HRW 7e Chapter 4 Page of 8 Halliday/Resnick/Walker 7e Chapter 4. The air inside pushes outard ith a force given by p i A, here p i is the pressure inside the room and A is the area of the indo. Similarly, the air on the outside pushes inard ith a force given by p o A, here p o is the pressure outside. The magnitude of the net force is F (p i p o )A. Since atm.0 0 5 Pa, 5 4 F (.0 atm 0.96 atm)(.0 0 Pa/atm)(.4 m)(. m).9 0 N.. The pressure increase is the applied force divided by the area: p F/A F/πr, here r is the radius of the piston. Thus This is equivalent to. atm. p (4 N)/π(0.0 m). 0 5 Pa. 5. Let the volume of the expanded air sacs be a and that of the fish ith its air sacs collapsed be. Then ρ m fish fish fish.08 g/cm and ρ.00 g/cm + a m here ρ is the density of the ater. This implies hich gives a /( + a ) 7.4%. ρ fish ρ ( + a ) or ( + a )/.08/.00, 7. (a) The pressure difference results in forces applied as shon in the figure. We consider a team of horses pulling to the right. To pull the sphere apart, the team must exert a force at least as great as the horizontal component of the total force determined by summing (actually, integrating) these force vectors. We consider a force vector at angle θ. Its leftard component is p cos θda, here da is the area element for here the force is applied. We make use of the symmetry of the problem and let da be that of a ring of constant θ on the surface. The radius of the ring is r R sin θ, here R is the radius of the sphere. If the angular idth of the ring is dθ, in radians, then its idth is R dθ and its area is da πr sin θ dθ. Thus the net horizontal component of the force of the air is given by π F πr p sin θ cos θdθ πr p sin θ πr p. h 0 π / 0 (b) We use atm.0 0 5 Pa to sho that p 0.90 atm 9.09 0 4 Pa. The sphere radius is R 0.0 m, so
HRW 7e Chapter 4 Page of 8 F h π(0.0 m) (9.09 0 4 Pa).6 0 4 N. (c) One team of horses could be used if one half of the sphere is attached to a sturdy all. The force of the all on the sphere ould balance the force of the horses. 8. Note that 0.05 atm equals 5065 N/m. Application of Eq. 4-7 ith the notation in this problem leads to d max 5065 ρ liquid g ith SI units understood. Thus the difference of this quantity beteen fresh ater (998 kg/m ) and ead Sea ater (500 kg/m ) is d max 5065 9.8 998-500 0.7 m. 0. Recalling that atm.0 0 5 Pa, Eq. 4-8 leads to atm ρ gh (04 kg/m ) (9.80 m/s ) (0.9 0 m).08 0 atm. 5.0 0 Pa. The pressure p at the depth d of the hatch cover is p 0 + ρgd, here ρ is the density of ocean ater and p 0 is atmospheric pressure. The donard force of the ater on the hatch cover is (p 0 + ρgd)a, here A is the area of the cover. If the air in the submarine is at atmospheric pressure then it exerts an upard force of p 0 A. The minimum force that must be applied by the cre to open the cover has magnitude F (p 0 + ρgd)a p 0 A ρgda (04 kg/m )(9.8 m/s )(00 m)(. m)(0.60 m) 7. 0 5 N.. In this case, Bernoulli s equation reduces to Eq. 4-0. Thus, p ρg h g 4 ( ) (800 kg/m )(9.8 m/s )(.5 m).6 0 Pa. 7. We can integrate the pressure (hich varies linearly ith depth according to Eq. 4-7) over the area of the all to find out the net force on it, and the result turns out fairly intuitive (because of that linear dependence): the force is the average ater pressure multiplied by the area of the all (or at least the part of the all that is exposed to the ater), here average pressure is taken to mean (pressure at surface + pressure at bottom). Assuming the pressure at the surface can be taken to be zero (in the gauge pressure sense explained in section 4-4), then this means the force on the all is ρgh multiplied by the appropriate area. In this problem the area is h (here is the 8.00 m idth), so the force is ρgh, and the change in force (as h is changed) is
