Buoyant force acting on an inverted glass in water

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erobz said:
Good... plug that in, solve for ##\delta(h)##. I would suggest evaluating that solution at ##h = 0## to ensure you have chosen the correct form.
Ok I will but why should I find this ?
 
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MatinSAR said:
Ok I will but why should I find this ?
Because you to eliminate ##\delta## as an independent variable and get everything in terms of the depth ##h##.
 
erobz said:
Because you to eliminate ##\delta## as an independent variable and get everything in terms of the depth ##h##.
If we consider h=0 :
1670960911717.png
 
MatinSAR said:
If we consider h=0 :
View attachment 318757
Thats isn't what I meant.

Did you solve for ##\delta## as a function of ##h## yet?

Remember, ##\delta## is part of both the pressure and the volume of the gas. When I say solve for ##\delta## I mean solve the following for ##\delta##:

$$P_{atm} A l = P(h,\delta) V\llap{-}(\delta)$$
 
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erobz said:
Thats isn't what I meant.

Did you solve for ##\delta## as a function of ##h## yet?

Remember, ##\delta## is part of both the pressure and the volume of the gas. When I say solve for ##\delta## I mean solve the following for ##\delta##:

$$P_{atm} A l = P(h,\delta) V\llap{-}(\delta)$$
Can I find δ using :

PatmL/(L -δ) = Patm+ρg(h-δ)
Frabjous said:
Think about what h=0 physically corresponds to.
The moment that the glass enters the water.
 
MatinSAR said:
Can I find δ using :

PatmL/(L -δ) = Patm+ρg(h-δ)
Give it a try.
 
erobz said:
Give it a try.
1670966257808.png

At h=0 delta should be zero. The above equation confirms this.
And we can see that if h goes to ∞ the delta goes to L.
 
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MatinSAR said:
View attachment 318764
At h=0 delta should be zero. The above equation confirms this.
And we can see that if h goes to ∞ the delta goes to L.
Seems good. I had to check whether that quadratic simplified, I didn't take it that far! Good Job!
 
erobz said:
Seems good. I had to check whether that quadratic simplified, I didn't take it that far! Good Job!
I appreciate your help. I didn't use PatmL/(L -δ) = Patm+ρg(h-δ) because
2nd degree equation was obtained ...

I have used the formula you suggested ...
 
MatinSAR said:
I appreciate your help. I didn't use PatmL/(L -δ) = Patm+ρg(h-δ) because
2nd degree equation was obtained ...

I have used the formula you suggested ...
What formula?
 
erobz said:
I don't understand! I solved it as a quadratic. However, your result indicates that my quadratic must be factorable I believe?
Maybe I made a mistake ...
was my answer correct ??
 
MatinSAR said:
Maybe I made a mistake ...
was my answer correct ??
It plots identically to my solution?
 
erobz said:
It plots identically to my solution?
Let me send you a picture of what I have done ...
 
erobz said:
It plots identically to my solution?
My mistake is here :
1670968816408.png


So to find delta we have to solve a quadratic equation ...

The question only asked about the increase and decrease of F force, I didn't want to waste your time at all... I'm sorry...
edit :

2022_12_14 1_55 AM Office Lens.jpg


I have solved that quadratic equation ...
 
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MatinSAR said:
My mistake is here :
View attachment 318768

So to find delta we have to solve a quadratic equation ...

The question only asked about the increase and decrease of F force, I didn't want to waste your time at all... I'm sorry...
edit :

View attachment 318770

I have solved that quadratic equation ...
What you did on accident the first time by omitting ##\delta## was actually a really good approximation for ##\delta \ll h##. For the parameters I used I couldn’t tell the apart.
 
MatinSAR said:
My mistake is here :
View attachment 318768

So to find delta we have to solve a quadratic equation ...

The question only asked about the increase and decrease of F force, I didn't want to waste your time at all... I'm sorry...
edit :

View attachment 318770

I have solved that quadratic equation ...
I wanted you to explore it. It wasn’t a waste of my time at all!
 
erobz said:
I wanted you to explore it. It wasn’t a waste of my time at all!
Thank you for your help and time ... 🙏🙏