Pressure and buoyancy in fluids •  FCQ’s for lecture and

tutorials will be next week. •  Buoyancy force today •  Fluid dynamics on Monday

(along with the loudest demonstration of the semester).

•  Review on Wednesday and Friday next week.

•  Final exam at 7:30am on Tuesday, May 5.

2

Pressure in a liquid

mg

Consider the bottom of a column of water with depth d and cross sectional area A inside a container open to the atmosphere.

d

p0A

pAp0A

pA

When you descend in a liquid, the weight of the liquid above you causes the pressure to increase.

Atmospheric pressure p0 pushes down with force of p0A.

The weight of the column pushes down with force mg. For a liquid with density ρ, m=ρV=ρAd.

Because the liquid is in static equilibrium, the upward force from pressure, pA, must equal the downward forces

pA = p0A+ ρAdg so p = p0 + ρgd (hydrostatic pressure)

3

Clicker question 1 Set frequency to BA

Three vessels are full of the same liquid and open to the same atmosphere. The pressure is measured in each at a distance of 3 m below the surface. What can we say about the pressures? A. only two are the same B. all three are different C. all three are the same 3 m

The hydrostatic pressure is p = p0 + ρgd

It only depends on the pressure on top and the amount of water in a column directly overhead so it is the same for all 3.

4

Pressure in a liquid In the problem, the liquids had the same height because they were filled that way.

If they were all connected (as shown), the liquid levels would have to be the same.

Why? Well, assume the first container had a higher level.

Then, since , pressure at A would be greater than at B. p = p0 + ρgd

A B

Assume a gradual decrease from A to B. Then at any point between them, pressure from the left is greater than from the right resulting in a net force to the right (not equilibrium). Therefore, fluid will flow to the right until equilibrium is reached.

5

Some rules for pressure Anywhere in a connected, static, uniform density fluid, the pressure at a given height is the same.

Pascal’s law: A pressure change at one point in an incompressible fluid appears undiminished at all points in the fluid.

0p

p = p0 + ρgd

p = p0 +FA d

Fp0

p = p0 +FA+ ρgd1

d1 d2

p = p0 + ρgd2

A force F is applied to a piston of area A, increasing the pressure by F/A.

Both pressures at same height so must be the same. Setting equal:

Clicker question 2 Set frequency to BA

Uma Thurman (mass of 60 kg) is standing on a piston, connected as shown to another piston on which a 6000 kg stretch Hummer is resting. How much bigger in area is the piston under the Hummer compared to the one under Uma? Try writing down two expressions for the pressure at the dotted line. A. same B. 10 times C. 100 times D. 1000 times E. 10000 times

pL = p0 +FLAL

On the left side we have FL =mUg

Right side: FR =mHg

pR = p0 +FRAR

p0 +FLAL

= p0 +FRAR

FLAR = FRAL AR =FRFLAL =100 ⋅ALso so

Buoyancy

mfgd1

p0A

pA

p0A+ ρfgAd2

p0A+ ρfgAd1 d2Take a volume of water with density ρf and area A extending from a depth of d2 to d1.

Summing the forces of the free body diagram gives

Fy∑ = p0A+ ρfgAd1 − p0A− ρfgAd2 −mfg= ρfgA(d1 − d2 )−mfg = 0

If we replace the fluid with an object, the only difference is mass.

Fy∑ = ρfgA(d1 − d2 )−mog = 0

Note that A(d1-d2) is the volume Vf of fluid displaced

Upward force equals the weight of displaced fluid ρfgA(d1 − d2 ) = ρfVfg

Archimedes principle A body partially or fully immersed in a fluid feels an upward force equal to the weight of the displaced fluid.

This force is called the buoyant force: FB = ρfVfg

As shown, it is due to the increase of pressure with depth in a fluid.

If the object is fully immersed then the volume of the displaced fluid is equal to the volume of the object: Vf =Vo

Note that volume is related to mass and density: mo = ρoVo

If an object is only partially submerged, the volume of the displaced fluid is less than the volume of the object: Vf <Vo

Buoyancy example A 2 cm by 2 cm by 2 cm cube of iron (ρ=8 g/cm3) is weighed with the iron outside, half in and fully in the water, as shown in the diagram. What is the measured weight in each case?

Iron mass: mo = ρoVo = 8 g/cm3 ⋅8 cm3 = 64 g = 0.064 kg

Out of the water: mog

T T −mog = 0T =mog = 0.064 kg ⋅10 m/s2 = 0.64 N

so

In the water:

FB +T −mog = 0 T =mog−FBso mogTFB

N 08.0m/s 10cm 8g/cm 1 233ff =⋅⋅== gVFB ρ

so T =mog−FB = 0.64 N− 0.08 N = 0.56 N

½ in the water:

FB +T −mog = 0 T =mog−FBso mogTFB

FB = ρfVfg =1 g/cm3 ⋅ 4 cm3 ⋅10 m/s2 = 0.04 Nso T =mog−FB = 0.64 N− 0.04 N = 0.60 N

Archimedes crown Archimedes lived from 287 BC to 212 BC. King Hiero II gave a goldsmith gold to make a crown. When the crown was made, it was found to weigh the same as the gold given but the king suspected that silver had been mixed in so he asked Archimedes to find out (without damaging the crown).

One day while getting into his tub, Archimedes noticed water was displaced in an amount equal to his volume and figure out a way to determine the density of the crown. He ran naked through the streets of Syracuse shouting ‘Eureka’, Greek for “I found it”. What had he found?

Balance the weight of the crown with pure gold in air.

Submerge both in water. If the crown is less dense than the gold, it must have a larger volume (in order to have the same weight).

In that case, the crown will displace more water than the gold, leading to a larger buoyant force on the crown, and a smaller apparent weight.

Clicker question 3 Set frequency to BA

A. rises. B. falls. C. stays the same.

An ice cube is floating in a glass of water. As the ice cube melts, the level of the water…

Since the ice cube is in equilibrium, the weight of water displaced is equal to the weight of the ice cube.

The buoyant force is always equal to the weight of the liquid displaced by the object.

When the ice cube melts it becomes water with the same weight as the ice cube so real water takes the place of the displaced water.