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PHYSICS QUESTION -- This is an extremely sophisticated physics

Customer Question

PHYSICS QUESTION -- This is an extremely sophisticated physics question; only answer if you are an EXPERT IN PHYSICS ------ A person in a boat floating in a small pond throws an anchor overboard. Does the level of the pond rise, fall or remain the same? ------

Please give an answer with formulas and numbers so i can understand the concept quantitatively as well as qualitatively.

I know the answer off the top of my head that the water level will decrease.

abozer :

Here is the proof.

abozer :

Let the volume below the level of water be our reference in order to decide if the water level will increase or decrease.

abozer :

Initial volume below the water level is the summation of the volume of water and the volume of the boat below the water level.

abozer :

So the initial volume is,

abozer :

Vi = Vwater + Vship1

abozer :

sorry, let's call it boat,

abozer :

Vi = Vwater + Vship1

abozer :

Vi = Vwater + Vboat1

abozer :

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.

abozer :

So the final volume can be written as follows,

abozer :

Vf = Vwater + Vboat2 + Vanchor

abozer :

We do not need to include the volume of the anchor in the initial volume because it is inside the boat.

abozer :

Now let's use the lifting force formula,

abozer :

F = V(below water level) * d(density of water)

abozer :

So for the initial case,

abozer :

The force is lifting both the boat and the anchor,

abozer :

Vboat1 * dw = mboat * g + manchor * g

abozer :

here g is the gravitational acceleration,

abozer :

from this equation we can get,

abozer :

Vboat1 = mboat * g / dw + manchor * g/dw

abozer :

Substitute this in the initial volume formula we got above,

Customer:

I don't know this formula F = V(below water level) * d(density of water)

Customer:

where is gravity?

abozer :

The force applied to the boat by the water,

abozer :

is equal to the volume of boat below water level times the density of water.

abozer :

The force must be equal to the weight of the boat and anchor,

abozer :

Weight is found by multiplying the mass and the gravity,

Customer:

this book I use doesn't provide that formula so I can't use it to solve the problem

abozer :

So the following equation must hold for the boat to float,

abozer :

Vboat1 * dw = mboat * g + manchor * g

abozer :

The symbols may be different,

abozer :

but that is the formula,

abozer :

Please check your book to see if there is a similar formula to find the force on an object in water.

Customer:

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

abozer :

ok. please post it.

Customer:

The anchor is in the raft and the raft is floating so F_{g }= F_{b }and p_{o}V_{o} = p_{f}V_{f}

The density of iron (the anchor) is greater than the density of water so

V_{o} < V_{f }

When the anchor is tossed overboard, the volume of water displaced by the anchor must now be equal to the volume of the anchor. Since the volume of the anchor is constant the volume of the water must now be less. In other words, since V_{f}is now equal to V_{o }then V_{f }must have decreased. Hence the water level falls.

Customer:

I let po = 5 and Vo = x , then pf = 1 and Vf = 10 so Vo = 5

Customer:

sorry, Vo = 2

abozer :

Could you tell me what the symbols, Fg, Fb, po, Vo, pf, and Vf are,

Customer:

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.

Customer:

Then since Vf is now 2 -- no longer 10 -- the volume of water has decreased and the pond level decreases.

Customer:

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

abozer :

ok. Let's see.

abozer :

So you are using the bouyant force.

Customer:

yes

abozer :

The bouyant force is equal to the mass of the displaced water.

abozer :

right?

Customer:

yes

abozer :

That's exactly what my formula means.

abozer :

Fbouyant = Vboat1 * dwater

abozer :

Vboat1 is,

abozer :

the volume of the displaced water in the initial case.

abozer :

right?

Customer:

no Fbouyant = Vwater * density of water * g

Customer:

Fg = V boat * density boat * g

abozer :

yes, sorry. there is a g,

abozer :

So in your approach,

abozer :

you are not considering the boat itself.

abozer :

and solving it just by using the anchor.

Customer:

right

abozer :

that also works but it is better to add it into analysis and it will cancel out later on.

Customer:

right i understand that

abozer :

and you say the volume of water changes but it does not.

abozer :

the level of water changes.

abozer :

Let's solve this problem algebraically,

abozer :

Here are my parameters,

abozer :

g: gravitational acceleration

abozer :

mboat = mass of boat

abozer :

manchor = mass of anchor

abozer :

Vboat1 = initial volume of the displaced water

abozer :

Vboat2 = final volume of the displaced water

Customer:

i don't want to use mass --- instead we use density x volume

abozer :

Vanchor: volume of the anchor

abozer :

ok.

abozer :

lets use density x volume

abozer :

dw : density of water

abozer :

da = density of anchor

abozer :

Initially the bouyant force is Vboat1*dw*g

abozer :

The total weight it carries is Wboat + da * Va * g

abozer :

Wboat is the weight of the boat, it will cancel out later on,

Customer:

slow down and wait

abozer :

let me know if you have any questions.

