Buoyant Force and Archimede's principle

In summary, for a balloon with a radius of 3.15m and a total mass of 15.2kg filled with helium, the buoyant force acting on the balloon is 149.112 N. When the balloon is released from the ground, the acceleration of the balloon can be calculated using the equation a = (Fb - Wb) / m, where m is the mass of the balloon. The density of the air plays a role in calculating the buoyant force using Archimede's principle. The tension in the rope holding the balloon down must also be taken into account in the free body diagram.
  • #1
Ering
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Homework Statement


A balloon used for a physics experiment has a radius of 3.15m and is filled with helium. The total mass of the balloon is 15.2kg and the density of the surrounding air is 1.28 kg/m3.
a. How much buoyant force is acting on the balloon?
b. When the balloon is released from the ground, what is the acceleration of the balloon?
c. As the balloon rises, what happens to the radius of the balloon? Explain.

Homework Equations


FB = WDis (buoyant force equals weight of the fluid displaced)
Archimede's principle = an immersed object is buoyed up by a force equal to the weight of the fluid it displaces

P = F/A (pressure equals force divided by area)

P2 = P1 + ρgh
P2 = pressure at top
P1 = pressure at bottom
ρ = density
g = gravity
h = depth/height

ρ = m/v (density equals mass divided by volume)

P1 + 1/2 ρV12 + ρgy1 = P2 + 1/2 ρV22 + ρgy2
V = speed
y = height/depth
P = pressure
ρ = density
g = gravity

The Attempt at a Solution


a. [/B]Buoyant force(Fb) = ?
I drew a free body diagram of the balloon, with Fb pointing up, and Wb, weight of the balloon, pointing down.
Summed forces:
ΣFy = may
Fb - Wb = 0 (since acceleration = 0)
Fb = Wb
Fb = Mb(g)
Fb = (15.2 kg)(9.81 m/s2)
Fb = 149.112 N

I'm not sure if this is correct or not? Are there any other forces acting on the balloon that I missed?

b. Acceleration of balloon (ab) = ?

Fb - Wb = ma
a = (Fb - Wb) / m

That's as far as I've gotten with this one - I'm not sure if this is the correct way to go about finding the acceleration. If so, for the m in the equation, would i be using the mass of the balloon?

I feel like there's something I need to add since the density of the air was given in the problem, but I'm not sure where that fits in?

Any help would be great,
thanks!
 
Last edited:
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  • #2
Did your FBD include the tension in the rope holding the balloon down? It's only because it's initially tethered that it doesn't accelerate. So your equation:

Fb - Wb = 0 (since acceleration = 0)

is not correct.

Use Archimede's principle to calculate the buoyant force. That's where the density of the air comes into play.
 

What is buoyant force?

Buoyant force is the upward force exerted by a fluid on an object that is partially or fully submerged in the fluid. It is equal to the weight of the fluid that the object displaces.

What is Archimede's principle?

Archimede's principle is a law of physics that states that the buoyant force exerted on an object immersed in a fluid is equal to the weight of the fluid that the object displaces. This principle is named after the Greek mathematician and scientist Archimedes.

How does the density of an object affect its buoyant force?

The density of an object affects its buoyant force by determining how much fluid it displaces. Objects with higher densities will displace more fluid and experience a greater buoyant force, while objects with lower densities will displace less fluid and experience a smaller buoyant force.

What is the difference between weight and buoyant force?

Weight is the force exerted on an object by gravity, while buoyant force is the force exerted by a fluid on an object. Weight is always directed towards the center of the Earth, while buoyant force is always directed upwards.

How does the shape of an object affect its buoyant force?

The shape of an object affects its buoyant force by changing how much fluid it displaces. Objects with larger surface areas will displace more fluid and experience a greater buoyant force, while objects with smaller surface areas will displace less fluid and experience a smaller buoyant force.

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