Slightly harder physics question
Discussion
A barge laden with pig iron is in a lock on a canal. The lock gates and sluices are closed so there is no in or outflow of water. The water level is exactly a metre from the top of the edge of the lock. The bargee decides to heave the pig iron over the side into the water. Does the water level in the lock change and, if so, is it up or down? 
Remembering what I can from school, because the iron is on the barge and so has already displaced the water with it's weight, dropping the iron over the side of the barge will not alter the water level at all. Because the gates are closed, its only a matter of moving from the barge to the water, nothing new is intoduced and so should stay as was.
The barge was displacing the water-weight-equivalent of the iron thrown overboard. While it was in mid-air 
the barge rose and the water-level dropped. Once the iron hit the water and sank (assuming no splashing over the sides) the water level rose by the displaced volume of the iron and the barge rose a little bit in relation to the lock but remained at a constant level in relation to the free surface.
Short answer = Water level dropped
the barge rose and the water-level dropped. Once the iron hit the water and sank (assuming no splashing over the sides) the water level rose by the displaced volume of the iron and the barge rose a little bit in relation to the lock but remained at a constant level in relation to the free surface. Short answer = Water level dropped
imo the water level would decrease. The volume displaced by the iron in the boat is going to be greater than on it's own (otherwise it wouldn't float). By chucking it in the water, there will be an overall reduction in displacement as the displacement volume of the iron in the water will be less than the reduction in displacement of the barge.
probably wrong though
probably wrong though
Kiwi XTR2 said:
The barge was displacing the water-weight-equivalent of the iron thrown overboard. While it was in mid-air
Short answer = Water level dropped
this is how i thought it would work, if the barge held tanks of water on board which it could empty then the equal displacement before and after emptying would hold true otherwise the water level would change depending on the lead or any other load on the barge being heavier or lighter than the water its actual mass displaces.
The absolute water level should remain the same due to no change in displacement. However, wouldn't the barge rise in the water through shedding the weight? Not that that would help with the lock problem of course...
On the other hand, if off-loading the weight doesn't change the height of the barge in the water then its boyancy must be modifying the net displacement, so shedding the weight would make the water rise....
School seems so long ago now.
On the other hand, if off-loading the weight doesn't change the height of the barge in the water then its boyancy must be modifying the net displacement, so shedding the weight would make the water rise....
School seems so long ago now.
Actually, depending on the SHAPE of the pig iron, the water level could fall quite dramatically.
Make it dart shaped.
It will puncture the bottom of the canal and let all the water out!
On the other hand, if it is a 'cunning trick question' and the pig iron is in fact a four legged pig-like grunty pig thing, it might clamber out of the canal and the water level would again fall, but not by as much.
Make it dart shaped.
It will puncture the bottom of the canal and let all the water out!
On the other hand, if it is a 'cunning trick question' and the pig iron is in fact a four legged pig-like grunty pig thing, it might clamber out of the canal and the water level would again fall, but not by as much.
When the iron is floating (i.e. on the barge) it will displace exactly its own weight in water. When it is resting on the bottom it will displace it's own volume in water. This means the water level will go down. It is exactly analagous to jacking the barge up out of the water i.e. putting weight on the bottom of the canal instead of supporting it by bouyancy; the barge will rise relative to the water, and the water level will go down relative to the ground.
Hmmm.
Surely 1 tonne of steel will required 1 tonne of bouyancy, which is generated by displacing one tonne of water (1 cubic metre of water ish)
If you throw that one tonne of steel from the boat, the boat will displace one cubic metre less of water, which was used to keep it floating, BUT, the steel being of higher density than water (if it has no pockets of air inside) will now displace less volume than the boat had to do to support it.
The boat will rise out of the water as the steel is thrown, making the level drop, then as the steel enters the water, the level of water will rise again, but not as much as it dropped when the boat rose.
Dave
Surely 1 tonne of steel will required 1 tonne of bouyancy, which is generated by displacing one tonne of water (1 cubic metre of water ish)
If you throw that one tonne of steel from the boat, the boat will displace one cubic metre less of water, which was used to keep it floating, BUT, the steel being of higher density than water (if it has no pockets of air inside) will now displace less volume than the boat had to do to support it.
The boat will rise out of the water as the steel is thrown, making the level drop, then as the steel enters the water, the level of water will rise again, but not as much as it dropped when the boat rose.
Dave
GreenV8S said:
When the iron is floating (i.e. on the barge) it will displace exactly its own weight in water. When it is resting on the bottom it will displace it's own volume in water. This means the water level will go down. It is exactly analagous to jacking the barge up out of the water i.e. putting weight on the bottom of the canal instead of supporting it by bouyancy; the barge will rise relative to the water, and the water level will go down relative to the ground.
you know, I think you might have a point! 
GreenV8S said:
When the iron is floating (i.e. on the barge) it will displace exactly its own weight in water. When it is resting on the bottom it will displace it's own volume in water. This means the water level will go down. It is exactly analagous to jacking the barge up out of the water i.e. putting weight on the bottom of the canal instead of supporting it by bouyancy; the barge will rise relative to the water, and the water level will go down relative to the ground.
That was kind of my point, except you put it across more intelligently.
And I got the outcome wrong...doh...
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