Solving the Equation: Finding h in a He-filled Balloon

[Touchme]

Homework Statement

A helium-filled balloon at atmospheric pressure is tied to a 3.1 m long, 0.100 kg string. The balloon is spherical with a radius of 0.40 m. When released, it lifts a length (h) of the string and then remains in equilibrium as in Figure P9.78. Determine the value of h. When deflated, the balloon has a mass of 0.25 kg. (Hint: Only that part of the string above the floor contributes to the load being held up by the balloon.)

Homework Equations

P2 = P1 + dg(y1 - y2)
d = density

The Attempt at a Solution

First I determine the pressure of the balloon by using P = mg/A, and I got 4.874 Pa
Then I use the equation P2 = P1 + dg(h)
4.874 = (0) + (1.29)(9.8)(h). The 1.29 is density of air.
I have no idea why I set p1 equal 0.
The answer I got is wrong and I am stuck. Help please.

 

[berkeman]

I don't think the pressure is relevant, but I could be wrong.

How much does a sphere of air weigh when r=0.4m?
How much does a sphere of helium weigh at the same radius?

After adding in the weight of the fabric of the balloon, what is the net lift of the balloon?

 

[Touchme]

density of air is 1.29 and density of He is 0.179

mass of air = 2161.4
mass of He = 299.9

299.9 + 0.25 = 300.15 the mass of the balloon + He

Im still not sure what to do here. How is the net lift relative to finding h? (I thought it was a pressure problem because it came from a chapter with Bernoulli equation and fluid flow).

 

[berkeman]

It's good to get in the habit of including units in all of your equations and work (and check to see that the units agree all along through your calcs).

The lift of the balloon is what lifts some length of the rope up. It can only lift up the amount of rope that has a weight equal to the net lift of the balloon, right? You are given enough info to calculate the linear density of the rope...

 

[Touchme]

ooo thanks for your help, I managed to solve it using B=pVg. I found all the forces acting on the balloon and set that equal to the weight of air. Thanks