HRW 7e Chapter 4 Page of 8 ρg ( h f hi ) (998 kg/m )(9.80 m/s )(8.00 m)(4.00.00 )m 4.69 0 5 N. 0. The gauge pressure you can produce is ( )( 000 kg m 9.8m s )( 4.0 0 m) p ρ gh 5.0 0 Pa atm.9 0 atm here the minus sign indicates that the pressure inside your lung is less than the outside pressure.. (a) According to Pascal s principle F/A f/a F (A/a)f. (b) We obtain f a (.80 cm) F A (5.0 cm) (0.0 0 N) 0 N. The ratio of the squares of diameters is equivalent to the ratio of the areas. We also note that the area units cancel. 5. (a) The anchor is completely submerged in ater of density ρ. Its effective eight is W eff W ρ g, here W is its actual eight (mg). Thus, W W ρ g 00 N eff.04 0 m. ( 000 kg/m ) ( 9.8 m/s ) (b) The mass of the anchor is m ρ, here ρ is the density of iron (found in Table 4-). Its eight in air is ( ) ( ) W mg ρg 7870 kg/m (.04 0 m ) 9.80 m/s.57 0 N. 6. (a) The pressure (including the contribution from the atmosphere) at a depth of h top L/ (corresponding to the top of the block) is p p + ρ gh + 5 5 top atm top.0 0 (00)(9.8)(0.00) Pa.04 0 Pa here the unit Pa (Pascal) is equivalent to N/m. The force on the top surface (of area A L 0.6 m ) is F top p top A.75 0 4 N. (b) The pressure at a depth of h bot L/ (that of the bottom of the block) is p p + ρ gh + 5 5 bot atm bot.0 0 (00)(9.8)(0.900) Pa.0 0 Pa
HRW 7e Chapter 4 Page 4 of 8 here e recall that the unit Pa (Pascal) is equivalent to N/m. The force on the bottom surface is F bot p bot A.96 0 4 N. (c) Taking the difference F bot F top cancels the contribution from the atmosphere (including any numerical uncertainties associated ith that value) and leads to F F ρg h h A ρgl bot top ( bot top).8 0 N hich is to be expected on the basis of Archimedes principle. To other forces act on the block: an upard tension T and a donard pull of gravity mg. To remain stationary, the tension must be T mg F F ( bot top) (450 kg)(9.80 m/s ).8 0 N. 0 N. (d) This has already been noted in the previous part: F.8 0 N, and T + Fb mg. b 9. (a) Let be the volume of the block. Then, the submerged volume is s /. Since the block is floating, the eight of the displaced ater is equal to the eight of the block, so ρ s ρ b, here ρ is the density of ater, and ρ b is the density of the block. We substitute s / to obtain ρ b ρ / (000 kg/m )/ 6.7 0 kg/m. (b) If ρ o is the density of the oil, then Archimedes principle yields ρ o s ρ b. We substitute s 0.90 to obtain ρ o ρ b /0.90 7.4 0 kg/m. 5. The volume cav of the cavities is the difference beteen the volume cast of the casting as a hole and the volume iron contained: cav cast iron. The volume of the iron is given by iron W/gρ iron, here W is the eight of the casting and ρ iron is the density of iron. The effective eight in ater (of density ρ ) is W eff W gρ cast. Thus, cast (W W eff )/gρ and W W W 6000 N 4000 N 6000 N (9.8 m/s )(000 kg/m ) (9.8 m/s )(7.87 0 kg/m ) eff cav gρ gρiron 0.6 m. 4. We use the equation of continuity. Let v be the speed of the ater in the hose and v be its speed as it leaves one of the holes. A πr is the cross-sectional area of the hose. If there are N holes and A is the area of a single hole, then the equation of continuity becomes A R va v NA v v v ( ) NA Nr here R is the radius of the hose and r is the radius of a hole. Noting that R/r /d (the ratio of diameters) e find
