Archimedes
- 2. ARCHIMEDES PRINCIPLE IMMERSED BODY COMPLETELY OR PARTIALLY IN A FLUID IS AFFECTED BY BUOYANT FORCE UPWARD EQUALS THE WEIGHT OF DISPLACED FLUID.
- 5. CASE ONE A BODY IS IMMERSED COMPLETELY IN A FLUID. relative=Vol water/V ol substance In the case of a floating body Fb=w air-w liquid=volg relative =w air/Fb water for body relative=Fb liquid/Fb water for liquid
- 6. EX:1 A body its weight in air 7.8 N, and its weight in water is 6.5 N calculate the density of a body.
- 7. ANS: Fb=w air w water =7.8 - 6.5 =1.3 N relative = w air/Fb =7.8/1.3 =6 =relativeX1000
- 9. CASE TWO A FLOATING BODY
- 10. Fb=w 痢l(Vol)lg =s(Vol)sg l /s =Vols/Voll Note that (Vol)l = displaced liquid or immersed part (Vol)s = whole body Important note In the opposite laws you can calculate the relative density by divided s/l(water) Also equals (Vol) l / (Vol) s
- 11. so relative = s/w = ms/mw = wair/Fbw = Fbl/Fbw = Volw/Vols
- 12. EX:1 A BODY ITS VOLUME IS 60CM3 HALF IMMERSED IN WATER CALCULATE ITS DENSITY
- 15. CASE THREE FLOATING BODY CARRYING A MASS
- 17. A body its volume is 50 cm3 carrying a mass =10 g in water its density is 1000 kg /m3 calculate the volume immersed part knowing that the density of an object is 600 kg/m3 Answer Fb=vsg+mg 痢vlg=vsg+mg 1000xvl=600x50x10-6+.01 Vl=40cm3
- 19. CASE FOUR A SUSPENDING BODY
- 20. Hollow body Body not contain space(massive) Fb=w lvlg =svlg l=s Fb=w lvlg=svlg l(vbody+vhallow)g=vlg body space
- 21. Ex: A hollow ball its mass is 0.5 kg placed in water and becomes suspended calculate the volume of a space inside the ball considering that Density of water=1000 kg/m3 Density of ball =6600 kg/m3
- 22. Ans Fb=w lvlg=svlg l(vlball - vlspace)g=svlg 1000(.5/6600 Vlspace) = .5 Vl space = .0004m3
- 24. Case five ` A suspending body between two different liquids
- 25. oil water Voloil Volwater Fb1+Fb2=ws oil (Vol)oil g + water(Vol)water g = s (vol)s g
- 26. Ex A wooden ball its mass is 800 g and its density is 700 kg /m3 floating on a surface of water then poured a mount of oil its density is 800 kg/m3 above the water calculate the volume of ball in oil considering that density of water is 1000 kg/m3 .
- 27. ANS 800X( Vol) oil+ 1000(.8/800 (Vol) oil) = .8 (Vol) oil = 1.71 X 10-3 m3
- 29. Case six Immersed body connected with a strange
- 30. Roped up Roped down Ft Fb w Fb W Ft Ft + Fb = w Ft = W Fb Ft = s Vol g lVol g Fb = w + Ft Ft = Fb W Ft = L Vol g SVol g
- 31. EXAMPLE A BODY IS IMMERSED COMPLETELY IN WATER ITS VOLUME IS 0.01 CM3 AND ITS DENSITY IS 600 KG/M3 ROPED DOWN IN A BACKER CONSIDERING THAT THE DENSITY OF WATER IS 1000 KG/M3 AND GRAVITY IS 10 M/S2
- 32. ANS Ft = L Vol g SVol g = 0.01 X 10 X 1000 + 0.01 X 10 X 600 = 160 N
- 34. Case seven Balloon filed by gas
- 35. (Fb)lift = (Fb )air -(W gas + Wballoon) (Fb)lift = (Vol g)air -(mg gas + mg balloon) When a balloon stopped to move up (Case of equilibrium) Fb air = w (gas+balloon) Fb air = mg balloon + mg gas airVol g = mg balloon + gasVol g W gas W ballon Fb air
- 36. Ex An empty balloon its mass is 150 kg and its volume is 400 m3 filled by hydrogen gas its density is 0.09 kg/m3 what is the lifting force . Considering that the average density for air is 1.29 kg/m3 and gravity is 10 m/s2.
- 37. Ans (Fb)lift = (Fb )air -(W gas + Wballoon) (Fb)lift = (Vol g)air -(mg gas + mg balloon) =(1.29X400X10) (150X10 + .09 X400 X10 = 3300 N
- 38. 悋忰惆