# Questions on Archimedes principle

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?

Doc Al
Mentor
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

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.

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 [Broken]
[/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

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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

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.

Doc Al
Mentor
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?

sophiecentaur