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Vegas1200C
27th January 2008, 00:25
A person in a boat drops a cannonball overboard; does the water level change?

wabiker
27th January 2008, 00:28
.......No.:smoke

Mr Jimi
27th January 2008, 00:31
Somebody watched CSI last nite :shhhh

I did too
:tour

mid30
27th January 2008, 00:35
depends on how big a body of water it is.

Vegas1200C
27th January 2008, 00:41
depends on how big a body of water it is.

That is irrelevant to the question. No matter how large a body of water or how small it will still change it the same direction or not change it at all. If it makes it easier for you, imagine it is a small lake.

68B_Body
27th January 2008, 00:44
no it wont rize the displacement is the same as if it was in the boat. toss the ball overboard the boat has less displacement, but the ball in the water adds displacement in the water

wabiker
27th January 2008, 00:46
The only thing that changes is how high/low in the water the boat sits.

68B_Body
27th January 2008, 00:52
take a big cup put it in a sink full of water push down on the cup in the water with no water getting in the cup, the water will rise further you push it down. far a a cannonball. its would be roughly 8 cubic inches of space, but the weight of it in a boat is probably 25-50 pounds. toss it off the boat, the boat will rise slightly, the water level will go down, but probably not even noticable. and the ball will take up 8 cubic inches of space in the water. causing the level of the body of water to rise probably close to what the level was with the ball on the boat:D:smoke

Vegas1200C
27th January 2008, 00:57
So what is your answer? Rise, Fall or Stay exactly the same?

68B_Body
27th January 2008, 00:59
it will fall as soon as the ball left the hands of the sailor, rise as soon as its submerged in the water and stay exactly the same as it was on the boat

Davemandu
27th January 2008, 01:03
my answer is: 4

68B_Body
27th January 2008, 01:15
it will stay the same if the balls weight displaces the same amount of water in the boat as size of the ball takes up space in the water... until i know those 2 variables no one really knows. i don't know what a cannon ball weighs in this experiment nor the size in cubic inches.

Moved On
27th January 2008, 01:21
Depends ... is the tide coming in or going out???

Gazza

dagsportster
27th January 2008, 01:23
If I recall this one correctly, the water level will fall because the weight of the bowling ball in the boat displaces a larger volume (of air) when it is in the boat (that's how boat's stay afloat) than it does (of water) when it is dropped. Since water is denser than air, so the level of the water will drop.

Moved On
27th January 2008, 01:33
Seriously though... if it's a small enough body of water to actually
make a measurement ... I'm gonna say the water level will go down.

It's a mass vs. density vs. volume thing. If you just put a cannon ball
in the water with no boat around, the water will rise by the volume of
the ball divided by the surface area of the water.

When the cannon ball is in the boat, it's causing the boat hull to draft
more than after the cannon ball is out out of the boat, because of the
reduced weight. When the boat has less draft the displacement will
become smaller and the water level will go down.

Since a cannon ball is very dense it will likely cause a significant amount
of draft in a small boat. Since the area of the hull is probably larger than
the diameter of the cannon ball the water displacement by the weighted
hull will likely be more volume than the volume of the cannon ball itself.
Resulting in a lowering of the water.

Gazza

Moved On
27th January 2008, 01:36
Now if you have a helium ballon in the boat and drop it
into the water what will happen?

Gazza

68B_Body
27th January 2008, 01:37
Now if you have a helium ballon in the boat and drop it
into the water what will happen?

Gazza

lol!:shhhh

Vegas1200C
27th January 2008, 01:38
I have a better helium balloon riddle for later. :geek

wabiker
27th January 2008, 01:57
Ummmm..... were ALL assuming the water is NOT frozen.

Vegas1200C
27th January 2008, 03:00
Ummmm..... were ALL assuming the water is NOT frozen.

A nice try but then it is not water it is ice.

wabiker
27th January 2008, 03:18
A nice try but then it is not water it is ice.

True.... but, It doesnt state that it was dropped in the Water... it says it was dropped Overboard.
..beside what the hell do I know about Ice... I live in Az:D

Blaze
27th January 2008, 03:33
no it wont rize the displacement is the same as if it was in the boat. toss the ball overboard the boat has less displacement, but the ball in the water adds displacement in the water

That's what I was thinking....

