PHY PHYSICS 231 Lecture 24: Buoyancy and Fluid Motion Remco Zegers Walk-in hour: Monday 9:15-10:15 am Helproom

PHY Previously Solids: Young’s modulus Shear modulus Bulk modulus Also fluids P=F/A (N/m 2 =Pa) F pressure-difference =  PA  =M/V (kg/m 3 ) General: Pascal’s principle: a change in pressure applied to a fluid that is enclosed is transmitted to the whole fluid and all the walls of the container that hold the fluid.

PHY Pressure vs Depth Horizontal direction: P 1 =F 1 /A P 2 =F 2 /A F 1 =F 2 (no net force) So, P 1 =P 2 Vertical direction: F top =P atm A F bottom =P bottom A-Mg=P bottom A-  gAh Since the column of water is not moving: F top -F bottom =0 P atm A=P bottom A-  gAh P bottom =P atm +  gh

PHY Pressure and Depth: P depth=h =P depth=0 +  gh Where: P depth=h : the pressure at depth h P depth=0 : the pressure at depth 0  =density of the liquid g=9.81 m/s 2 h=depth P depth=0 =P atmospheric =1.013x10 5 Pa = 1 atm =760 Torr From Pascal’s principle: If P 0 changes then the pressures at all depths changes with the same value.

PHY A submarine A submarine is built in such a way that it can stand pressures of up to 3x10 6 Pa (approx 30 times the atmospheric pressure). How deep can it go?

PHY Does the shape of the container matter? NO!!

PHY Pressure measurement. The open-tube manometer. The pressure at A and B is the same: P=P 0 +  gh so h=(P-P 0 )/(  g) If the pressure P=1.01 atm, what is h? (the liquid is water) h=(1.01-1)*(1.0E+05)/(1.0E+03*9.81)= =0.1 m

PHY Pressure Measurement: the mercury barometer P 0 =  mercury gh  mercury =13.6E+03 kg/m 3  mercury,specific =13.6

PHY Pressures at same heights are the same P0P0 P=P 0 +  gh h P0P0 h h

PHY Buoyant force: B h top h bottom P top =P 0 +  w gh top P bottom =P 0 +  w gh bottom  p=  w g(h top -h bottom ) F/A=  w g  h F=  w g  hA=  gV B=  w gV=M water g F g =w=M obj g If the object is not moving: B=F g so:  w gV=M obj g P0P0 - Archimedes (287 BC) principle: the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object

PHY Comparing densities B=  fluid gV Buoyant force w=M object g=  object gV Stationary: B=w  object =  fluid If  object >  fluid the object goes down! If  object <  fluid the object goes up!

PHY A floating object h A B w w=M object g=  object V object g B=weight of the fluid displaced by the object =M water,displaced g =  water V displaced g =  water hAg h: height of the object under water! The object is floating, so there is no net force (B=w):  object V object =  water V displaced h=  object V object /(  water A) only useable if part of the object is above the water!!

PHY An example ?? N 1 kg of water inside thin hollow sphere A) ?? N 7 kg iron sphere of the same dimension as in A) B) Two weights of equal size and shape, but different mass are submerged in water. What are the weights read out?

PHY Another one An air mattress 2m long 0.5m wide and 0.08m thick and has a mass of 2.0 kg. A) How deep will it sink in water? B) How much weight can you put on top of the mattress before it sinks?  water =1.0E+03 kg/m 3

equation of continuity A1,1A1,1 A2,2A2,2 v1v1 v2v2 1 2 the mass flowing into area 1 (  M 1 ) must be the same as the mass flowing into area 2 (  M 2 ), else mass would accumulate in the pipe).  M 1 =  M 2  1 A 1  x 1 =  2 A 2  x 2 (M=  V=  Ax)  1 A 1 v 1  t =  2 A 2 v 2  t (  x=v  t)  1 A 1 v 1 =  2 A 2 v 2 if  is constant (liquid is incompressible) A 1 v 1 =A 2 v 2 x1x1 x2x2

PHY Bernoulli’s equation W 1 =F 1  x 1 =P 1 A 1  x 1 =P 1 V W 2 =-F 2  x 2 =-P 2 A 2  x 2 =-P 2 V Net Work=P 1 V-P 2 V m: transported fluid mass same  KE=½mv 2 2 -½mv 1 2 &  PE=mgy 2 -mgy 1 W fluid =  KE+  PE P 1 V-P 2 V=½mv 2 2 -½mv mgy 2 -mgy 1 use  =M/V and div. By V P 1 -P 2 =½  v 2 2 -½  v  gy 2 -  gy 1 P 1 +½  v  gy 1 = P 2 +½  v  gy 2 P+½  v 2 +  gy=constant P: pressure ½  v 2 :kinetic Energy per unit volume  gy: potential energy per unit volume Another conservation Law

PHY Moving cans P0P0 P0P0 Top view Before air is blown in between the cans, P 0 =P 1 ; the cans remain at rest and the air in between the cans is at rest (0 velocity) P 1 +½  v  gy 1 = P o P1P1 Bernoulli’s law: P 1 +½  v  gy 1 = P 2 +½  v  gy 2 P 0 =P 2 +½  v 2 2 so P 2 =P 0 -½  v 2 2 So P 2 <P 0 Because of the pressure difference left and right of each can, they move inward When air is blown in between the cans, the velocity is not equal to 0. P 2 +½  v 2 2 (ignore y) case: 1: no blowing 2: blowing

PHY Applications of Bernoulli’s law: moving a cart No spin, no movement V air Spin and movement P 1 V 1 =V air +v V 2 =V air -v P 2 Near the surface of the rotating cylinder: V 1 >V 2 P 1 +½  v 1 2 = P 2 +½  v 2 2 P 1 -P 2 = ½  (v 2 2- v 1 2 ) P 1 -P 2 = ½  [(v air -v) 2 -(v air + v) 2 ] P 2 >P 1 so move to the left

PHY hole in a tank P depth=h =P depth=0 +  gh h x y If h=1m & y=3m what is x? Assume that the holes are small and the water level doesn’t drop noticeably. P0P0

PHY h x1x1 y If h=1 m and y=3 m what is X? A P0P0 B