Why Does Smoke Settle at the Bottom in a Smoke Waterfall Experiment?

AI Thread Summary
The discussion centers on a physics presentation involving a Smoke Waterfall experiment in an airtight container. The phenomenon occurs because the smoke, being cooler and denser than the surrounding air, settles at the bottom instead of rising. The burning cigarette consumes oxygen, creating a low-pressure area that allows smoke to escape from the container's exit. Participants question the physics understanding behind the demonstration and suggest consulting a friend for insights. The inquiry highlights curiosity about the physical principles at play in this visually striking experiment.
MXCD
Messages
3
Reaction score
0

Homework Statement



I have a presentation on our physics class and I have decided to demonstrate the phenomenon of Smoke Waterfall inside an airtight container. I saw my friend do this tons of times. It's really amazing but how can I explain the physics behind this trick? In general knowledge hot air is supposed to go up right? why is it the smoke coming out of the cigarette butt settle at the bottom? Help pls! :smile:

here's a video.
Smoke Waterfall

Homework Equations



-

The Attempt at a Solution



-
 
Physics news on Phys.org
How much air is there in the container to begin with? Where does that air go when the smoke comes in? If you really don't know the physics why would you choose it as a presentation in a class?

Have you considered asking your friend?
 
HallsofIvy said:
How much air is there in the container to begin with?
*Default value? just air...in an empty container...with a hole on the side where you can insert a rolled paper to burn.

Where does that air go when the smoke comes in?
*this is how i understand this: the lit end needs oxygen to burn. I don't know why the other end(w/c is inside the container) serves as an exit for the smoke from the lit end. maybe the lit end i blocked because it's burning that's why the smoke comes out of the other end? and to answer your question, i guess it goes the same with the lit end w/c is outside of the container. fire sucks the air inside the container out.

If you really don't know the physics why would you choose it as a presentation in a class?
Have you considered asking your friend?
*Yes, I have. They agreed with my proposal so (this is a silly reason) we could have a reason to smoke inside the classroom.

I'd like to know what physical principal laws were shown in my proposed demonstration.
And I guess out of curiosity too, I just want to know how this works.

thanks btw. your question made me think :smile:
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top