HRW 7e Chapter 4 Page 5 of 8 v (.9cm) ( ) ( ) v 0.9m s 8.m s. Nd 4 0.cm 4. We use the equation of continuity and denote the depth of the river as h. Then, ( 8.m)(.4m)(.m s) + ( 6.8m)(.m)(.6m s) h( 0.5m)(.9m s) hich leads to h 4.0 m. 4. Suppose that a mass m of ater is pumped in time t. The pump increases the potential energy of the ater by mgh, here h is the vertical distance through hich it is lifted, and increases its kinetic energy by mv, here v is its final speed. The ork it does is W mgh+ mv and its poer is W m P gh+ v. t t No the rate of mass flo is m/ t ρ Av, here ρ is the density of ater and A is the area of the hose. The area of the hose is A πr π(0.00 m).4 0 4 m and Thus, ρ Av (000 kg/m ) (.4 0 4 m ) (5.00 m/s).57 kg/s. ( ) 5.0m s P ρ Av gh+ v (.57 kg s) ( 9.8m s )(.0m) + 66 W. 45. (a) We use the equation of continuity: A v A v. Here A is the area of the pipe at the top and v is the speed of the ater there; A is the area of the pipe at the bottom and v is the speed of the ater there. Thus v (A /A )v [(4.0 cm )/(8.0 cm )] (5.0 m/s).5m/s. (b) We use the Bernoulli equation: p+ ρv + ρgh p + ρv + ρgh, here ρ is the density of ater, h is its initial altitude, and h is its final altitude. Thus p p+ ρ( v v) + ρg( h h) + + 5.6 0 Pa. 46. We use Bernoulli s equation: 5.5 0 Pa (000 kg m ) (5.0 m s) (.5m s) (000 kg m )(9.8m/s )(0 m)
HRW 7e Chapter 4 Page 6 of 8 p pi ρg+ ρ v v ( ) here ρ 000 kg/m, 80 m, v 0.40 m/s and v 9.5 m/s. Therefore, e find p.7 0 6 Pa, or.7 MPa. The SI unit for pressure is the Pascal (Pa) and is equivalent to N/m. 47. (a) The equation of continuity leads to r va va v v r hich gives v.9 m/s. (b) With h 7.6 m and p.7 0 5 Pa, Bernoulli s equation reduces to 4 p p ρgh+ ρ( v v) 8.8 0 Pa. 54. (a) The volume of ater (during 0 minutes) is π ( ) ( )( )( ) ( ) vt A 5m s 0 min 60s min 0.0m 6.4 m. 4 (b) The speed in the left section of pipe is A d.0cm v v v ( 5m s) 5.4 m s. A d 5.0cm (c) Since pressure, p+ ρv + ρgh p + ρv + ρgh and h h, p p 0, hich is the atmospheric p p v v 5 ( ).0 0 Pa (.0 0 kg m ) ( 5m s) ( 5.4m s) 0 + ρ + 5.99 0 Pa.97 atm. Thus the gauge pressure is (.97 atm.00 atm) 0.97 atm 9.8 0 4 Pa. 66. The normal force F N exerted (upard) on the glass ball of mass m has magnitude 0.0948 N. The buoyant force exerted by the milk (upard) on the ball has magnitude F b ρ milk g
HRW 7e Chapter 4 Page 7 of 8 here 4 π r is the volume of the ball. Its radius is r 0.000 m. The milk density is ρ milk 00 kg/m. The (actual) eight of the ball is, of course, donard, and has magnitude F g m glass g. Application of Neton's second la (in the case of zero acceleration) yields F N + ρ milk g m glass g 0 hich leads to m glass 0.044 kg. We note the above equation is equivalent to Eq.4-9 in the textbook. 70. To be as general as possible, e denote the ratio of body density to ater density as f (so that f ρ/ρ 0.95 in this problem). Floating involves an equilibrium of vertical forces acting on the body (Earth s gravity pulls don and the buoyant force pushes up). Thus, F F ρ g ρ g b g here is the total volume of the body and is the portion of it hich is submerged. (a) We rearrange the above equation to yield ρ ρ f hich means that 95% of the body is submerged and therefore 5% is above the ater surface. (b) We replace ρ ith.6ρ in the above equilibrium of forces relationship, and find ρ f.6ρ.6 hich means that 59% of the body is submerged and thus 4% is above the quicksand surface. (c) The anser to part (b) suggests that a person in that situation is able to breathe. 76. The donard force on the balloon is mg and the upard force is F b ρ out g. Neton s second la (ith m ρ in ) leads to ρ ρ ρ ρ. out out g in g in a g a ρ in The problem specifies ρ out / ρ in.9 (the outside air is cooler and thus more dense than the hot air inside the balloon). Thus, the upard acceleration is (.9.00)(9.80 m/s ).8 m/s. 79. (a) From Bernoulli equation p+ ρv + ρgh p + ρv + ρgh, the height of the ater extended up into the standpipe for section B is related to that for section by
HRW 7e Chapter 4 Page 8 of 8 h h + v v g ( ) B B Equation of continuity further implies that v A vba B, or here A R B vb v v 4v AB RB v R R /( π ) (.0 0 m /s) /( π(0.040 m) ) 0.40 m/s. With h 0.50 m, e have h B + (9.8 m/s ) 0.50 m ( 5)(0.40 m/s) 0.8 m. (b) From the above result, e see that the greater the radius of the cross-sectional area, the greater the height. Thus, h > h > h > h. C B A