Pressure and Buoyancy

The building blocks of ordinary matter are atoms and molecules.  Molecules are two or more atoms held together by a chemical bond.  Atoms themselves are made up of more fundamental particles, i.e. protons, neutrons, and electrons.  In a solid, the atoms and molecules are densely packed and held in place by intermolecular forces.  The atoms in a solid can be modeled as being held together by tiny springs that permit them to vibrate back and forth about their equilibrium position, but not to exchange positions with other atoms.  Solids are nearly incompressible.  In a liquid the atoms and molecules are also densely packed.  They cannot easily escape from one another, but they are free to move with respect to each other.  Liquids are nearly incompressible.  In a gas, intermolecular forces are weak and short ranged, and the atoms and molecules can move about nearly independently.  Gases are compressible.  Gases and liquids are fluids, i.e. collections of atoms or molecules that are free to move with respect to each other.

Assume we have a collection of gas molecules in a container in gravity-free space.  Then each molecule moves along with constant velocity in a straight line, until it hits another molecule, or a container wall.  A collision between two molecules is similar to a collision between two balls.  The molecules exchange momentum, but the total momentum of the two molecules is conserved.  When a molecule hits a wall, it bounces back.  Its momentum changes.  To change the molecule's momentum, the wall must exert a force on the molecule.  Newton's third law tells us that the molecule exerts a force on the wall.  The greater the number of molecules hitting a wall, the greater is the force on the wall.  In a container with different size walls, the bigger walls will receive more hits that the smaller walls and therefore experience a greater force.  We define the pressure P in the container as the magnitude of the normal force F on a wall divided by the surface area A of the wall.  In gravity free space the pressure is the same everywhere inside the container.

P = F/A

In SI units, the units of pressure are N/m2 = Pa (pascal).

The faster the molecules move in the container, the greater is the change in momentum when they bounce off a wall, and the more often do they hit the walls.  

Assume a molecule moves horizontally with speed v back and forth between two infinitely-massive walls, which are a distance L apart.  When it hits the right wall its momentum changes from p1 = +mv to p2 = -mv.  The change in the molecule's momentum is Dpmol = p2 - p1 = -2mv.  The time interval between successive hits on the right wall is Dt = 2L/v.  So the average force the wall exerts on this molecule is Fmol = Dpmol/Dt = -2mv/(2L/v) = -mv2/L.  By Newton's third law, the average force that the molecule exerts on the wall is Fwall = mv2/L, it is proportional to the square of the speed of the molecule.

The force exerted by a gas on the walls of a container is proportional to the square of the speed of the molecules, the pressure is therefore proportional to the kinetic energy of the molecules.  Not all the molecules have the same kinetic energy.  The pressure in a container is proportional to the average kinetic energy of the molecules.  If all the molecules in a container would be at rest, their kinetic energy would be zero and the pressure would be zero.  The pressure is also proportional to the particle density rparticle of the molecules in the container.  The particle density is the number of particles (molecules) in the container, divided by the volume of the container.  If we double the number of molecules in the container, the pressure will double.

The density r of a substance is defined as its mass per unit volume, r = M/V.

Our general definition of the average pressure exerted by a normal force F on an area A is P = F/A.  If the pressure varies from point to point, then the pressure at a specific point exerted on a surface element dA is given by P = dF/dA, where dF is the normal force acting on dA.

Problems:

A 50kg woman balances on one heel of a high-heel shoe.  If the heel is circular with radius 0.5cm,  what pressure does she exert on the floor?
Solution:
P = F/A = (50kg9.8m/s2)/(p(0.005m)2) = 6.2´106N/m2 = 6.2MPa
A king orders a gold crown having a mass of 0.5kg.  When it arrives from the metal smith, the volume of the crown is found to be 185 cm3.  Is the crown made of solid gold?
Solution:
The density of the crown is (0.5kg/185cm3)´(106cm3/1m3) = 2.7´103kg/m3.  The density of gold is 1.9´103kg/m3.  The crown is not made of solid gold.

Liquids in a constant gravitational field

Consider a large pool of water on the surface of the earth and a box-shaped volume of water at some depth in the pool.  Imagine it enclosed by some weightless container.

box.gif (2354 bytes)

The volume of water is in equilibrium and stays in place.  It does not rise and it does not fall.  The net force on it must be zero.  The vertical component of the net force is

Fnet = PbottomA - PtopA - Mg = 0.
Fnet = PbottomA - PtopA - rhAg = 0.
Pbottom - Ptop = rhg.

The pressure in the pool increases with depth.  If we let h denote the vertical distance of a point below the surface of the water, then we can write the pressure at this point as

P = P0 rgh.

P is the pressure at depth h and P0 is the pressure at the surface.  Very often this pressure is atmospheric pressure.  The atmospheric pressure at sea level at room temperature is approximately

1 atmosphere = 101 kPa = 14.7 pounds per square inch (psi).

The pressure at a point below the surface of a liquid in a constant gravitational field depends only on the depth of that point and the pressure at the surface.  Any change in the pressure at the surface is therefore transmitted to every point in the liquid.  This is called Pascal's law.

