# Buoyant force of air

#### Tokspor

A helium-filled balloon (p = 0.5 kg/m^3) is released into the air. The balloon has a diameter of 1 m. Neglecting air resistance, what will be the balloon's initial upward acceleration when released from rest? (density of air is 1.2 kg/m^3)

A. 3.65 m/s^2
B. 5.34 m/s^2
C. 13.73 m/s^2
D. 19.45 m/s^2

The correct answer is C. 13.73 m/s^2.

I tried to solve it originally by starting with...
1. F = ma.
2. Plugging in F(buoyant) = m*a
3. F(buoyant) = (density of balloon)*(volume of balloon)*a
4. (density of air)*(volume of balloon)*g = (density of balloon)*(volume of balloon)*a

Then I solved for a. But this is wrong?

According to the text, you are supposed to solve this by finding F(buoyant). Then find F(gravity). Then, F(buoyant) - F(gravity) = F(remaining).

Then, solve for a in F(remaining) = m(balloon)*a.

How does this make sense? F(remaining) is just a value indicating how much F(buoyant) exceeds F(gravity).

#### RogueUnicorn

##### rawr.
7+ Year Member
A helium-filled balloon (p = 0.5 kg/m^3) is released into the air. The balloon has a diameter of 1 m. Neglecting air resistance, what will be the balloon's initial upward acceleration when released from rest? (density of air is 1.2 kg/m^3)

A. 3.65 m/s^2
B. 5.34 m/s^2
C. 13.73 m/s^2
D. 19.45 m/s^2

The correct answer is C. 13.73 m/s^2.

I tried to solve it originally by starting with...
1. F = ma.
2. Plugging in F(buoyant) = m*a
3. F(buoyant) = (density of balloon)*(volume of balloon)*a
4. (density of air)*(volume of balloon)*g = (density of balloon)*(volume of balloon)*a

Then I solved for a. But this is wrong?

According to the text, you are supposed to solve this by finding F(buoyant). Then find F(gravity). Then, F(buoyant) - F(gravity) = F(remaining).

Then, solve for a in F(remaining) = m(balloon)*a.

How does this make sense? F(remaining) is just a value indicating how much F(buoyant) exceeds F(gravity).
which would generate the NET upwards force. draw a force diagram

#### Geekchick921

##### Achievement Unlocked: MD
Moderator Emeritus
10+ Year Member
Imagine this as a free body diagram, we have the weight of the balloon, about 20N, as a vector pointing down, and the buoyant force, about 48N, as a vector pointing up. The different (28N) will be the net force acting on the balloon, and since the balloon has a mass of about 2kg, the upwards acceleration is about 14 m/s^2.

#### PhilIvey

10+ Year Member
5+ Year Member
A helium-filled balloon (p = 0.5 kg/m^3) is released into the air. The balloon has a diameter of 1 m. Neglecting air resistance, what will be the balloon's initial upward acceleration when released from rest? (density of air is 1.2 kg/m^3)

A. 3.65 m/s^2
B. 5.34 m/s^2
C. 13.73 m/s^2
D. 19.45 m/s^2

The correct answer is C. 13.73 m/s^2.

I tried to solve it originally by starting with...
1. F = ma.
2. Plugging in F(buoyant) = m*a
3. F(buoyant) = (density of balloon)*(volume of balloon)*a
4. (density of air)*(volume of balloon)*g = (density of balloon)*(volume of balloon)*a

Then I solved for a. But this is wrong?

According to the text, you are supposed to solve this by finding F(buoyant). Then find F(gravity). Then, F(buoyant) - F(gravity) = F(remaining).

Then, solve for a in F(remaining) = m(balloon)*a.

How does this make sense? F(remaining) is just a value indicating how much F(buoyant) exceeds F(gravity).
Minimal math is involved. Fb=Pair(V)(g) Mg=Pballoon(V)(g)
So since everything is the same except for density, you have the following:
1.2K-.5K=.7K K=(V*g)
.7*V*g=.5(V) Net FB=ma m=Pballon(Volume(
7/.5=A = 70/5=14.

#### Tokspor

Thanks for the responses. I got it now.