BWP 5p
27th January 2008, 04:19
Nice Try Vegas............I ain't biting on this VERY OBVIOUS one:doh:doh:doh:laugh:smoke:D

milmat1
27th January 2008, 04:56
Question :

Does the boat Displace the same amount of Additional Water when the Cannon ball is onboard, As the Cannon ball would displace on it's own ??

Point Being:

The added weight of the Cannon ball may not cause the boat to displace the same amount of Additional water as the Cannon ball would displace on it's own !!

If you answer this question, You will answer yours !

If the answer is Yes, Then the water level will not change !
If the answer is no, The water level will change !

glh
27th January 2008, 05:08
Why do you have a cannonball in your boat?

Vegas1200C
27th January 2008, 05:12
Question :

Does the boat Displace the same amount of Additional Water when the Cannon ball is onboard, As the Cannon ball would displace on it's own ??

Point Being:

The added weight of the Cannon ball may not cause the boat to displace the same amount of Additional water as the Cannon ball would displace on it's own !!

If you answer this question, You will answer yours !

If the answer is Yes, Then the water level will not change !
If the answer is no, The water level will change !

Close but not quite. If the answer is yes then the water level will not change. If the answer is no then you have to look at whether the boat without the cannon ball plus the cannon ball alone displaces more or less than the boat with the cannon ball on board.

GrumpyCoyote
27th January 2008, 05:13
Why do you have a cannonball in your boat?

For the cannon - duh.

milmat1
27th January 2008, 05:24
Close but not quite. If the answer is yes then the water level will not change. If the answer is no then you have to look at whether the boat without the cannon ball plus the cannon ball alone displaces more or less than the boat with the cannon ball on board.

I Thought thats what I said,, ?? Anyway, Point well taken....

Urrell
27th January 2008, 06:20
it will fall as soon as the ball left the hands of the sailor, rise as soon as its submerged in the water and stay exactly the same as it was on the boat

:banana:banadanc:banana

Moved On
27th January 2008, 06:30
Question :
Does the boat Displace the same amount of Additional Water when the Cannon ball is onboard, As the Cannon ball would displace on it's own ??

The answer has to be no. Because iron is denser (weighs more) than water.

Consider a test tube shaped boat the same diameter as the cannon ball, with a cannonball in the bottom, the boat will hang into the water with the cannonball completely below the surface and a large volume of air within the test tube shaped boat also under the water surface. Once the cannon ball is taken out the boat will float with very little air under the surface.

You can try this yourself... I just did in the bathroom sink down stairs. Fill up the sink with water. Put a plastic cup into the sink and look at how much displacement (draft) you have. I had a half an inch with my cup alone. Then put some water in the cup and measure the air displacement. I had an inch of water in the cup but still had only a half an inch of air below the outside water inside the cup. Clearly if I bailed the water out of my cup the sink level would not have changed. This proves that if the object taken out of the boat has the same mass density of water, the water level will not change.

Then I grabbed a small rock and put it into the bottom of the plastic cup, it was a fairly flat rock just less than the diameter of the cup and about half an inch thick..... I put the cup in the water and now there was about 3/4" of draft from the surface to the top of the rock not 1/2". So clearly the higher density rock caused more air displacement than the rock itself would have, because the cup was down 1.25" into the water not 1" into the water. If I had put the rock into the water the cup would have displaced 3/4" time the area of the cup less sink water... but the rock would have displaced the sink water by 1/2" times the diameter of the cup.

Gazza

MadMax25
27th January 2008, 06:56
Yes… an interesting physics question.
Here is the answer and explanation.
As long as the boat is floating, the presence of the ball in the boat
helps the boat displace an amount of water equal to the WEIGHT of the ball.
If the metal ball weighs 10 pounds, then 10 pounds of water ( larger volume )
will be displaced and the ocean level will rise accordingly.
When the ball is tossed overboard, the instant the ball leaves
the tossers hands, the boat will rise slightly and the water level
in the ocean will drop slightly, as less ocean water is now displaced by the
now reduced weight of the floating boat.
When the ball enters the ocean, it’s a different story.
The ball is more dense than the water, so it sinks.
The amount of water in the ocean is now displaced ONLY by the physical
VOLUME of the ball.
There lies the difference. When something is floating it displaces water
according to the weight of the object. When it is NOT floatable,
then the volume of the sinking object decides how much water is displaced.

