# Archimedes principle

1. Oct 27, 2006

### gunblaze

Alright.. I have a problem.

I have 2 liquids, mercury and water together in the beaker.

Assuming both liquids are non mixable, the water, with a lower density, will be at the top while mercury will be at the bottom.

Now, I push a block into the water and it floats at halfway between the mercury and the water. I need to find the height of X1, which is the height of the block submerged in the mercury.

Now, my friend's solution to find X1 is to equate the upthrust by the mercury to the weight of the block and the upthrust by the water. since the block is floating.

However, the solution gives the upthrust by both the water and the mercury = to the weight of the block.

The difference here is that my friend actually think that the upthrust by the water is acting downwards same as the weight while the upthrust by the mercury is acting upwards.

Well, for me, I told my friend that if we were to think of archimedes principle, upthrust is equals to the weight of the water displaced by the block and so, the solution is right.

However, i can't deny the fact that my friend is also right to a certain extent. Since the pressure by the water molecules at the top will exert a downward pushing force on the block.

So who's right, who's wrong. Appreciate it if anyone can help clarify this.

2. Oct 28, 2006

### OlderDan

Your block will experience a downward force on top from the water, and an upward force at the bottom from the mercury. In each case, the force will be the pressure in the fluid times the horizontal area of the block. I am assuming all other surfaces are vertical, so there are no other vertical forces acting. This is no different than the situation that exists when the block is submerged in only one liquid. If the upward force from the liquid below is not sufficient to overcome the downward force fom the liquid above plus the weight of the object, the object will sink. If the upward force is in excess of the downward force plus weight, the object will rise to the surface.

Archimedes principle is applicable as long as the object is surrounded by fluids. A block that floats partially submerged in water is being pushed down by the air and upward by the water. Technically, to calculate the buoyant force due to this pressure difference you should calculate the weight of both the water and the air displaced by the object. Since the density of the air is so small compared to that of water, the weight of the air is ignored.

For your object, you cannot ignore the weight of the displaced water compared to the weight of the displaced mercury. You must add the two weights to find the buoyant force. The object will float partially submerged in mercury at the point where the total weight of displaced liquid equals the weight of the object.

The reason Archimedes principle is valid is because the pressure difference between two levels in a fluid or multiple fluids is just the weight of a column of fluid(s) divided by the area at the base of the column. There is therefore just the right relationship between pressure difference and weight of displaced fluid to account for the difference between the upward and downward forces on the object from fluid pressure.

3. Oct 28, 2006

### Andrew Mason

I am having trouble understanding what this means. Can you pin point where the block is exactly? Is it touching the mercury? Is is completely submerged in the water?

AM

4. Oct 28, 2006

### gunblaze

So, is the upthrust by the water acting downwards?..

or upwards?

I still dun get it.

5. Oct 28, 2006

### gunblaze

Erm. Okay, to simplify things, its a beaker of half mercury and water. Water above and mercury below. The block is submerged halfway between the liquids. The upper half of the block is in the water and the lower half of the block is in the mercury.

Tried uploading the pic but failed. Will try again, sorry for inconvenience caused.

6. Oct 28, 2006

### OlderDan

Heights are measured relative to the bottom of the beaker.
Let
ρ_w = the density of water
ρ_m = the density of mercury
h_t = the height of the top of your block
h_b = the height of the bottom of your block
h_w = the height of the top of the water layer in your beaker
h_m = the height of the top of the mercury layer in your beaker
A = area of the top or bottom of the block
d_w = h_t - h_m = height of displaced water layer
d_m = h_m - h_b = height of displaced mercury layer
V_w = A*d_w = Volme of water displaced
V_m = A*d_m = Volme of mercury displaced
y = an arbitrary position above the bottom of the beaker

Assume the block is floating with horizontal top and bottom surfaces (of area A) and all other surfaces vertical. The vertical forces will cancel. The pressure at the top surface of the water (y = h_w) is 1atm. The pressure at other points in the fluid is

p = ρ_w*g*(h_w - y) + 1atm <== pressure anywhere in the water

p_top = ρ_w*g*(h_w - h_t) + 1atm <== pressure at the top of the block

p = ρ_w*g*(h_w - h_m) +1 atm <== pressure at the water/mercury boundary

p = ρ_w*g*(h_w - h_m) + ρ_m*g*(h_m - y) +1 atm <== pressure anywhere in the mercury

p_bottom = ρ_w*g*(h_w - h_m) + ρ_m*g*(h_m - h_b) +1 atm <== pressure at the bottom of the block

Water is pushing downward on the top of the block, but it is also pushing downward on the mercury, increasing the pressure in the mercury. The force on the bottom of the block is greater with the water present than it would be if the water was not there.

The pressure at the top of the block times the area of the top surface of the block is the force acting downward on the top of the block.

The pressure at the bottom of the block times the area of the bottom surface of the block is the force acting upward on the bottom of the block.

The net force acting upward on the block is

F = A{p_bottom - p_ top}

F = A{ρ_w*g*(h_w - h_m) + ρ_m*g*(h_m - h_b) +1 atm - [ρ_w*g*(h_w - h_t) + 1atm]}

F = A{ρ_w*g*(h_w - h_m) + ρ_m*g*(h_m - h_b) - ρ_w*g*(h_w - h_t)} <== 1atm cancels

F = A{ρ_w*g*[(h_w - h_m) - (h_w - h_t)] + ρ_m*g*d_m } <== regroup and substitute d_m

F = A{ρ_w*g*(h_t - h_m) + ρ_m*g*d_m} <== h_w cancels

F = A{ρ_w*g*d_w + ρ_m*g*d_m} <== substitute d_w

F = ρ_w*g*V_w + ρ_m*g*V_m <== identify volumes

F = Weight of displaced water + Weight of displaced mercury <== Archemedes principle

Last edited: Oct 28, 2006