1 / 16

PHYSICS 231 Lecture 23: Buoyancy and fluid motion

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

cree
Download Presentation

PHYSICS 231 Lecture 23: Buoyancy and fluid motion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. PHYSICS 231Lecture 23: Buoyancy and fluid motion Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom PHY 231

  2. 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

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

  4. Unstable system This system is not stable; the pressure at different heights is not 0. PHY 231

  5. The pressure at A and B is the sameif the fluid column is not moving PHY 231

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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 PHY 231

  12. When air is blown in between the cans, the velocity is not equal to 0. P2+½v22 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 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

  13. 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

  14. If h=1m and y=3m what is X? P0 B Use Bernoulli’s law h PA+½vA2+gyA= PB+½vB2+gyB At A: PA=P0 vA=? yA=y=3 At B: Pb=P0 vB=0 yB=y+h=4 P0+½vA2+g3=P0+g4 vA=(g/2)=2.2 m/s y A x1 PHY 231

  15. 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

  16. Fluid flow P1+½v12+gy1= P2+½v22+gy2 P1+½v12 = P2+½v22 (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

More Related