PHYSICS 231 Lecture 24: Buoyancy and Fluid Motion

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

Previously Young’s modulus Solids: Shear modulus Bulk modulus Also fluids General: P=F/A (N/m2=Pa) Fpressure-difference=PA =M/V (kg/m3) 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 231

Pressure vs Depth Horizontal direction: P1=F1/A P2=F2/A F1=F2 (no net force) So, P1=P2 Vertical direction: Ftop=PatmA Fbottom=PbottomA-Mg=PbottomA-gAh Since the column of water is not moving: Ftop-Fbottom=0 PatmA=PbottomA-gAh Pbottom=Patm+ gh PHY 231

Pressure and Depth: Pdepth=h =Pdepth=0+ gh Where: Pdepth=h: the pressure at depth h Pdepth=0: the pressure at depth 0 =density of the liquid g=9.81 m/s2 h=depth Pdepth=0=Patmospheric=1.013x105 Pa = 1 atm =760 Torr From Pascal’s principle: If P0 changes then the pressures at all depths changes with the same value. PHY 231

A submarine A submarine is built in such a way that it can stand pressures of up to 3x106 Pa (approx 30 times the atmospheric pressure). How deep can it go? PHY 231

Does the shape of the container matter? NO!! PHY 231

Pressure measurement. The open-tube manometer. The pressure at A and B is the same: P=P0+gh so h=(P-P0)/(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 231

Pressure Measurement: the mercury barometer P0= mercurygh mercury=13.6E+03 kg/m3 mercury,specific=13.6 PHY 231

Pressures at same heights are the same P0 P0 h h h P=P0+gh P=P0+gh P=P0+gh PHY 231

Buoyant force: B P0 Ptop =P0+ wghtop Pbottom =P0+ wghbottom p = wg(htop-hbottom) F/A = wgh F = wghA=gV B =wgV=Mwaterg Fg=w=Mobjg If the object is not moving: B=Fg so: wgV=Mobjg - htop hbottom Archimedes (287 BC) principle: the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object PHY 231

Comparing densities B =fluidgV Buoyant force w =Mobjectg=objectgV Stationary: B=w object= fluid If object> fluid the object goes down! If object< fluid the object goes up! PHY 231

A floating object A w=Mobjectg=objectVobjectg B=weight of the fluid displaced by the object =Mwater,displacedg = waterVdisplacedg = waterhAg h: height of the object under water! h B w The object is floating, so there is no net force (B=w): objectVobject= waterVdisplaced h= objectVobject/(waterA) only useable if part of the object is above the water!! PHY 231

?? N A) An example ?? N B) 7 kg iron sphere of the same dimension as in A) 1 kg of water inside thin hollow sphere Two weights of equal size and shape, but different mass are submerged in water. What are the weights read out? PHY 231

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/m3 PHY 231

equation of continuity x2 x1 2 v2 v1 1 A1,1 A2,2 the mass flowing into area 1 (M1) must be the same as the mass flowing into area 2 (M2), else mass would accumulate in the pipe). M1= M2 1A1x1= 2A2x2 (M=V=Ax) 1A1v1t=2A2v2t (x=vt) 1A1v1 =2A2v2 if  is constant (liquid is incompressible) A1v1 =A2v2

Bernoulli’s equation same W1=F1x1=P1A1 x1=P1V W2=-F2x2=-P2A2 x2=-P2V Net Work=P1V-P2V m: transported fluid mass KE=½mv22-½mv12 & PE=mgy2-mgy1 Wfluid= KE+ PE P1V-P2V=½mv22-½mv12+ mgy2-mgy1 use =M/V and div. By V P1-P2=½v22-½v12+ gy2- gy1 P1+½v12+gy1= P2+½v22+gy2 P+½v2+gy=constant P: pressure ½v2:kinetic Energy per unit volume gy: potential energy per unit volume Another conservation Law PHY 231

When air is blown in between the cans, the velocity is not equal to 0. P2+½v22 (ignore y) Moving cans P0 Before air is blown in between the cans, P0=P1; the cans remain at rest and the air in between the cans is at rest (0 velocity) P1+½v12+gy1= Po Top view P1 case: 1: no blowing 2: blowing Bernoulli’s law: P1+½v12+gy1= P2+½v22+gy2 P0=P2+½v22 so P2=P0-½v22 So P2<P0 Because of the pressure difference left and right of each can, they move inward P0 PHY 231

V2=Vair-v P2 Spin and movement P1 V1=Vair+v Near the surface of the rotating cylinder: V1>V2 P1+½v12= P2+½v22 P1-P2= ½(v22- v12) P1-P2= ½[(vair-v)2-(vair+ v)2] P2>P1 so move to the left Applications of Bernoulli’s law: moving a cart No spin, no movement Vair PHY 231

hole in a tank P0 Pdepth=h =Pdepth=0+ gh h y If h=1m & y=3m what is x? Assume that the holes are small and the water level doesn’t drop noticeably. x PHY 231

If h=1 m and y=3 m what is X? P0 B h y A x1 PHY 231

PHYSICS 231 Lecture 24: Buoyancy and Fluid Motion