Urrell
27th January 2008, 07:11
Yes… an interesting physics question.
Here is the answer and explanation.
As long as the boat is floating, the presence of the ball in the boat
helps the boat displace an amount of water equal to the WEIGHT of the ball.
If the metal ball weighs 10 pounds, then 10 pounds of water ( larger volume )
will be displaced and the ocean level will rise accordingly.
When the ball is tossed overboard, the instant the ball leaves
the tossers hands, the boat will rise slightly and the water level
in the ocean will drop slightly, as less ocean water is now displaced by the
now reduced weight of the floating boat.
When the ball enters the ocean, it’s a different story.
The ball is more dense than the water, so it sinks.
The amount of water in the ocean is now displaced ONLY by the physical
VOLUME of the ball.
There lies the difference. When something is floating it displaces water
according to the weight of the object. When it is NOT floatable,
then the volume of the sinking object decides how much water is displaced.

Got you - It will only displace it's volume when submerged. If its specific gravity is more than one there will be less displaced than when in the boat. :doh

Moved On
27th January 2008, 07:15
Yes… an interesting physics question.
Here is the answer and explanation.

That's an eloquent and understandable explanation... much better
than the ones I tried.. kudos to you.. you should be a teacher.

Gazza

CT1200
27th January 2008, 08:35
Anyone ever wonder how much water a aircraft carrier displaces?

Is it spring yet? ;-)

dagsportster
27th January 2008, 12:12
Anyone ever wonder how much water a aircraft carrier displaces?

Is it spring yet? ;-)

Nimitz class displaces 102,000 tons.

Now, it's a good thing Vegas specified a cannonball, rather than a bowling ball, since some bowling balls float. Because the size (volume) of a bowling ball is regulated, an 8lb ball and a 16lb ball have the same volume, but manufacturers vary the core weight (mass) to vary the weight. So how would the answer change if a floating bowling ball was dropped off the boat?

mikgordon
27th January 2008, 12:47
This is a little more complicated then it seems. A few questions that need to be answered first. How big is the boat? What size cannon ball are we talking? How big in inches, and how much does it way? Does the cannon ball explode or is it a dud? (Figuring if it explodes, many fish will be killed and begin to float to the surface.)

shinallb
27th January 2008, 15:16
...but will it fly!?

Vegas1200C
27th January 2008, 15:26
This is a little more complicated then it seems. A few questions that need to be answered first. How big is the boat? What size cannon ball are we talking? How big in inches, and how much does it way? Does the cannon ball explode or is it a dud? (Figuring if it explodes, many fish will be killed and begin to float to the surface.)

The size of the boat doesn't matter unless it is sinking under the weight of the cannon ball and then it isn't a boat anymore. The size of the cannon ball doesn't matter because it will be more dense than water and therefore will displace more water as mass than volume. Cannon balls do not explode. They are designed to fly through an object, not explode.

The question is, will the next 10 people read any of this or will they continue to vote "No it will not change". As it stands, the vote is 9 rise, 9 fall and 24 no change so almost 80 percent are incorrect and 20 percent are right.

hoosier xlc
27th January 2008, 15:32
If you drop it through the bottom of the boat then none of it is relevant is it?

Vegas1200C
27th January 2008, 15:41
Nimitz class displaces 102,000 tons.

Now, it's a good thing Vegas specified a cannonball, rather than a bowling ball, since some bowling balls float. Because the size (volume) of a bowling ball is regulated, an 8lb ball and a 16lb ball have the same volume, but manufacturers vary the core weight (mass) to vary the weight. So how would the answer change if a floating bowling ball was dropped off the boat?

If the bowling ball is exactly equal to the density of the water then there is no change in the level. Since it is floating then it is probably less dense so it's volume would displace more water than its mass and the level would go up.

MadMax25
27th January 2008, 17:07
[QUOTE=Gary;1072923]That's an eloquent and understandable explanation... much better than the ones I tried.. kudos to you.. you should be a teacher.

