Calculating Motion of a Test Balloon on a Methane Planet

[robax25]

total Buoyant force= m(Ballon)g-m(Box)g=15kg*0.425m/s² - 4kg*0.425m/s²=4.675 N

 

[haruspex]

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.

 

[robax25]

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

 

[haruspex]

Buoyant force on the immersed box is 0.17 N

No. How are you calculating that?
Please state Archimedes' Principle.

 

[robax25]

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

 

[haruspex]

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]

Buoyant force for BallonFb=170N

 

[haruspex]

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.

 

[robax25]

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.

 

[haruspex]

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?

 

[robax25]

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

 

[haruspex]

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]

how to proceed now?I need to draw st, vt, at graph?

 

[haruspex]

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.

 

[robax25]

which kind of equation? can you give me some hints please.

 

[robax25]

should I use terminal velocity equation?

 

[robax25]

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

 

[haruspex]

-40.97N/4kg

How much mass is accelerating?

 

[robax25]

Fnet = mg-Fb= 19kg*0.425 m/s² - (0.17N+42.5N)= -34.595N a=-34.595N/19kg=-1.8 m/s²

 

[haruspex]

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]

yes but it is not a matter

 

[robax25]

Now i need to find out the time to fall

 

[haruspex]

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.

 

[robax25]

yes

 

[robax25]

y=0.5 at² = 0.5* 1.8m/s²*t²
t²= 10m/(0.5*1.8m/s²)=11.11s²
t=3.33s.

 

[haruspex]

y=0.5 at² = 0.5* 1.8m/s²*t²
t²= 10m/(0.5*1.8m/s²)=11.11s²
t=3.33s.

Looks right

 

[robax25]

the graph would be like that but I need to change the value.

 

[haruspex]

Yes.

the graph would be like that but I need to change the value.