I know the answer off the top of my head that the water level will decrease.
Here is the proof.
Let the volume below the level of water be our reference in order to decide if the water level will increase or decrease.
Initial volume below the water level is the summation of the volume of water and the volume of the boat below the water level.
So the initial volume is,
Vi = Vwater + Vship1
sorry, let's call it boat,
Vi = Vwater + Vboat1
I used the index 1 next to the boat because the Volume of boat below the water level will change after the anchor is removed.
So the final volume can be written as follows,
Vf = Vwater + Vboat2 + Vanchor
We do not need to include the volume of the anchor in the initial volume because it is inside the boat.
Now let's use the lifting force formula,
F = V(below water level) * d(density of water)
So for the initial case,
The force is lifting both the boat and the anchor,
Vboat1 * dw = mboat * g + manchor * g
here g is the gravitational acceleration,
from this equation we can get,
Vboat1 = mboat * g / dw + manchor * g/dw
Substitute this in the initial volume formula we got above,
I don't know this formula F = V(below water level) * d(density of water)
where is gravity?
The force applied to the boat by the water,
is equal to the volume of boat below water level times the density of water.
The force must be equal to the weight of the boat and anchor,
Weight is found by multiplying the mass and the gravity,
this book I use doesn't provide that formula so I can't use it to solve the problem
So the following equation must hold for the boat to float,
The symbols may be different,
but that is the formula,
Please check your book to see if there is a similar formula to find the force on an object in water.
no, this isn't working. let's try something different. I got the answer you proposed in the beginning doing it my way. I want you to check my work and reasoning and see if it makes sense as a quantitative solution to this problem
ok. please post it.
I let po = 5 and Vo = x , then pf = 1 and Vf = 10 so Vo = 5
sorry, Vo = 2
Could you tell me what the symbols, Fg, Fb, po, Vo, pf, and Vf are,
Once the anchor is thrown overboard Vo is still 2 (constant volume) but Vf must now ALSO be 2 since the anchor is submerged in the water and the anchor and the volume of water displaced must be the same.
Then since Vf is now 2 -- no longer 10 -- the volume of water has decreased and the pond level decreases.
Fg is weight of the object, Fb is bouyant force, po is density of the anchor (iron, simplified), pf is the density of the water (simplified), Vo is the volume of the object, Vf is the volume of the water
ok. Let's see.
So you are using the bouyant force.
The bouyant force is equal to the mass of the displaced water.
That's exactly what my formula means.
Fbouyant = Vboat1 * dwater
the volume of the displaced water in the initial case.
no Fbouyant = Vwater * density of water * g
Fg = V boat * density boat * g
yes, sorry. there is a g,
So in your approach,
you are not considering the boat itself.
and solving it just by using the anchor.
that also works but it is better to add it into analysis and it will cancel out later on.
right i understand that
and you say the volume of water changes but it does not.
the level of water changes.
Let's solve this problem algebraically,
Here are my parameters,
g: gravitational acceleration
mboat = mass of boat
manchor = mass of anchor
Vboat1 = initial volume of the displaced water
Vboat2 = final volume of the displaced water
i don't want to use mass --- instead we use density x volume
Vanchor: volume of the anchor
lets use density x volume
dw : density of water
da = density of anchor
Initially the bouyant force is Vboat1*dw*g
The total weight it carries is Wboat + da * Va * g
Wboat is the weight of the boat, it will cancel out later on,
slow down and wait
let me know if you have any questions.
lets go back to my original premise:
The anchor is in the raft and the raft is floating so Fg = Fb and poVo = pfVf
ok. let's stick with your method if you want.
po is density of anchor and pf is the density of the water. Vo is the volume of the anchor and Vf is the volume of water displaced
this equation is not right.
density of anchor times volume of anchor gives mass of anchor.
density of water times density of water gives mass of water.
why do you assume they are equal.
for floating objects Fg = Fb and poVo = pfVf
i left out the g
anchor is not floating itself
on both sides since it cancels
it would sink if the boat is not there
oh, that's right -- I messed up then
I think we need to follow a logical way to solve the problem.
do not ignore the boat.
let's make clear this part.
do you have any question about these two.
alright slow down
These two forces must be equal so that the boat and the anchor floats.
I think of the bouyant force as the Vwater*dw*g
but Vwater here means the displaced water when the boat is put into the pond.
not the overall volume of the water.
just the amount displaced.
ok when you say Vboat1 I can think of this as Vwater-displaced-by-the-boat, right
so the initial equation is,
ok let's go back to "Initially the bouyant force is Vboat1*dw*g"
Vboat1*dw*g = Wboat + da * Va * g
OK, total weight of boat Wboat = boat + anchor (which is da*Va*g)
Ok hold on
so now bouyant force = gravitational force
Let's write a similar equation for the final case,
Anchor is not in the boat anymore.
Vboat2*dw*g = Wboat
Now subtract the second equation from the first one side by side,
Vboat1*dw*g - Vboat2*dw*g = Wboat + da*Va*g - Wboat
simplify by removing Wboat,
just to be sure: Vboat1 is water displaced by boat with anchor and Vboat2 is water displaced by boat without anchor, correct?
Vboat1*dw*g - Vboat2*dw*g = da*Va*g
don't get it -- how do we justify removing Wboat
see the right hand side of the equation,
ok stupid of me i see
Wboat is added and subtracted. So cancels,
divide both sides of the equation by g.
g's cancel out,
Vboat1*dw* - Vboat2*dw* = da*Va*
Vboat1*dw - Vboat2*dw = da*Va
Now divide both sides by dw,
Vboat1 - Vboat2 = da*Va/dw
Vboat1 - Vboat2 = da*Va*g/dw -- right
Vboat1*dw*g - Vboat2*dw*g = da*Va*/dw -- this is better
sorry, should be Vboat1 - Vboat2 = da*Va*/dw -- right
shit i forgot what Va is
Now remember that the anchor is in the water and it will also displace water.
Va is the volume of anchor.
volume of anchor, of course -- proceed
displacement of water by anchor is,
sorry just Va,
water displaced by the anchor is Va.
equal to the volume of the anchor
let's go back to Va*dw*g
i lost the connection to what we're doing and where we're at
We were trying to find the difference between the initial case and the final case, in terms of the amounts of water displaced.
ok final case is when we throw the anchor in the water
so ... hang on, let me think about this
Vboat1 - Vboat2 is the difference,
now the anchor is in the water and the volume of the anchor must equal the volume of the water -- but what does that get us
we need to add that to our final result.
Vboat1 - Vboat2 gave us a positive value. This means that the initially displaced water was greater than the final case.
So the water level must decrease. However we need to take into account the volume of the anchor in the water, too.
That will increase the level.
back a little.
Vboat1 - Vboat2 = da*Va*/dw
da is way greater than dw,
like a 1000 times,
so the volume of the water displaced by the anchor must be A LOT less and so it doesn't really affect the final result with any significance ???
so we can ignore it
The final result without ignoring is,
Decrease in the water level = da*Va*/dw - Va
The second term is insignificantly small
excellent explanation -- but I need to copy this down and i don't know how to do that with this cumbersome interface. any suggestions
I will do it for you.
please keep everything so i can follow both are thought processes
can you email it to me
I cannot email but I will put a link here so you can download.
Click the above link.
go it -- thanks so much for your patience -- will include bonus -- excellent instruction