ChatterBank0 min ago
Floatation
If I had two floats, one rigid (like a glass sphere) and one flexible (like a balloon) both containing 1 litre of air and hung from both a 1.2 kg weight, then dropped them into the sea, what would happen?
I imagine the flexible one would gradually collapse with the water pressure and become less and less buoyant (displacing less volume) and continue to sink whereas the rigid one would reach an equilibrium at a certain depth and stay at that depth.
Am I right or am I missing something?
I imagine the flexible one would gradually collapse with the water pressure and become less and less buoyant (displacing less volume) and continue to sink whereas the rigid one would reach an equilibrium at a certain depth and stay at that depth.
Am I right or am I missing something?
Answers
Best Answer
No best answer has yet been selected by Graham-W. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.Firstly ascertain whether they would float or sink.
1 litre of air will displace 1 litre of seawater
1 litre of seawater weighs about 1.03 kg
An attached weight of 1.2 kg will therefore cause both to sink.
You are correct with your assumptions about the flexible one - it would sink with increasing acceleration as the air gets compressed. It will of course reach a terminal velocity because of drag.
You are not correct about the solid one. Seawater is virtually incompressible so the density of the seawater will not change enough with depth to counter the weight of 1.2 kg plus the weight of the rigid sphere. It will accelerate slowly at first until drag causes it to reach a terminal velocity. A sphere is the best shape to withstand external pressure but eventually (if the sea is deep enough) it will catastrophically collapse and accelerate to a higher terminal velocity and continue to the bottom.
HOWEVER:
All that only applies if the weights have zero volume.
You don't say what the weights are made of or what their volumes are.
Assuming iron weights:
Based on a density of iron (7.874) and seawater (1.03) I calculate that a 1.2 kg iron weight would weigh 1.043 Kg when immersed in seawater, which is only just in excess of the 1.03 Kg of seawater displaced by the balloon.
The net downwards force will be just 13g
1 litre of air will displace 1 litre of seawater
1 litre of seawater weighs about 1.03 kg
An attached weight of 1.2 kg will therefore cause both to sink.
You are correct with your assumptions about the flexible one - it would sink with increasing acceleration as the air gets compressed. It will of course reach a terminal velocity because of drag.
You are not correct about the solid one. Seawater is virtually incompressible so the density of the seawater will not change enough with depth to counter the weight of 1.2 kg plus the weight of the rigid sphere. It will accelerate slowly at first until drag causes it to reach a terminal velocity. A sphere is the best shape to withstand external pressure but eventually (if the sea is deep enough) it will catastrophically collapse and accelerate to a higher terminal velocity and continue to the bottom.
HOWEVER:
All that only applies if the weights have zero volume.
You don't say what the weights are made of or what their volumes are.
Assuming iron weights:
Based on a density of iron (7.874) and seawater (1.03) I calculate that a 1.2 kg iron weight would weigh 1.043 Kg when immersed in seawater, which is only just in excess of the 1.03 Kg of seawater displaced by the balloon.
The net downwards force will be just 13g