Problems:

Determine the absolute pressure at the bottom of a lake that is 30m deep.
Solution:
P = P0 + rgh.  P0 = 101kPa = 1atm.  rwater = 1000kg/m3rgh = 1000´9.8´30N/m2 = 294kPa » 3atm.  P » 4atm.  The water pressure increases by approximately 1atm for every 10m increase in depth.

 

What is the hydrostatic force on the back of Grand Coulee Dam if the water in the reservoir is 150m deep and the width of the dam is 1200m.

Solution:
Let us denote the distance below the surface by y.  The pressure on the side of the dam facing the water at depth y is P0 + rgy.  The force on a surface element of width w and height dy at depth y is 
dF = (P0+rgy)dA = (P0+rgy)wdy.  
The total force on this side of the dam is  Image631.gif (1674 bytes), outward, if all quantities are measured in SI units.  The total force on the side of the dam facing air is wP0150, inward.  The net force on the dam is F = wrg1502/2, outward.  With w = 1200m and r = 1000kg/m3, F = 1.32´1011N.

 

The figure below shows an aerial view from directly above two dams.  Both dams are equally long and equally deep.  The dam on the left holds back a very large lake and the dam on the right holds back a narrow river.  Which dam has to be build more strongly?

Solution:
The net force on both dams is the same, the strength of both dams must be the same.

Buoyancy

Now consider again the box-shaped volume of water in equilibrium at some depth in the pool.  The upward force provided by the surrounding water must exactly balance the force of gravity acting on the water in the box.  The upward force provided by the surrounding water must be equal to the weight of water in the box.

If we replace the volume of water with a box of the same shape containing some other material, then the net upward force provided by the surrounding water does not change.  It depends only on the difference in the pressure at the top and at the bottom of the box.  But the weight of the box changes, and therefore the net force on the box changes.  If the weight is greater than that of the corresponding volume of water, the net force is downward and the box will accelerate downward and fall.  If the weight is less than that that of the corresponding volume of water, the net force is upward and the box will accelerate upward and rise.  This is Archimedes' principle.  It holds for all fluids, i.e. it holds for all liquids and gases.

An object partially or wholly immersed in a gas or liquid is acted upon by an upward buoyant force B equal to the weight w of the gas or liquid it displaces.

B = w

Links:

Buoyancy Brainteasers
Buoyant Force

Problems:

A Styrofoam slab has a thickness h and a density robject.  What is the area of the slab, if it floats with its top surface just awash in fresh water when a swimmer of mass m is on top?
Solution:
When an object floats, the magnitude of the buoyant force is equal to the magnitude of its weight.  The magnitude of the buoyant force is equal to the magnitude of the weight wwater of the displaced water.  wwater = rwaterAhg, where A is the area of the slab. 
The the magnitude of the weight of the object is  wobject = robjectAhg + mg.  
We need wwater = wobject.
rwaterAhg = robjectAhg + mg.
rwaterAh - robjectAh = m.
A = m/(rwaterh - robjecth)

 

A frog in a hemispherical pod finds that he just floats without sinking into a sea of blue-green ooze with density 1.35 g/cm3.  If the pod has radius 6cm and negligible mass, what is the mass of the frog?

Solution:
When an object floats, the magnitude of the buoyant force is equal to the magnitude of its weight.  The magnitude of the buoyant force is equal to the the magnitude of the weight wliquid of the displaced liquid.
wliquid = rliquidVg.
The volume V of the displaced liquid is the volume of one half sphere, 
V = 2pr3/3 = 2p(6cm)3/3 = 452cm3.
The magnitude of the weight of the object is wobject = mfrogg.  (We are neglecting the weight of the air-filled pod.)
rliquidVg = mfrogg. 
mfrog = 1.35(g/cm3)452cm3 = 610g.

 

A barge is carrying a load of gravel along a river.  It approaches a low bridge, and the captain realizes that the top of the pile of gravel is not going to make it under the bridge.  The captain orders the crew to quickly shovel gravel from the pile into the water.  Is this a good decision?
Solution:

Assume an object has a weight w and a density r greater than that of water.  When the object floats in a boat, the weight of the water displaced because of this object is equal to the weight of the object.  When the object sinks when thrown overboard, the weight of the displaced water is less than the weight of the object.  An object with r > rwater of a given weight displaces more water when floating than when being submerged.  When a given volume of gravel is shoveled into the water, a larger volume of the ship will rise out of the water.  But that does not necessarily mean that the maximum height h of the load above the water's surface increases.  This maximum height h depends on how the load is distributed.  If the load is evenly spread over the entire deck of the ship, the shoveling sand into the water is not a good idea.  But if the load is a pyramid-shaped pile, then removing the top of the pyramid is a good idea.

Exercise  (You can earn up to 5 points extra credit by completing this exercise.)

Link:

The Cartesian Diver

(This is an experiment you can try at home.)

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Pressure and Buoyancy
Buoyancy

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