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SphygmomanometerWhen a person‛s blood pressure is taken with a device known as a sphygmomanometer, it ismeasured on the arm, at approximately the same level as the heart. If the measurement were tobe taken on the patient‛s leg instead, would the reading on the sphygmomanometer be greaterthan, less than, or the same as when the measurement is made on the arm? Explain. Ans: Gravity pulls the blood down, so greater pressure below the heart and higher pressureabove the heart. 15–3 #25 A patient is to receive an intravenous injection of

medication. In order to work properly, the pressure offluid containing the medication must be 109 kPa at theinjection point. (a) If the fluid has a density of 1020 kg/m3

find the height at which the bag of fluid must be suspendedabove the patient. Assume that the pressure inside thebag is one atmosphere. (b) If a less dense fluid is usedinstead, must the height of suspension be increased ordecreased? Explain.

P + ½ρv2 + ρgh = Pinject + ½ρvinject2 + ρghinject Let hlower be the zero point

Patm + 0 + ρgh = 109000 + 0 + ρg(0) velocity is very low at both points

101300 + ρgh = 109000 h = 7700 / ρ gh = 7700 / 1020 (9.81)h = 0.770 meters 15–5 #35A person weighs 756 N in air and has a body-fat percentage of 28.1%. (a) What is the overalldensity of this person‛s body? (b) What is the volume of this person‛s body? (c) Find the apparentweight of this person when completely submerged in water.

ρfat = 900 kg/m3 ρbody = 1100 kg/m3

ρeffective = 28.1% ρfat + 71.9%ρbodyρeffective = 28.1% 900 + 71.9%1100ρeffective = 1044 kg/m3

FW = m g756N = m(9.8)m = 77.1 kg

V = m / ρ ρ = m/V

V = 77.1 / 1044V = 0.0738 m3

Fscale + FB = mg FB = ρwaterVg

Fscale = m g - ρwater V g Fscale = 756 – (1000) 0.0738 m3 (9.81 m/s2)Fscale = 31.7 N

15–7 Syringe

A hypodermic syringe contains water. The barrel of thesyringe has a cross-sectional area of 10-5 m2, and theneedle has a cross-sectional area of 10-8 m2. In theabsence of a force on the plunger, the pressure

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throughout is 1 atm. A force of magnitude 3.00 N actson the plunger, making water squirt horizontally from theneedle. What is the speed of the water as it leaves theneedle‛s tip.

A1v1 = A2v2

10-5 v1 = 10-8 v2v2 = 1000 v1So v1

2 is neglible (below)

P + ½ρv2 + ρgh = P2 + ½ρv22 + ρgh2

(Patm + F/A) + 0 + 0 = Patm + ½ρv22 + 0

ΔP = ½ρv22

3/(10-5) = ½ 1000 v22

v2 = 24.5 m/s

 

#3A 50 kg woman balances on one heel of a pair of high heeled shoes. If the heel is circular and hasa radius of 0.500 cm, what pressure does she exert on the floor?P = F/A

P = m g / πr2

P = 50kg(9.8m/s2) / π(0.005)2

P = 6.24 x 106 Pascals

DamIn the example on our class notes, let‛s say a hatch 5 meters inwidth was placed near the bottom of a 30 meter dam. The hingedhatch is located 20 to 22 meters below the surface. The hinge isalong the top of the hatch at 20 meters. What is the torqueexerted by the water about the hinge?FOR FUN…not a 122 problem

dF = P dA dA = w dh

dF = ρgh (5 dh) w = 5 meters

F = 5ρg ∫h dhdτ = F • (h – 20)dτ = 5ρg h dh • (h – 20)

∫dτ = 5ρg (∫h2 dh - 20∫h dh)τ = 5 1000 9.8 [(h3/3 - 20 h2/2)]τ = 50000 [(223–203)/3 – 10(222–202)]τ = 2,090,000 Nm CCW

Lever arm is h - 20 dh is from 20 to 22meters deep.

#20A U-tube of uniform cross sectional area, open to the atmosphere,is partially filled with mercury. Water is then poured into botharms. If the equilibrium configuration of the tube is with h2 = 1.00cm determine the value of h1.Pleft = Pright (Aleft = Aright)

Fleft Aright = Fright Aleft Fleft = Fright

1“g”(h+h1+h2) = 1“g”h + 13.6“g”h2h1+h2 = 13.6h2h1 = 12.6h2 h2 = 1.00 cm

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ρ g h = ρ g h + ρ g h1“g”(h+h1+h2) = 1“g”h + 13.6“g”h2

h1 = 12.6 cm

BallAt the beach by the ocean, what is the force required to submerge a 30.0 Newton beach ball withradius of 20.0 cm under the salt water?

FNet = Fdown +FW

Fdown + mgFdown + 30Fdown + 30

= FB= msaltwater g= ρ saltwater (V) g= 1030(0.0335) g

ρ saltwater ≈ 1030 kg/m3

V = (4/3)πr3

V = (4/3)π(0.2)3

V = 0.0335 m3

Fdown = 308 Newtons #41Water flows through a fire hose of diameter 6.35 cm at a rate of 0.0120 m3/sec. The fire hoseends in a nozzle of inner diameter (id) of 2.20 cm. What is the speed with which the water exitsthe nozzle?

We know mass/time is constant…soV1 / t = V2 / t Initial diameter (and radius) is irrelevant

A1 Δx1/t = V2 / tπr2 (v) = 0.0120 m3/sv = 0.012/ π(0.011)2

v = 31.6 m/s #45Through a pipe 15.0 cm in diameter, water is pumped from the Colorado River up to Grand CanyonVillage, located on the rim of the canyon. The river is at an elevation of 564 m and the village isat an elevation of 2096 m. (a) What is the minimum pressure at which the water must be pumpedif it is to arrive at the village? (b) If 4500 m3 are pumped per day, what is the speed of thewater in the pipe? (c) What additional pressure is necessary to deliver this flow?

