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.
How so? I already solved them in physics class this afternoon. Well, actually my professor did.
In hell we party, in heaven you pray. What's more fun?
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?
Am I in the ball park?
... in heaven you pray. What's more fun?
Man, it's been a long time since I've done physics...
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?
and , how come they haven't changed much in 20 years?
No
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.
awwww...... I thought this would be fun physics; not highschool simple formula junk -.-
Ok, here is the proper answer. Assuming the pigs don't fly,Weight of pig = m gNormal force = m g cos(theta)Force along the frictionless slide is given by m g sin(theta)With friction it ism 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 isa_{nf} = g sin(theta)with friction, it isa_{f} = g (sin(theta) – mu cos(theta)),where mu is the coefficient of frictionFormula for constant acceleration, by integratingv = a tand distancex = ½ a t^2The slides have the same length2 t_{nf}= t_{f} Equating the lengths½ a_{nf} t_{nf}^2 = ½ a_{f} t_{f}^2substituting the relavant formulae we get¼ a_{nf} t_{f}^2 = a_{f} t_{f}^2Substituting the accelerations,¼ sin(theta)=sin(theta)-mu cos(theta).The coefficient of friction is thereforemu = 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.
Why don't I get a sexy cookie???
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!
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.
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,..
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.
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?
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. 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.
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
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.