A helium-filled balloon at atmospheric pressure is tied to a 2.7 m long..?

Question

A helium-filled balloon at atmospheric pressure is tied to a 2.7 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.)

Answer 1

Total weight acting downward:
Balloon = 0.25 kg
Helium = 4/3 π (0.4)^3 * (0.1786)
String = h*0.1/2.7
Total weight downward = 0.2979 + 0.0370h

Upthrust = 4/3 π (0.4)^3 * (1.2)
= 0.3217

Equate the two and get h = 0.643 m

Answer 2

A helium filled balloon wit r=0.4 m will lift 205 grams. If the balloon weighs 0.25 kilo it means there is no lift to get the piece of string of the ground. So h=0.

Answer 3

There are a lot of things to figure out here. But what it boils down to is this: If it "remains in equlibrium," the upward forces exactly counterbalance the downward forces.

The downward forces are:
1. Weight of the balloon material ( = (0.25kg)(g))
2. Weight of the helium (= (mass of helium)(g) = (density of helium)(volume of balloon)(g))
3. Weight of length "h" of string (note that every 2.7meters of string has a mass of 0.1kg. Therefore, the weight of length "h" is: (mass of length "h")(g) = (h × 0.1kg/2.7m)(g))

The only upward force is:
1. Buoyancy (= weight of displaced air = (mass of displaced air)(g) = (density of air)(volume of balloon)(g))

So, some numbers you need to get (probably from a table in your book) are:

density of helium (at atmospheric pressure);
density of air (at atmospheric pressure).

Then you need to plug in all the numbers, and write an equation that sets the upward force equal to the sum of the downward forces. From that equation you should be able to solve for "h".