# Using the Archimedes Principle to do calculations for this floating buoy

• Engineering
• Ben_Walker1978
In summary, A spherical buoy with a diameter of 2m and weight of 3kN is designed to be used in various circumstances by being floated in water and anchored with a chain. Using the equation Fb=Pf * Vd * g, where Pf is the density of sea water, Vd is the volume of the sphere, and g is the acceleration due to gravity, the upthrust or buoyant force (Fb) in sea water can be calculated. The specific gravity of sea water at 4 degrees is 1.042 and the volume of a sphere is given by Volume = πD3/6. By plugging in the values, we get a buoyant force of 42.8N.
Ben_Walker1978
Homework Statement
Yes
Relevant Equations
Buoyant Force
A Spherical buoy has a diameter of 2m and weighs 3kN. It is designed to be used in a variety or circumstances by it being floated in water and anchored to the floor with a chain. Draw the free body diagram and determine the following given that the volume of a sphere is,
Volume = πD3/6
1. The upthrust or buoyant force (Fb) in sea water.
2. The tension in the chain (T)
Assume at 4Degrees the specific gravity of sea water is 1.042, g = 9.81m/s^2

1.

Equation
Pf * Vd * g

Fb=Pf * Vd * g=(1.042kg/m3)(4.188m3)(9.81m/s2)

=42.8N
Is this correct? Trying to learn this. I think it is correct. But wanted to check so i understand how to do it correctly.

Then i am going to find question 2.

Question 2:

T = W - Fb

W = Weight
Fb = Buoyant Force

T = 3 - 42.8 = 39.8N

Tension = 39.8N

Is the Tension in the chain correct?

Last edited:
Your density is too low by a factor 1000 (can you lift a cubic meter of water with one hand?). Apart from that it is correct.

Ben_Walker1978
mfb said:
Your density is too low by a factor 1000 (can you lift a cubic meter of water with one hand?). Apart from that it is correct.

I have changed the density to 1.042 now.

So now all is correct including the answer?

Last edited:
Could you also look at my tension in the chain? Is this correct? Thanks.

The density didn't change and it's still way too low.

All the Newton values should be kiloNewtons.

Ben_Walker1978
What density do you mean? I am unsure. i changed the density of sea water. Could you tell me what density please? Sorry for not knowing i am trying to learn this.

Ben_Walker1978 said:
Homework Statement:: Yes
Relevant Equations:: Buoyant Force

Fb=Pf * Vd * g=(1.042kg/m3)(4.188m3)(9.81m/s2)
1.042kg/m3 is not the density of sea water. It's below the density of air!

Check numbers for plausibility. 1 kg is something you can lift easily. 1 cubic meter of water is not. It can't have a mass anywhere close to 1 kg.

Ben_Walker1978
This is the question i am trying to complete:

Maybe they have sea water as just an example?

So with this question is it correct?

In order to calculate a tension on the chain, don't you have to know the depth that the chain pulls the buoy below its normal floating depth? If the chain allows the buoy to float at its normal depth, then I don't think there will be any tension on the chain. Maybe the question wants you to assume that the buoy is completely submerged.

I though Tension was. T = w - Fb

The amount the buoy floats which is Fb let's us calculate it?

I don't know if this is what the problem expects you to do, but here is my two cents:
As long as the maximum buoyant force is greater than the weight, then the buoy can float on the water while partially submerged and the actual buoyant force equals the weight. In that case, I don't think there is any tension on the chain.
I suspect that this answer is not what the problem wants because it may not be requiring you to use what you are learning in the class. Your calculation is for a completely submerged buoy, held underwater by the chain. That is probably what the problem wants.

Ben_Walker1978 said:
This is the question i am trying to complete:

[ATTACH type="full" alt="Part 2 Archimedes.png"]262088[/ATTACH]

Maybe they have sea water as just an example?

So with this question is it correct?
1.042kg/m3 is certainly not the correct density.
Most likely 1042kg/m3 is the correct density.

@Ben_Walker1978 , What is the density of water (NOT sea water), and what is the definition of "specific gravity"? You need to get this right to calculate the density of sea water.

But then it ends up as this. Is this correct then? When i change it to 1024.Pf * Vd * g

Fb=Pf * Vd * g=(1042kg/m3)(4.188m3)(9.81m/s2)

=42,809N
Then i am going to find question 2.

Question 2:

T = W - Fb

W = Weight
Fb = Buoyant Force

T = 3 - 42,809 = -39,806N

Tension = -39,806N

Better.

Tension must be positive, but you can take the absolute value.
Ben_Walker1978 said:
T = 3 - 42,809 = -39,806N
That notation is really sloppy.

Why?
Should i do:
T = w - Fb
T = 3 - 42,809 = -39.806

Is this correct though? Can it be minus

Why what?
3 - 42,809 = -42,806
What you have is 3000, not 3. And then it's missing units:
3000 N - 42,809 N = -39,809 N. Note the last digit, you accidentally subtracted 3 and 3000 from the number.

By definition, tension is positive. Here it's a force downwards, its magnitude in calculations will depend on the (arbitrary) choice which direction is positive, but the tension as final answer must be positive.

Ben_Walker1978
I understand now. So change 3kN to 3000N so it is the same unit at 42,809 N

so even though the answer is -39,809N. I put 39,809N As the answer?

Thank you for keep helping me. Much appreciated

-39,809N is the force on the buoy.
Next line: 39,809 N tension in the chain.

## 1. How does the Archimedes Principle apply to floating buoys?

The Archimedes Principle states that any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid it displaces. This means that the weight of the buoy is equal to the weight of the water it displaces, allowing it to float.

## 2. What factors affect the buoyancy of a floating buoy?

The buoyancy of a floating buoy is affected by the density of the fluid it is floating in, the volume of the buoy, and the weight of the buoy itself. The greater the density of the fluid, the greater the buoyancy force. A larger volume of the buoy will also increase its buoyancy, while a heavier buoy will decrease it.

## 3. How can the Archimedes Principle be used to calculate the buoyancy force of a floating buoy?

To calculate the buoyancy force of a floating buoy, you will need to know the density of the fluid it is floating in, the volume of the buoy, and the weight of the buoy. Then, using the formula Fb = ρVg, where Fb is the buoyancy force, ρ is the density of the fluid, V is the volume of the buoy, and g is the acceleration due to gravity, you can calculate the buoyancy force.

## 4. Can the Archimedes Principle be used to determine the stability of a floating buoy?

Yes, the Archimedes Principle can also be used to determine the stability of a floating buoy. If the buoy is floating in a stable position, the buoyancy force will be greater than or equal to the weight of the buoy. If the buoy is in an unstable position, the weight of the buoy will be greater than the buoyancy force.

## 5. Are there any limitations to using the Archimedes Principle for buoyancy calculations?

While the Archimedes Principle is a useful tool for calculating buoyancy, it does have some limitations. It assumes that the fluid is incompressible and that the buoy is in a uniform environment. It also does not take into account any external forces acting on the buoy, such as wind or waves, which can affect its stability.

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