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 quiz (extra credit) A block of mass M is lying on the floor. The contact surface between the block and the floor is A. A second block, of mass 2M but with a contact surface of only 0.25A is also placed on the floor. What is the ratio of the pressure exerted on the floor by block 1 to the pressure exerted on the floor by block 2 (I.e. P 1 /P 2 )? a)1/8 b)¼ c)½ d)1 e)2 P 1 =F 1 /A 1 =Mg/A P 2 =F 2 /A 2 =2Mg/(0.25A)=8Mg/A P 1 /P 2 =(Mg/A)/(8Mg/A)=1/8

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? P depth=h =P depth=0 +  gh 3E+06=(1.0E+05)+(1.0E+03)(9.81)h h=296 m.

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 weight read out? B=  water V displaced g w=  sphere V sphere g A) B=  water V sphere g w=  water V sphere g so B=w and 0 N is read out! B) B=  water V sphere g=M water sphere g w=  iron V sphere g=M iron sphere g=7M water sphere g T=w-B=6*1*9.8 =58.8 N

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 A) h=  object V object /(  water A) h=M object /(1.0E+03*2*0.5)=2.0/1.0E+03=2.0E-03m=2mm B) if the objects sinks the mattress is just completely submerged: h=thickness of mattress. 0.08=(M weight +2.0)/(1.0E+03*2*0.5) So M weight =78 kg

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? Use Bernoulli’s law P A +½  v A 2 +  gy A = P B +½  v B 2 +  gy B At A: P A =P 0 v A =? y A =y=3 At B: P b =P 0 v B =0 y B =y+h=4 P 0 +½  v A 2 +  g3=P 0 +  g4 v A =  (g/2)=2.2 m/s A P0P0 B

PHY Each water element of mass m has the same velocity v A. Let’s look at one element m. v A =  (g/2)=2.2 m/s x1x1 3m vAvA In the horizontal direction: x(t)=x 0 +v 0x t+½at 2 =2.2t In the vertical direction: y(t)=y 0 +v 0y t+½at 2 =3-0.5gt 2 = 0 when the water hits the ground, so t=0.78 s so x(0.78)=2.2*0.78=1.72 m 0