close it, this is not the case. As you say, there will be an equilibrium. It's all about pressure differences. Objects often move in unexpected directions because of them. And we haven't even considered eddies yet that form behind the door once it is open.

I actually disagree with this.  I think that if your car doors are fully open, and you are travellinq very fast they will close because of the air pressure against the door.  The doors now act as a foil much like that on a race car.  The foil on the race car acts to push the car downwards when it is moving fast.  This is so that it still grips the road and does not slip at fast speeds.  Breaks/flaps on the plane act in the same way.  There is an enormous amount of force on the breaks when they are up at high speeds, which is used to slow the plane down or to steer it.

This situation is different to one in the question that follows.  In that, the car doors are closed. The air molecules tend to want to stay in the same place, due to inertial.  So as the car accelerates there is an accumulation of air molecules at the back of the car.


There might only be one way to solve this conundrum -- who has a car to spare?

al.

[pianonut]

ok.  if there's an accumulation of air molecules at the back of the car, and helium is lighter than air - the balloon gravitates toward the front (torp and i are on agreement on this) slowly along the celing and ending up at the back of your head electrically charging itself on your hair. (or is this the air filled balloon that attaches to your hair?)  guess the helium is always the one that as soon as you open the car door - zip, off it goes.

my son, being the curious one, always let go of his helium balloons.

ps  thanks for the encouragment!  i've found lately that i can do a lot more things than i thought possible.  still not ready to become a pilot, but for those that are interested, anchorage university has teamed up with the airport and has a world class whatdoyoucallit? practice plane that lets you practice in all kinds of weather, looking at different terrain, etc. landing at merrill field.

 

[xvimbi]

I actually disagree with this.  I think that if your car doors are fully open, and you are travellinq very fast they will close because of the air pressure against the door.  This is becuase the ambient pressures in and outside the car is in equilibrium.   The doors now act as a foil much like that on a race car.  The foil on the race car acts to push the car downwards when it is moving fast.  This is so that it still grips the road and does not slip at light speeds.  Breaks/flaps on the plane act in the same way.  There is an enormous amount of force on the breaks when they are up at high speeds, which is used to slow the plane down or to steer it.

I am not so sure about this. Flaps, airfoils and spoilers are primarily designed to give stability and some traction. In the case of race cars, the front spoilers are designed to lift the car (i.e. they have a slight convex surface at the top, the bottom surface is flat), whereas the rear spoiler is turned upside down to create downward force. The car itself is shaped such that it experiences lift when it is driving. The speed of a plane is probably mostly determined by its engines, not so much by the flaps.

In any case, the situation with the door is different. Air cannot flow freely around it, because it is fixed at the front to the fuselage (if we assume a car-like door). There is only air flowing on its outside, the air on its inside (when the door is open) is not moving (in the absence of eddies). Air that is flowing around a body results in a lower pressure compared to air that is stationary. Therefore, the door will move outwards to a point where there are simply too many "air molecules" bumbing right into the door. Those molecules bumping into the door determine the pressure that is acting on the door (if there are no molecules bumping into it, there is no pressure acting on it). So, there will be an equilibrium position for the door, which I guess is much closer to 90 degrees than to zero degrees.

[allchopin]

Seems hard to believe that the speed of the air leaving the plane would be more than the wind speed on the door - you'll have to try it sometime and let us know how it goes

[xvimbi]

Seems hard to believe that the speed of the air leaving the plane would be more than the wind speed on the door

No, the speed of the wind leaving the plane is not higher than the speed of the wind at the door, but that is irrelevant. It is the pressure that is the important aspect. The pressure is higher inside the plane than outside (although there is wind flowing by at 400 mph; actually, it's BECAUSE there is wind flowing by at 400 mph), therefore objects will move from the inside of the plane to the outside.

Here is a different way to look at it: The pressure acting on a plane's fuselage depends on the number of "air molecules" hitting the fuselage, their kinetic energy and the angle at which they are hitting the fuselage (it's admittedly a somewhat simplistic and mechanical view, but that should suffice). Let's consider three scenarios:
1. the plane is stationary, and air is streaming at the fuselage at an angle of 90 degrees, that is straight at it, with 400 mph. This will result in a certain pressure acting on the fuselage.
2. now, lets move the plane forward, while not disturbing the air. The air will now effectively be hitting the fuselage at an angle, which will reduce the resulting pressure. This is because the momentum transfer from the air molecules to the fuselage (and therefore the pressure) is highest at 90 degrees and falls off with any angle not 90 degrees.
3. Let's now have the air streaming by the plane, without any air molecule hitting the fuselage. The resulting pressure on the fuselage will be zero. For what the fuselage is concerned, it is in complete vacuum. In reality, this doesn't really happen, of course. There will always be some air molecules that bump into the fuselage and will create pressure, but it will be very low compared to scenario 1.

Let's assume, that the pressure inside the plane is 1 atmosphere. As soon as there is a hole in the fuselage, we will have a severe pressure gradient from 1 atmosphere inside the plane to well below 1 atmosphere outside the plane. A pressure gradient generates force that acts on any object within that gradient, such that the object moves from a point of higher pressure towards the point of lowest pressure (that is why there is wind and moving clouds). If that force is large enough, it will even move entire people, and that's exactly what happens when you have a hole in a plane at 30,000 ft.

Even if the plane stood still at 30,000 ft, there would be a great pressure difference, because the air pressure decreases with the distance from the surface of the earth. This difference is not great enough, though, to move people.

Here is a little story to show you just how large the forces are that are generated by pressure gradients: I was once climbing Mt. Rainier (14410', 4392m). At the top, I drank from my sturdy Nalgene plastic bottle and closed the lid because it was empty. When I came down again and unpacked by backpack, I found the Nalgene bottle completely crushed. That was simply because the pressure of the air when I closed the bottle was much lower than the pressure at the bottom of the mountain. The force resulting from that pressure difference was great enough to completely squoosh the bottle, in fact it was far greater than the force anyone in our climbing party could generate, as we tried to crush another bottle with our hands, but failed.

you'll have to try it sometime and let us know how it goes

I have done that many times. Next time you ride in a car, open the door and try to close it again. Do that first at 20 mph. If you are brave enough, do it at 60 mph, but make sure you are buckled up

[allchopin]

The car scenario is a bit different than the plane because there is no cabin pressure to deal with.  The car drives at ground level, so the air pressure in and out of the body will be equal.  And so if the car door were open at an 89 degree angle, it there would not be substantial force behind it to keep it open at such an angle (depending on how fast you go).  I am getting up the nerve to try this at 60mph.

[xvimbi]

The car scenario is a bit different than the plane because there is no cabin pressure to deal with.  The car drives at ground level, so the air pressure in and out of the body will be equal.

Not at all. This is what I am trying to say all the time. The pressure around a moving object will be different compared to a stationary object, because there are fewer "air molecules" hitting the object.

Have you ever noticed the following: you are waiting in a left-turn lane, while traffic is zooming by next to you. Whenever there is a car driving by right next to you, there is a pull on both cars. The same situation happens when there are two trains passing each other at high speed. The two trains get pulled together somewhat.

And so if the car door were open at an 89 degree angle, it there would not be substantial force behind it to keep it open at such an angle (depending on how fast you go).  I am getting up the nerve to try this at 60mph.

I would try it first at considerably lower speeds...