To GARY & URRELL…
Thanks for the kind words. I work in research and development
of automated machinery… most call it robotics.
Besides designing electrical, mechanical, electronic and software stuff,
I’m also called upon to write patent applications.
In a patent application you need to select wording that exactly
and completely explains what is going on to someone who has
never worked on, or even seen, what it is you are trying to describe.
At the end of your explanation, they need to understand completely
how and why your contraption functions. Most importantly,
you need to explain why your device is an improvement over existing similar devices.
Properly explained insights or functional nuances will make or break
your attempt at a patent being granted.
To that end, I select my words very carefully.

As far as writing style is concerned, I take my hat off to KevM.
He’s a natural... keeping the topic moving and his audiences tuned in.
Thanks also to VEGAS1200C for initiating this interesting discussion.

:banana :banana :banana

wabiker
27th January 2008, 17:18
Cannon balls do not explode. They are designed to fly through an object, not explode.


....ermmmm....not entirely accurate.:smoke

dejomo
27th January 2008, 17:29
The total weight that is in the boat is reduced...so it(the boat) would displace less water and rise just a bit...
...but...
The overall level of the water outside the boat would stay the same because the total amount of water being displaced is equal(ball in the boat as compared to ball out of the boat):geek:D

dejomo
27th January 2008, 17:32
....ermmmm....not entirely accurate.:smoke


Exploding cannonballs did(maybe still do) excist... They were called "Grape Shot"....imagine a flying Claymore...

dejomo
27th January 2008, 17:34
How funny...a bunch of educated bikers...

Vegas1200C
27th January 2008, 17:37
Grape shot is not a cannon ball it is a bag of stuff designed to spread out like buck shot. Neither explodes, they just spread out.

Cannon balls are solid.

MadMax25
27th January 2008, 17:42
How funny...a bunch of educated bikers...

I guess we're going to take that as a compliment.

:banana :banana :banana

68B_Body
27th January 2008, 17:45
How funny...a bunch of educated bikers...

yea i can imagine em all in their smoking jackets and pipes at the local biker bar discussing physics and mechanical engineering and theories of relativity's

Gone
27th January 2008, 17:45
The boat and everything in it, including the cannon ball, already displaced all the water it's going to.

jharback
27th January 2008, 17:47
And if no one is around to hear it drop, did it really drop?

Vegas1200C
27th January 2008, 18:14
How funny...a bunch of educated bikers...

I actually don't find this suprising. I would imagine a decent percentage of biker have a mechanincal mind. Working on bikes might not be rocket science but it is closer to that than accounting. That said, the wrong answers are still coming in 4 to 1.

BWP 5p
27th January 2008, 18:23
If you drop it through the bottom of the boat then none of it is relevant is it?

:clap:clap:clap:laugh:laugh:laugh:clap:clap:clap

wabiker
27th January 2008, 18:23
Cannon balls are solid.
....not all of them :D

Vegas1200C
27th January 2008, 18:28
Cannon ball, strictly, a round solid missile of stone or
iron made to be fired from a cannon, but now often applied
to a missile of any shape, whether solid or hollow, made
for cannon. Elongated and cylindrical missiles are
sometimes called bolts; hollow ones charged with
explosives are properly called shells.

wabiker
27th January 2008, 18:40
Cannon ball, strictly, a round solid missile of stone or
iron made to be fired from a cannon, but now often applied
to a missile of any shape, whether solid or hollow, made
for cannon. Elongated and cylindrical missiles are
sometimes called bolts; hollow ones charged with
explosives are properly called shells.
Nice try.... But, Early experiments using hollow balls filled with gunpowder and plugged with a Fuze. At first they thought they needed someone to light the fuze prior to loading the BALL. Until they discovered that the Fuze would ignite of its own during the firing. If you look at old paintings of Blackpowder aged Battle scenes.. airbursts were quite common.

They were Hollow spherical shaped projectiles... Generally used in muzzleloaded cannons.....sounds like a Cannon BALL to me.:D

scottgearman
27th January 2008, 18:42
NO, as the ball displaces water, it's weight in the boat did the same.....if the ball is in the boat or at the bottom it displaces the same amount of water.....

