Fluids-1 Fluids (= liquids or gases) Two definitions: Fforcepressure = , parea A⊥≡ , F⊥ = force perpendicular to area, [p] = N / m2 = pascal (Pa) mass mdensity = , , ( = greek letter "rho" )volume Vρ≡ ρ [ρ] = kg / m3 or gram / cm3 = g / cm3 33water336103=103101 gram 1 kg 100 cm kg10cm 10 g 1 m m⎛⎞⎟⎜ρ =×× =⎟⎜⎟⎟⎜⎝⎠3 , ρ 33iron7.9 g / cm 7900 kg / m== "specific gravity" = waterρρ = 1 water 7.9 iron 11.9 Pb 2.7 Al _____________________________ * ___________________________________ Consider a bucket of water. What is the pressure at the bottom of the bucket due to the weight of the water? total mass Fweightmp=AA⊥==gA Will now show that p = ρ g h mmVVρ = ⇒ = ρ , mg V g (A h) gpghAA Aρρ== = =ρh More generally, , where ∆h is the change in the depth below the surface of the water. This derivation assumes that density ρ = constant, which it is for the case of water, because water is incompressible. pg∆ = ρ∆area A m h = |∆y| A h V = A hFluids-2 We are at the bottom of an ocean of air! Notice that air is about 1000 times less dense than water. 3air1.29 kg / m (at sea level)ρ =atmospheric pressure at sea level = 52atmp1atm1.0110 Pa=14.7 lb/i==× nAir consists mostly of oxygen and nitrogen molecules. At room temperature, the molecules have thermal energy and are moving around rapidly (speed ≈ 400 m/s), colliding with each other and with every exposed surface. The pounding of the air molecules on a surface, like the pitter-pat of rain on the roof, adds up to a large force per area: patm = 14.7 psi. This is a big pressure! We are not ordinarily aware of this big pressure because the air pushes on us equally from all sides (even from our insides due to the air in our lungs). The big forces on us from all sides cancel out and there is no net force on us. At a given depth, a fluid exerts the same pressure in every direction. Consider a block of water (outlined by an imaginary box) within a bucket of water. Since the water is in equilibrium, the forces (and therefore the pressures) on opposite sides of the block must cancel. Example: At the surface of a swimming pool the pressure on a swimmer (due to the air) is patm = 1 atm. At what depth below the surface of the water is the total pressure on the swimmer = 2 atm? Answer: when the pressure due to the weight of the water alone is 1 atm, then the total pressure will be 2 atm. 533 2p 1.01 10 Pap g h h 10.3mg (10 kg / m )(9.8m /s )∆ ×∆ = ρ∆ ⇒∆ == =ρ When computing the total pressure at a depth h below the surface of a liquid, we must include the pressure due to the atmosphere above the liquid: tot atmpghρ=+p total p = p due to fluid above + p due to atmosphere above fluidvacuumairforces balancedforces unbalanced(Space Shuttle astronaut heroically plugs hole with head,)airsurfaceair moleculesFluids-3 The pressure at a given depth is the same regardless of the shape of the container. Same h ⇒ same p at bottom h Must be same level (due to same pressure at bottom) or else could have a perpetual water fall ⇒ free energy ⇒ impossible Archimedes' Principle A solid body, either partially or totally submerged in a fluid experiences an upward buoyant force = weight of the displaced fluid Before computing the buoyant force, let's first ask: Why is there an upward buoyant force? smaller p pgy∆ = ρ∆ net upward force on submerged object = buoyant force If buoyant force > mg, then object floats. If buoyant force < mg, then object sinks. Claim: An object of mass m and volume V, submerged in a fluid with density ρ fluid , will experience an upward buoyant force with magnitude = weight of displaced fluid = buoy fluid fluidFmg Vgρ= ⋅ = ⋅⋅ ∆h = |∆y| mass m larger pFluids-4 Proof: This is a simple but slightly subtle argument. We note the forces on a submerged object from the surrounding fluid (the buoyant force) are exactly the same as the forces on a block of fluid with the same size and depth as the submerged object. In equilibrium, the buoyant force on the block of fluid from the surrounding fluid must be equal in magnitude to the weight of the fluid, otherwise the block would not be in equilibrium. Therefore, the magnitude of the buoyant force on the submerged object is the same as the weight of the displaced fluid. Example: A block of copper (Cu) with mass m = 400 g and density ρ Cu = 8.9 g/cm3 is suspended by a string while under water. How does the tension in the sting compare to the weight of the copper block? Since the block is not moving, the net force on it is zero and we can write: TBFFm+= g ( since | upward forces | = | downward forces | ) So we have FT = mg – FB We must now compute the magnitude of the buoyant force FB. Archimedes says that FB is the weight of the displaced water = mwater g = ρwater V g where V is the volume of the displaced water = volume of the copper block. We get V from CuCummVVρρ= ⇒ = so waterB water waterCu CumFVg g mρρρρρ== =g water waterTBCu Cu1F mg F mg m g mg 1 mg 1 0.89 mg8.9ρρρρ⎛⎞⎛⎞⎛⎞⎟⎟⎜⎜⎟⎜⎟⎟= − = − = − = − =⎜⎜⎟⎜⎟⎟⎟⎜⎜⎜⎟⎟⎜⎜⎝⎠⎝⎠⎝⎠ tension = ? FT buoyant force FB weight mgforces on block : m object fluid fluid buoyant force weight of fluidFluids-5 So the tension in the string is only 89% of the weight of the copper block. The buoyant force is helping support the weight of the block, so the tension is less than the full weight of the block. This calculation can be turned around and used to compute the density of a block, given its mass and the tension in the string. Legend has it that Archimedes used this technique to determine the density of the king's crown (the king of Syracuse, a Greek colony in Sicily). The king was worried that the crown was not pure gold, and Archimedes was able to show that the density of his crown was considerably less than the density of gold (ρAu = 19.3 g/cm3 ), confirming the king's suspicion. Archimedes (Greek, 287-212 B.C.) was the greatest mathematician and scientist of antiquity. He was also a brilliant engineer and inventor. His accomplishments in math, especially in geometry, were unmatched until the work of Isaac Newton, nearly 2000 years later. Archimedes should not be confused with a lesser scientist of ancient times: