Fluid Mechanics - Floating Bodies

[blazin247nc]

Fluid Mechanics -- Floating Bodies

Homework Statement

A solid block with specific gravity 0.9 floats so that 75% of the body is submerged in water and the other 25% sits in an unknown fluid layered above the water. Find the specific gravity of the unknown fluid.

density of water = 1000 kg/m^3
density of the block = 900 kg/m^3 (from SG)

Homework Equations

For floating bodies W = Fb

W = d*g*V
Fb = d*g*Vdisplaced

The Attempt at a Solution

W = Wblock + Wfluid
Wblock = 900*9.8*(a+b) ----->length and width cancel when equations are set equal
Fb = 1000*9.8*b -------->a = height of block above water (.25) b= below water (.75)

This is where I get lost. I know I need to incorporate the weight of the unknown fluid pushing down on the block, but does the weight of the fluid only include the displaced weight?

where:

W = 900*9.8*(a+b) + d-fluid*9.8*a ? But when I do this I get a negative density :(

 

[alphysicist]

Hi blazin247nc,

Homework Statement

A solid block with specific gravity 0.9 floats so that 75% of the body is submerged in water and the other 25% sits in an unknown fluid layered above the water. Find the specific gravity of the unknown fluid.

density of water = 1000 kg/m^3
density of the block = 900 kg/m^3 (from SG)

Homework Equations

For floating bodies W = Fb

W = d*g*V
Fb = d*g*Vdisplaced

The Attempt at a Solution

W = Wblock + Wfluid
Wblock = 900*9.8*(a+b) ----->length and width cancel when equations are set equal
Fb = 1000*9.8*b -------->a = height of block above water (.25) b= below water (.75)

This is where I get lost. I know I need to incorporate the weight of the unknown fluid pushing down on the block,

No, that is not right since you are already using the buoyant force from the water. There are two ways you could deal with the fluid forces acting on the objects in these types of problems.

The first way is to find the force pushing down on the top of the object due to the liquid pressure, and the force pushing up on the bottom of the object due to the water pressure. In this problem, those sum of those two forces (one is positive, one is negative) would have to equal the weight of the object.

The second way (which is normally the best way) is to think in terms of the buoyant forces from both of the liquids. The buoyant forces are upwards, because the pressure is greater the deeper you go.

These are alternate approaches; the problem in your approach is that you are mixing them and using the buoyant force of the water, and the force downard from the unknown liquid. The total buoyant force already comes from adding together the individual forces on the top and bottom of the object. Does that help?

but does the weight of the fluid only include the displaced weight?

where:

W = 900*9.8*(a+b) + d-fluid*9.8*a ? But when I do this I get a negative density :(

 

[blazin247nc]

OK, I see what I was doing wrong now. I got 600 kg/m^3 as the density of the mystery fluid. Seems like a more reasonable answer than -200 :P