Moved On
27th January 2008, 18:48
If you drop it through the bottom of the boat then none of it is relevant is it?

Well actually that would be relevant too. If the boat and the cannon ball sank,
then their would be nothing displacing the air space that was originally holding
the boat and cannonball up.... so the water level would go down even more.

Gazza

Moved On
27th January 2008, 18:52
The total weight that is in the boat is reduced...so it(the boat) would displace less water and rise just a bit...

True.

The overall level of the water outside the boat would stay the same because the total amount of water being displaced is equal(ball in the boat as compared to ball out of the boat)

Wrong, your assumption that the word equal belongs in that sentence is
incorrect... didn't you read any of the previous posts??? A bucket of steel
weighs more than a bucket of water not equal.

Gazza

Moved On
27th January 2008, 18:56
NO, as the ball displaces water, it's weight in the boat did the same.....if the ball is in the boat or at the bottom it displaces the same amount of water.....

It displaces the same weight not volume.

Gazza

MadMax25
27th January 2008, 22:41
Cannon balls painted black are faster.

:smoke :smoke :smoke

glh
28th January 2008, 04:49
For the cannon - duh.

Then why did you throw it overboard?

Rascal
28th January 2008, 04:53
no it wont rize the displacement is the same as if it was in the boat. toss the ball overboard the boat has less displacement, but the ball in the water adds displacement in the water

:iagree Sounds right to me...

dejomo
28th January 2008, 05:32
yea i can imagine em all in their smoking jackets and pipes at the local biker bar discussing physics and mechanical engineering and theories of relativity's


:doh:laugh

Now that's funny...:laugh

dejomo
28th January 2008, 05:38
Nice try.... But, Early experiments using hollow balls filled with gunpowder and plugged with a Fuze. At first they thought they needed someone to light the fuze prior to loading the BALL. Until they discovered that the Fuze would ignite of its own during the firing. If you look at old paintings of Blackpowder aged Battle scenes.. airbursts were quite common.

They were Hollow spherical shaped projectiles... Generally used in muzzleloaded cannons.....sounds like a Cannon BALL to me.:D


GOD DAMN!!!

Go man go!!!!

ok that's enough strokin for one night...:D

Those type were still pretty popular in the settlement days...unfortunately the calvary(or more properly their cannon cockers) used them on the native americans...not the proudest of things our military has done...then again I can think if things I've seen that...well...:frownthre

dejomo
28th January 2008, 05:42
True.



Wrong, your assumption that the word equal belongs in that sentence is
incorrect... didn't you read any of the previous posts??? A bucket of steel
weighs more than a bucket of water not equal.

Gazza


AHHH SOOO...master not just snatched pebble from pupils palm but kicked honorable pupil in groin for snatching efforts...LOL

Ya I overlooked the volume to weight ratio(weight to area IS different for Steel than Water):p YOU got ME!!!

DRAWTOOL
28th January 2008, 05:57
eureka...:smoke

Fe Head
28th January 2008, 10:18
I have not read all the posts but I assume the correct answer has been figured by now.

Which is the water in the lake will be lower by approximately 7 times the volume of the cannon ball.

This is because when the cannon ball is in the boat its weight cause 8 times its volume in water to be displaced by the boat ( Steel is 8 times heavier than water of the same volume). When the cannon ball enters the water it only displaces its own volume no matter what its density is (that is why it sinks) and the other 7 times of its volume are replaced by the boat rising and allowing the level in the lake to drop or become lower by 7 times the cannon ball's volume.

The only way the level would stay the same is if you used a cannon ball with the same density as water which is always 1.
To make the water rise using the cannon ball you would need to forget the boat alltogether and just heave the cannon ball into the water from shore or drop it from a bridge.

Cheers;

Carl-04XL
28th January 2008, 18:04
The only way the level would stay the same is if you used a cannon ball with the same density as water which is always 1.
To make the water rise using the cannon ball you would need to forget the boat alltogether and just heave the cannon ball into the water from shore or drop it from a bridge.

Cheers;

Well, if you carried the cannonball and walked out into the water, holding the cannonball over your head, until the water was at lower lip level and then dropped the cannonball. You would drown.

