Interesting question... - page 8

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Choomazik >> :

Balloons go up in a parade. And at a kid's birthday party, too. (If they're filled with helium, of course, not parental fumes.)

They're breaking in.

You got me confused.

and I still think the balloon's going backwards in the carriage.

>> WHERE IS ANDREW?

 

I suggest you delete all the rambling posts.

Don't you have an insistent desire to stop fluttering? It was a normal topic, Lecha raised an actual question. And now you've already dragged contraception into it.

Let's zadachki in another thread. Maybe even in another forum.

 
sayfuji >> :

I suggest you delete all the rambling posts.

Don't you have an insistent desire to stop fluttering? It was a normal topic, Leha raised an actual question. And now you've already dragged contraception into it.

Let's zadachki in another thread. Maybe even in another forum.

Well - not too bad - the general hospital temperature is clear, plus while I was getting home and my wife's small eee pc found DDFedor. Three people interested. Progress. :) We can discuss further in private.

 
About the balloon: the problem statement says that it dangles. That's where it dangles, that's where it's going to go. Fucking physicists :) If the balloon hangs in the air, it has no positive buoyancy and will not fly forward. If the balloon hits the ceiling, it will not fly, but will roll. And there's no telling where.
 

In a problem of this type, it is necessary to carefully write down all the forces acting on the ball. There are not many of them, see fig:

The Archimedean force is directed against the force of gravity and the resulting force is fully compensated by the resistance of the ceiling. Therefore the ball is stationary in the vertical axis.

In the horizontal plane two oppositely directed forces act on the ball - the inertial force F=ma and the Archimedean force arising from the density gradient of the medium along the direction of the carriage. These forces are uncompensated and will cause the ball to roll along the ceiling towards the larger force. This uncompensated part is equal in absolute value to the product of the acceleration of the car by the difference V(air density-helium density)-m0, where V is the volume of the balloon, m0 is the mass of its shell. In order to understand the sign of this expression and consequently the direction of motion of the ball, it is sufficient to look at the expression for the Archimedean force in the vertical plane. They coincide exactly! And the ball, if it were not for the ceiling, would move against the direction of acceleration of free fall (see fig.). So, in the case of the train moving from right to left, the acceleration of the car will be in the opposite direction (when the car takes off, you are thrown back), and the horizontal Archimedean force will be directed by the movement of the car, i.e. from right to left, that is where the ball will roll.

The ball will roll across the ceiling in the direction of the train!

 

Neutron, wouldn't the support reaction force N

wouldn't it work?

 
How could it not? - It will! That's what I said above - "...and the resulting force is fully compensated for by the resistance of the ceiling."
 
Neutron >> :

The balloon will roll across the ceiling in the direction of the train!

Exactly! The explanation is exhaustive. Now we can talk about volatility.

Sorry to have distracted the public :) .

 

Something's not right here.

For example, why is the acceleration of the WAGON directed in the opposite direction to the movement of the wagon?

 
Newton's third law, F=-F the force of action, is equal to the force of counteraction with opposite sign! Therefore, the force acting on the body in the car is in the opposite direction from the force applied to the car (directed by U) and accelerating it. On the other hand, according to Newton's second law F=ma, the direction of acceleration a always coincides with the direction of force F.
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