The density of a material is defined as the mass per unit volume.


density  =  r  =  mass/Volume          (the Greek letter rho)


        The units are                      =  kilograms/meter3  =  kg/m3


It’s a measure of how tightly the atoms of a material are packed.   It has nothing to do with the hardness of the material.




Density (kg/m3)














        The specific gravity of a material is the ratio of its density to that of water.  For example, the specific gravity of aluminum would be 2.7.  This number is dimensionless.


        The pressure, P, is defined as the ratio of force to area:



        The units of pressure are:  Newtons/meter2  = N/m2 = Pascals  =  Pa




        A hammer supplies a force of 700 N.  The hammer head has an area of 7.1 x 10-4 m2.  What is the pressure?


                                                P  =  F/A

                                                P  =  (700 N/7.1 x 10-4 m2)

                                                P  =  9.86 x 105  N/m2

                                                P  =  9.86 x 105  Pa





        If a fluid is at rest, then all points at the same depth must be at the same pressure (otherwise it would be moving!).  However the pressure WILL vary with depth: it will have the weight of the fluid on top of it.


P = pressure,  Po = pressure due to the air (atmospheric pressure)



        Because the fluid is at rest, the net force will be  _________ ?


So we can write:


                                                Fnet  =  0

                                   PA – mg – PoA  =  0


However we know that:             

Mass  =  density x volume  =  density x (area x height)


     M  =  rV  =  rAh   


So we can substitute:


                        PA – (rAh)g - PoA  =  0


cancel out the Area, A:


                                P – (rh)g - Po  =  0


and the pressure at any depth will be:


                                P  =  Po + rgh



In words this says: the pressure at a depth ‘h’ will be the atmospheric pressure(14.7 lbs/in2) + (rgh)


Example  1:  Calculate the pressure at a depth of 3.0 m in a swimming pool.

rwater  =  (1000 kg/m3)


P  =  Po + rgh


P  =  (1.013 x 105 Pa)  +  (1000 kg/m3)(9.8 m/s2)(3.0 m)


P  =  (1.013 x 105 Pa)  +  2.94 x 104  Pa


P  =  1.31 x 105 Pa  =  131,000 Pa  =  19 lb/in2.



at a depth of 30 m (100 ft) the pressure would be:             57 lb/in2.


at a depth of 300 m (1000 ft) the pressure would be:         440 lb/in2.


Example 2:

You are a CIA operative in the jungles of Burma escaping from a rebel force.  You see a stream and reeds and you cut off a reed, jump into the stream and lie about half a meter under water breathing thru the reed.  Can you breathe?


Answer:  let’s determine the pressure on your chest…

P  =  rgh


P  =  (1000 kg/m3)(9.8 m/s2)(0.5 m)


P  =  4900  Pa  =  0.71 psi


If your chest is 12 in. by 12 in.  (approximately) then the area = 144 sq. in.


Force = Pressure x Area  =  0.7 psi  x  144  =  100 pounds


So, yeh, probably…but don’t go much deeper!





Pressure applied to an enclosed fluid is transmitted equally to all points of the fluid (and therefore to all walls of the vessel containing the fluid)



        What good is this? 


        We can turn this phenomenon to our advantage if we alter the areas exposed to equal pressures: as in an hydraulic lift:



Since the pressure must be the same everywhere:


Pat 1  =  Pat 2


But Pressure = Force/Area, so


F1/A1  =  F2/A2


F2  =  (F1/A1)A2 



This says that the Force at the outlet (2) is augmented by the size of the area of the outlet!


So if we make the area 1000 times larger, we can lift 1000 times the force we apply at F1!!!