WHY?

johnnysquire
28th January 2008, 18:25
Well, if you carried the cannonball and walked out into the water, holding the cannonball over your head, until the water was at lower lip level and then dropped the cannonball. You would drown.

WHY?

Assuming you weren't wearing a helmet, it's because you'd be unconscious.:clap

Fe Head
28th January 2008, 18:40
Walking into the water and dropping the cannon ball is different than using a boat but exactly the same as either throwing it off a bridge or heaving it from shore - the water would rise (a little).

I suppose whether one drowns or not is a matter of the degree the water rises and if it obstructs both airways, assuming one stays fixed in place.

Mattbastard
28th January 2008, 23:01
It's a volume thing.

Say you have a brick of concrete weighing the same as the cannonball, but with twice the volume. The boat will be forced down into the water the same amount, but when dropped into the water it will displace more liquid than the cannonball and RAISE the level of water.

Same goes for a similar weight but smaller volume, the level will lower.

No I didn't read the entire post, but none of those answers are correct because we need to know the specific gravity of the ball.
One could assume it's greater than 1, but therein lies the assumption.

wabiker
28th January 2008, 23:53
The water level will drop...From the perspective of the Boat.

The boat will sit Higher in the water without the added weight of the cannonball, thus *lowering the level*

glh
29th January 2008, 01:58
If you used the cannon to shoot a low density cannonball into the water, you could raise the water level for a time.

First it would go down, as the recoil from the downward pointed cannon lifted the boat.

Then it would go up, with the lighter than water cannonball submerged and the boat back displacing its full weight.

Then it would go back down, when the cannonball shoots back out of the water like it was ... shot out of a cannon.

When the cannonball falls back into the water ...
:smoke

Fe Head
18th March 2008, 09:27
Wow after all this time and only 16 respondents have it figured out correctly! Amazing. As Sherlock Holmes used to to say " Elementary my dear Watson".

glh
19th March 2008, 06:18
Wow after all this time and only 16 respondents have it figured out correctly! Amazing. As Sherlock Holmes used to to say " Elementary my dear Watson".

Nobody is going to respond to your thread resurrection, 'cause there is too much thinking involved. Or perhaps the folk will surprise me.

I wish I could juggle the equations in my head that would allow me to instantly answer such questions, no matter what the density, weight, or volume of the boat, the cannonball, or the "water" is at a given point in time. I know how to frame the question, and did answer correctly, but it took a whole lot of brainpower to simply set up the question framework in my head.

It would be interesting to know why those who chose wrong do so.

Fe Head
19th March 2008, 09:01
Nobody is going to respond to your thread resurrection, 'cause there is too much thinking involved. Or perhaps the folk will surprise me.

I wish I could juggle the equations in my head that would allow me to instantly answer such questions, no matter what the density, weight, or volume of the boat, the cannonball, or the "water" is at a given point in time. I know how to frame the question, and did answer correctly, but it took a whole lot of brainpower to simply set up the question framework in my head.

It would be interesting to know why those who chose wrong do so.

Opps glh I saw the thread sitting on the entry portal page and yet when I checked into it your Jan 28 reply was the latest ? I did not intentionally resurrect it, however, I maybe should have taken its appearance as a glitch or something and past over it. Your correct about having the framework setup properly otherwise it is all water under the bridge. ]
Cheers;

Vegas1200C
19th March 2008, 12:56
Since it came back around and no one will read any of the earlier threads I will explain this one again.

The cannon ball will weigh more than the same size ball of water therefore it will push down the boat more than the same volume of water. When you drop that cannon ball over the side it can only displace its size, not its weight. The boat goes up because it lost the weight of the cannon ball which send the water level down. The size (volume) of the cannon ball raises the water level but not as much as it's weight (mass) did when it was forcing the boat down.

If the cannon ball weighed exactly the same as the water it wouldn't change level at all. If it weighed less than water, the water level would rise as the ball would float meaning its volume is displacing less then its mass.

Fe Head
19th March 2008, 13:08
Perfect Score Vegas - you can go to the front of the class:clap and also the boat will rise about 7 times the volume of the perverbial cannonball leaving the water level lower by an equal amout.:) Thanks
Cheers;

milmat1
29th April 2010, 22:52
YES it will Rise !!

