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PHYSICS 231 Lecture 23: Buoyancy and fluid motion. Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom. P 0. Previously. P depth=h =P depth=0 + gh. Pascal’s principle: a change in pressure applied to a fluid that is enclosed is transmitted to the whole
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PHYSICS 231Lecture 23: Buoyancy and fluid motion Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom PHY 231
P0 Previously... Pdepth=h =Pdepth=0+ gh 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
Pressures at same heights are the same P0 P0 h h h P=P0+gh P=P0+gh P=P0+gh PHY 231
Unstable system This system is not stable; the pressure at different heights is not 0. PHY 231
The pressure at A and B is the sameif the fluid column is not moving PHY 231
Buoyant force: B P0 Ptop =P0+ wghtop Pbottom =P0+ wghbottom p = wg(htop-hbottom) F/A = wgh F = wghA=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
B= waterVdisplacedg w= sphereVsphereg A) B= waterVsphereg w= waterVsphereg so B=w and 0 N is read out! B) B= waterVsphereg=Mwater sphere w= ironVsphereg=Miron sphereg=7Mwater sphereg T=w-B=6*1*9.8 =58.8 N ?? 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 weight 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 A) h= objectVobject/(waterA) h=Mobject/(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=(Mweight+2.0)/(1.0E+03*2*0.5) So Mweight=78 kg PHY 231
Bernoulli’s equation same W1=F1x1=P1A1 x1=P1V W2=-F2x2=-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=½v22-½v12+ gy2- gy1 P1+½v12+gy1= P2+½v22+gy2 P+½v2+gy=constant P: pressure ½v2:kinetic Energy per unit volume gy: potential energy per unit volume PHY 231
When air is blown in between the cans, the velocity is not equal to 0. P2+½v22 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+½v12+gy1= Po Top view P1 Bernoulli’s law: P1+½v12+gy1= P2+½v22+gy2 P0=P2+½v22 so P2=P0-½v22 So P2<P0 Because of the pressure difference left and right of each can, they move inward P0 PHY 231
Last lecture... 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=1m and y=3m what is X? P0 B Use Bernoulli’s law h PA+½vA2+gyA= PB+½vB2+gyB At A: PA=P0 vA=? yA=y=3 At B: Pb=P0 vB=0 yB=y+h=4 P0+½vA2+g3=P0+g4 vA=(g/2)=2.2 m/s y A x1 PHY 231
Each water element of mass m has the same velocity vA. Let’s look at one element m. vA=(g/2)=2.2 m/s vA In the horizontal direction: x(t)=x0+v0xt+½at2=2.2t In the vertical direction: y(t)=y0+v0yt+½at2=3-0.5gt2 = 0 when the water hits the ground, so t=0.78 s so x(0.78)=2.2*0.78=1.72 m 3m 0 x1 PHY 231
Fluid flow P1+½v12+gy1= P2+½v22+gy2 P1+½v12 = P2+½v22 (v22-v12)=2(P1-P2)/ If P1=4.0*105 Pa, P2=2.0*105 Pa and by counting the amount of water coming from the right v2 is found to be 30 m/s, what is v1? (=1E+03 kg/m3) 900-v12=2*(2.0E+5)/(1E+03) v1=22.3 m/s PHY 231