Calculating Motion of a Test Balloon on a Methane Planet

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how can I get motion of the box?
 
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robax25 said:
how can I get motion of the box?
The starting state is that the box is 10m down in the methane and the balloon is at (on) the surface. You can suppose the balloon is not immersed at all in the methane, since it will not be for long.
What are the forces on the balloon? What are the forces on the box?
(You may be tempted to consider the balloon, rope and box as a single system and only consider the external forces on that, but that is not safe. Do you see why?)
 
the box has force F(box)=mg and balloon is full of Helium. It goes up.The atmospheric density changes the motion of the box.I need to use Archimedes formula
 
I can use this formula to get Drag force=Cd ρ v² *A/2 and we don't know the value of Cd, we can skip it? does It ok?
 
robax25 said:
need to use Archimedes formula
Quite so. What is the total buoyant force?
robax25 said:
I can use this formula to get Drag force=Cd ρ v² *A/2 and we don't know the value of Cd, we can skip it? does It ok?
I wouldn't worry about drag. There is nowhere near enough information to quantify that.
 
total Buoyant force= m(Ballon)g-m(Box)g=15kg*0.425m/s² - 4kg*0.425m/s²=4.675 N
 
robax25 said:
total Buoyant force= m(Ballon)g-m(Box)g=15kg*0.425m/s² - 4kg*0.425m/s²=4.675 N
The buoyant force has nothing to do with the masses of the objects.
Quote Archimedes' principle.
 
The box is immersed in methane liquid . So Buoyant force on the immersed box is 0.17 N and Weight of the box is = 1.7 N
 
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robax25 said:
Buoyant force on the immersed box is 0.17 N
No. How are you calculating that?
Please state Archimedes' Principle.
 
I uploaded the pdf file. Archimedes principle state that the upward force acts on an object is equal to weight of the fluid that the body displaces
 
robax25 said:
I uploaded the pdf file. Archimedes principle state that the upward force acts on an object is equal to weight of the fluid that the body displaces
Sorry, my mistake. (Earlier you had a term that equated to 1.7N, and did not notice you were now writing 0.17N.) But you could have got there a bit more simply using Archimedes' principle.

So now, what is the buoyant force on the balloon? You will have to use Archimedes for that.
 
robax25 said:
Buoyant force for BallonFb=170N
The balloon is not immersed in the liquid methane. Take it as sitting on or a bit above the surface.
 
but Buoyant force does not depend on either it is submerged or immersed. It will be same if object is submerged or immersed. Buoyant force is telling you either object will float or not. In this case, Buoyant force is greater than the Force of gravity of the Ballon.
 
robax25 said:
but Buoyant force does not depend on either it is submerged or immersed. It will be same if object is submerged or immersed. Buoyant force is telling you either object will float or not. In this case, Buoyant force is greater than the Force of gravity of the Ballon.
Whether you want to call it immersed or submerged, the balloon is not in the liquid, so does not get any buoyant force from the liquid. What is exerting a buoyant force on the balloon, and what is its density?
 
I get it now. Buoyancy force is exerted by atmosphere of the planet which has density 100kg/m³. so Fb= 100kg*0.425 m/s² =42.5 N
 
robax25 said:
I get it now. Buoyancy force is exerted by atmosphere of the planet which has density 100kg/m³. so Fb= 100kg*0.425 m/s² =42.5 N
Yes.
 
robax25 said:
how to proceed now?I need to draw st, vt, at graph?
Yes, but first you need to find the actual equation for the motion.
 
which kind of equation? can you give me some hints please.
 
I do not consider Terminal velocity. So total net force Fnet = mg-Fb= 4kg*0.425 m/s² - (0.17N+42.5N)= -40.97N so ma= -40.97N; a= -40.97N/4kg= -10.24 m/s²
 
Fnet = mg-Fb= 19kg*0.425 m/s² - (0.17N+42.5N)= -34.595N a=-34.595N/19kg=-1.8 m/s²
 
robax25 said:
Fnet = mg-Fb= 19kg*0.425 m/s² - (0.17N+42.5N)= -34.595N a=-34.595N/19kg=-1.8 m/s²
Looks better.
Why the minus sign? Are you taking down as positive?
 
robax25 said:
Now i need to find out the time to fall
Fall? Do you mean time for the box to reach the surface?
Since you are ignoring drag, you can use the relevant equation you posted in post #1.
 
y=0.5 at² = 0.5* 1.8m/s²*t²
t²= 10m/(0.5*1.8m/s²)=11.11s²
t=3.33s.
 

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