Customer:

lets go back to my original premise:

Customer:

The anchor is in the raft and the raft is floating so F_{g }= F_{b }and p_{o}V_{o} = p_{f}V_{f}

abozer :

ok. let's stick with your method if you want.

Customer:

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

abozer :

this equation is not right.

abozer :

density of anchor times volume of anchor gives mass of anchor.

abozer :

density of water times density of water gives mass of water.

abozer :

why do you assume they are equal.

Customer:

for floating objects F_{g }= F_{b }and p_{o}V_{o} = p_{f}V_{f}

Customer:

i left out the g

abozer :

anchor is not floating itself

Customer:

on both sides since it cancels

abozer :

it would sink if the boat is not there

Customer:

oh, that's right -- I messed up then

abozer :

I think we need to follow a logical way to solve the problem.

Customer:

ok

abozer :

do not ignore the boat.

Customer:

ok

abozer :

Initially the bouyant force is Vboat1*dw*g

The total weight it carries is Wboat + da * Va * g

abozer :

let's make clear this part.

abozer :

do you have any question about these two.

Customer:

alright slow down

abozer :

These two forces must be equal so that the boat and the anchor floats.

Customer:

I think of the bouyant force as the Vwater*dw*g

abozer :

?

abozer :

you can.

abozer :

but Vwater here means the displaced water when the boat is put into the pond.

abozer :

not the overall volume of the water.

abozer :

just the amount displaced.

Customer:

ok when you say Vboat1 I can think of this as Vwater-displaced-by-the-boat, right

abozer :

correct.

abozer :

so the initial equation is,

Customer:

ok let's go back to "Initially the bouyant force is Vboat1*dw*g"

abozer :

Vboat1*dw*g = Wboat + da * Va * g

Customer:

OK, total weight of boat Wboat = boat + anchor (which is da*Va*g)

abozer :

correct

Customer:

Ok hold on

Customer:

so now bouyant force = gravitational force

Customer:

proceed

abozer :

yes.

abozer :

Let's write a similar equation for the final case,

abozer :

Anchor is not in the boat anymore.

abozer :

Vboat2*dw*g = Wboat

Customer:

good

abozer :

Now subtract the second equation from the first one side by side,

OK, sorry to be so tardy in responding, but hope the offer still stands. Here's the question:

When an ice cube floating on the water's surface melts, does the water level rise, fall or stay the same?

I think it stays the same since the product of d(water) x v(water) = d(ice) x v(ice) in the beginning and must also be true at the end (since the ice turned into water can be considered to be "floating" on the water). Hence no change. Is this correct?

The answer to the question is "No Change". But your solution approach is not correct. You are calculating the mass of ice and water and saying that these are equal. This is conservation of mass. This would give you erronous results if the object was hollow.

A more general approach is by using the buoyant force.

Density of ice is around 0.9 g/cm^3 and that of water is 1 g/cm^3.

This means 9/10 of the volume of ice will be below the water level when floating.

When ice melts its volume decreases by a factor of 9/10 and exactly fills the volume below the water level.

So the water level doesn't change.

Let me show you how we know the following "This means 9/10 of the volume of ice will be below the water level when floating".

Buoyant force = ρ_water x Volume of displaced water x gravity

Buoyant force also equals the weight of the object so that it floats.

Buoyant force = m x gravity

Here we can replace m by ρ_object x V object

So,

Buoyant force = ρ_object x V object x gravity

Now, substitute this into the first equation,

ρ_object x V object x gravity = ρ_water x Volume of displaced water x gravity

Divide both sides by gravity,

ρ_object x V object = ρ_water x Volume of displaced water

Divide both sides by ρ_water x V object,

ρ_object / ρ_water = Volume of displaced water / V object

This is a general equation,

For your question,

Substitute,

ρ_water = 1 and ρ_object = 0.9

0.9/1 = Volume of displaced water / V object

Volume of displaced water is the volume of ice cube that is below the water level.

0.9 = 9/10

So we can say,

Volume of ice below water level is 9/10 of its total volume.

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