Because when in the boat it is displacing it's weight. When in the water it is displacing it's Mass. And the ball is more dense than the water. The level will rise..

Bone
29th April 2010, 23:04
YES it will Rise !!

Because when in the boat it is displacing it's weight. When in the water it is displacing it's Mass. And the ball is more dense than the water. The level will rise..

You've got it backwards Mil...

The Density has nothing to do with how much water it displaces when UNDERWATER. A balloon the same size as the cannonball will displace the same amount (VOLUME) of water IF you hold it down just under the surface of the water (let's not complicate things by considering that pressure will cause the balloon to shrink at any real depth which changes things).

However when the cannonball is in the boat then it's WEIGHT is the issue, as the weight causes MORE water to be displaced than would it's VOLUME alone.

Vegas said it right

Since it came back around and no one will read any of the earlier threads I will explain this one again.

The cannon ball will weigh more than the same size ball of water therefore it will push down the boat more than the same volume of water. When you drop that cannon ball over the side it can only displace its size, not its weight. The boat goes up because it lost the weight of the cannon ball which send the water level down. The size (volume) of the cannon ball raises the water level but not as much as it's weight (mass) did when it was forcing the boat down.

If the cannon ball weighed exactly the same as the water it wouldn't change level at all. If it weighed less than water, the water level would rise as the ball would float meaning its volume is displacing less then its mass.

drvsafe
29th April 2010, 23:10
no it wont rize the displacement is the same as if it was in the boat. toss the ball overboard the boat has less displacement, but the ball in the water adds displacement in the water


I like this answer the best...makes the most sense.

chardhin
29th April 2010, 23:22
it will fall as soon as the ball left the hands of the sailor, rise as soon as its submerged in the water and stay exactly the same as it was on the boat

yup.

Bone
29th April 2010, 23:32
Unfortunately I believe there are fatal flaws in both those statements you guys like.


Originally Posted by 68B_Body
no it wont rize the displacement is the same as if it was in the boat. toss the ball overboard the boat has less displacement, but the ball in the water adds displacement in the water

I like this answer the best...makes the most sense.

The bold statement is only correct IF the VOLUME (Size) and the MASS (Weight) are the same (with regards to amount of water they displace), and I believe that is an incorrect ASSumption for a cannonball.



Originally Posted by 68B_Body
it will fall as soon as the ball left the hands of the sailor, rise as soon as its submerged in the water and stay exactly the same as it was on the boat
yup.

Again, see above, I believe this is incorrect when discussing something as dense as a cannonball as it takes MUCH more than it's volume of water to displace it when floating in a boat then it physically displaces when it sinks in the water.

Vegas and Fe Head both explained it well.

Bone
29th April 2010, 23:35
Leave it to Fe Head to answer with the facts

I have not read all the posts but I assume the correct answer has been figured by now.

Which is the water in the lake will be lower by approximately 7 times the volume of the cannon ball.

This is because when the cannon ball is in the boat its weight cause 8 times its volume in water to be displaced by the boat ( Steel is 8 times heavier than water of the same volume). When the cannon ball enters the water it only displaces its own volume no matter what its density is (that is why it sinks) and the other 7 times of its volume are replaced by the boat rising and allowing the level in the lake to drop or become lower by 7 times the cannon ball's volume.

The only way the level would stay the same is if you used a cannon ball with the same density as water which is always 1.
To make the water rise using the cannon ball you would need to forget the boat alltogether and just heave the cannon ball into the water from shore or drop it from a bridge.

Cheers;

Tin Man 2
30th April 2010, 00:02
Okay lets try this, If the Ball was made of material with the Density of a collapsed star its easy to see the outcome. A boat the size of a Freighter would displace a very large amount of water holding the Star Ball afloat, Drop the Star Ball overboard and only its volume would raise the water leval, not its weight.

woodywoodwick
30th April 2010, 00:03
Does the cannon ball create a splash in the water? The splash could cause some water to splash into the boat, ultimately causing the water level to drop because it will be in the boat which equals less water in the lake.

xtremraptor
30th April 2010, 01:08
wow i just got me sum edgamacationing stuff.

i also got a great laugh. if you didn't you should read the whole thread, it's hilarious