Questions on Archimedes principle

In summary, the conversation discusses the calculation of buoyant force in relation to a suspended hollow ball in water and how it is affected by the volume of the displaced water and the density of the ball. It is clarified that the buoyant force equals the weight of the displaced fluid and does not change as the object is lowered. The conversation also touches on the topic of neutral buoyancy and how an uneven distribution of mass or irregular shape can affect the position of a floating object. The conversation ends with a question about whether the water level would change once the object is completely submerged.
  • #1
Misr
385
0
1-A hollow ball is suspended in water,has a larger density than the density of the water..how?

Fb=Fg (okay)
ρ(of water)* (Vol of ball substance+Vol.of the cavity)=ρ(of the ball) * vol of the ball substance)
(not okay)
I don't understand how we calculate the buoyant force
The buoyant force should be calculated according to the equation: Fb=ρ(of water)*Vol of the substance*g
not Fb=ρ(of water)* (Vol of ball substance+Vol.of the cavity)*g



2-when a body is suspended in water,does the level of water rise up when we try to move this body to the bottom?and does the buoyant force increases as we go down?

Thanks in advance..
 
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  • #2
Misr said:
1-A hollow ball is suspended in water,has a larger density than the density of the water..how?
What do you mean by 'suspended'? Is it being held by a string or other support?

Fb=Fg (okay)
ρ(of water)* (Vol of ball substance+Vol.of the cavity)=ρ(of the ball) * vol of the ball substance)
(not okay)
I don't understand how we calculate the buoyant force
The buoyant force should be calculated according to the equation: Fb=ρ(of water)*Vol of the substance*g
not Fb=ρ(of water)* (Vol of ball substance+Vol.of the cavity)*g
The buoyant force equals the weight of the displaced fluid. In this case: ρ(of water)*(Vol of ball)*g
 
  • #3
Misr said:
The buoyant force should be calculated according to the equation: Fb=ρ(of water)*Vol of the substance*g
No, it should be "Volume of water displaced by submerging the object". See Archimedes' law.
If the object is homogenous and completely submerged, then your formula works too. But is a quite special case. Here you do not have a homogenous object.
 
  • #4
What do you mean by 'suspended'? Is it being held by a string or other support?
Nope, but Fg=Fb
see this image

[PLAIN]http://img193.imageshack.us/img193/5597/unlewd.jpg
[/PLAIN]
The buoyant force equals the weight of the displaced fluid. In this case: ρ(of water)*(Vol of ball)*g
so you mean that wheather the ball is hollow or not,it displaces the same amount of water-since the external volumes is the same?

No, it should be "Volume of water displaced by submerging the object". See Archimedes' law.
yeah,okay so is the volume of the displaced water equal to the volume of the substance?since there's a cavity in the ball?

don't forget my second question,i really need to know the answer

Thanks in advance
 
Last edited by a moderator:
  • #5
Misr said:
yeah,okay so is the volume of the displaced water equal to the volume of the substance?since there's a cavity in the ball?
on't forget my second question,i really need to know the answer

Thanks in advance

The volume of the displaced water in your case is the volume (external) of the ball, when the ball is completely submerged. If by substance you mean only the walls of the ball, then no. The volume of "substance" is much less than the total volume of the ball. If the ball has a hole in it so that the water can enter and the air can exit, then the volume displaced will be equal to the volume of "substance".

2. The level will rise until the ball is completely submerged. After that it does not change.
 
  • #6
Misr said:
Nope, but Fg=Fb
see this image
So you have something of neutral buoyancy, that remains in equilibrium when submerged.

so you mean that wheather the ball is hollow or not,it displaces the same amount of water-since the external volumes is the same?
Right.

yeah,okay so is the volume of the displaced water equal to the volume of the substance?since there's a cavity in the ball?
The fact that the ball contains a cavity is irrelevant to the amount of fluid it displaces.

don't forget my second question,i really need to know the answer
The buoyant force equals the weight of the displaced fluid. Does that change as the object is lowered? Once the object is completely submerged, why would the water level change?
 
  • #7
If the ball cannot deform (which would introduce another level of complication) then only its volume would count towards the buoyancy force. However, an uneven distribution of mass or irregular shape can cause the floating / suspended object to rotate, if it is free, so that its centre of mass is at a point which minimises the gravitational potential. This is why a stick tends to lie on its side in the water and not upright - despite experiencing exactly the same upthrust all the time.
 
  • #8
The buoyant force equals the weight of the displaced fluid. Does that change as the object is lowered?
No,Mr Doc Al

Once the object is completely submerged, why would the water level change?
I don't know :( would the water level change?
 

Related to Questions on Archimedes principle

1. What is Archimedes principle?

Archimedes principle states that when an object is fully or partially submerged in a fluid, the buoyant force acting on the object is equal to the weight of the fluid that is displaced by the object.

2. Who was Archimedes and what was his contribution to this principle?

Archimedes was a Greek mathematician, physicist, engineer, inventor, and astronomer who lived in the 3rd century BC. He is credited with discovering and formulating the principle of buoyancy, now known as Archimedes principle.

3. How is Archimedes principle applied in everyday life?

Archimedes principle has many practical applications, such as understanding why objects float or sink in water, how ships stay afloat, and how submarines can control their depth. It is also used in the design of boats, submarines, and other watercraft.

4. Is Archimedes principle only applicable to water?

No, Archimedes principle is applicable to any fluid, including liquids and gases. It can also be applied to objects submerged in air, such as hot air balloons.

5. Can Archimedes principle be used to calculate the density of an object?

Yes, Archimedes principle can be used to calculate the density of an object by measuring the volume of fluid displaced when the object is submerged and comparing it to the weight of the object. The density of the object can be calculated using the formula: density = mass/volume.

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