# Homework Help: Buoyancy and Height

1. Dec 1, 2009

### bolivartech

1. The problem statement, all variables and given/known data

A helium filled balloon is tied to a 2.00-m-long, 0.050 0-kg uniform string. The balloon is spherical with a radius of 0.400 m. When released, it lifts a length h of string and then remains in equilibrium as shown in figure P14.51. Determine the value of h. The envelope of the balloon has a mass of 0.250 kg.

2. Relevant equations

Code (Text):

ρ=  m⁄v

v=  4/3 pi r^3

B=ρ[SUB]air[/SUB] * g * V[SUB]obj[/SUB]

3. The attempt at a solution

Code (Text):

B=(1.29 kg⁄m^3 )(9.8 m⁄s^2 )(.26808 m^3)

B=3.39 N

m_he=.047986

F_g=Mg

F_g=(.347986 kg)(9.8 m⁄s^2 )

F_g=3.41 N

This is something the prof. said to start with, I'm not sure how this helps aside from proving it has pretty much reached equillibrium. I really just don't know what formula that involves height will work.

2. Dec 1, 2009

### mgb_phys

Think height in terms of weight of string not altitude

3. Dec 1, 2009

### bolivartech

Ok so here is the thought process I am having. Find the change in force, use F=ma to find how much mass is lifted (string). Use that to determine how much in h is lifted.

(
Code (Text):
mass total / height total = mass lifted / height lifted)
So do I use the small difference in force between my normal force and the buoyent force? .02N I know it doesn't move a significant amount.

4. Dec 2, 2009

### bolivartech

I haven't received a response but from what I can tell that is right. I did get the same answer as the prof, now I just have to find out if it was luck or if the process was correct. Thanks for the help!

5. Dec 2, 2009

### mgb_phys

At equilibrium bouyancy = weight = mass of length of string h + mass of envelope.

ps. remember the people here are in different time zones