(a) P + ½ρv2 + ρgh = P2 + ½ρv22 + ρ g h2

(1atm + Pg) + ½ρv22 + 0 = 1atm + ½ρv2

2 + 1000(9.8)(2096-564) (vinit ≈ vfinal…if 10 gpm at one point in the hose…must be 10 gpm at any other point)I‛m also assuming the 564 meter elevation is the bottom so the 2096 meter elevation is the top, so h2 = (2096-564).

(1atm + Pg) + 0 + 0 = 1atm + 0 + 1000(9.8)(2096-564)P = 1atm + 1000(9.8)(2096-564)P = 1atm + 150 x 105 Pa

(b) 4500 m3/d (1d/24h)(24h/3600s)flow rate = 0.052083 m3/secflow rate = A v.052083 = π¼d2 v

(c)This time we need give some extra velocity…so

(1atm + Pg) = 1atm + ½ρv22 + 1000(9.8)1532

P = 1 atm + ½ 1000(2.95)2 + 150 x 105 PaP = 1 atm + 4340 Pa + 150 x 105 Pa

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.052083 = π¼ .152 vv = 2.95 m/s

P = 1.013 x 105 Pa + 4340 Pa + 150 x 105 PaP = 151.05 x 105 Pa

#53A hypodermic syringe contains water. The barrel of the syringe has a cross-sectional area of2.50 x 10-5 m2, and the needle has a cross-sectional area of 10-8 m2. In the absence of a forceon the plunger, the pressure everywhere is 1 atm. A force of magnitude 2.00 N acts on theplunger, making water squirt horizontally from the needle. Determine the speed of the medicineas it leaves the needle‛s tip. ans 12.6m/sA1v1 = A2v2

2.50 x 10-5 v1 = 10-8 v2v2 = 2500 v1So v1

2 is neglible (below)

P + ½ρv2 + ρgh = P2 + ½ρv22 + ρgh2

P-P2 + 0 + n/a = + ½ρv22 + n/a

ΔP = ½ρv2

2/(2.50 x 10-5) = ½ 1000 v2 v = 12.6 m/s

#8The small piston in a hydraulic lift has a cross-sectional area of 3.00 cm2, and its large piston hasa cross-sectional area of 200 cm2. What force must be applied to the small piston for the lift torise of a load of 15.0 kN?P1 = P2 F1 A2 = F2 A1

F1 200 = 15kN (3) F1 = 225 N

#14A cubic tank is filled with water 2.00 meters deep. At thebottom of one side wall is a rectangular hatch 1 m high and2 meters wide, which is hinged at the top of the hatch.(a) Determine the force the water exerts on the hatch. (b) Find the torque exerted by the water about the hinges.Ans: F = 29,400 N to the right,  τ = 16300 Nm CCW FOR FUN…similar to the Hinge-Dam problem dF = P dAdF = ρgh (width dh)dF = ρgh (2 dh)dF = 2ρg h dhF = 2ρg ∫h dh (dh varies from 1 to 2 m)

F = 2ρg ½(22 – 12)F = 2 ρ g ½ (3)

dτ = F • (h – 1)dτ = 2ρg h dh • (h – 1)∫dτ = 2ρg (∫h2 dh - ∫h dh) (dh varies from 1 to 2 m)

τ = 2 ρ g (h3/3 - h2/2)τ = 2 ρ g (23–13)/3 – (22–12)/2τ = 2 ρ g (7/3 – 3/2)τ = 2(1000)(9.8) (5/6)

Lever arm is h - 1

h is from 1 to 2 meters deep.

When h is half way down, say 1 ½meters deep…the lever from thepivot is ½ meter

…when it‛s 2 meters deep…thelevel arm is now 1 meter from thepivot

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F = 2(1000)(9.8) ½ (3)F = 29,400 N to the right

τ = 16300 Nm CCW …thus the lever arm is h-1

#23A ping pong ball has a diameter of 3.80 cm and average density of 0.0840 g/cc. What force isrequired to hold it completely submerged under water?FB = (ρH2O Vobject)g FW = mgFB = FW + Fpush

Fpush = FB - FWFpush = (ρH2O Vobject)g - mgFpush = (ρH2O Vobject)g -(ρobjectVobject)g

Fpush = ((ρH2O - ρobject) Vobject)g

Fpush = (1-0.0840) (4/3)π (3.8/2)3*980Fpush = 25,800 dynes = 0.258 Newtons

#29A cube of wood having an edge dimension of 20.0 cm and a density of 0.650 g/cc floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? (b) Howmuch lead weight has to be placed on top of the cube so that its top is just level with the water?FB = ρH20g(Ah) FW = ρg(Ah) FW = FBρg(Ah) = ρH20g(Ah1)ρ(h) = ρH20(h1).65(20) = 1 h1 h1 = 13 cm is submergedor 7cm is above the water

FB = Fcube + FPb ρH20g(V) = ρcubeg(V) + FPb FPb = gV (ρH20 - ρcube)FPb = 9.8 (0.2m)3 (1000-650)FPb = 27.4 N (converted 0.650 g/cc to 650 kg / m3)if you notice…the answer in the back of the book isincorrect. Weight is not measured in kg.

#30A spherical aluminum ball of mass 1.26 kg containsan empty spherical cavity that is concentric withthe ball. The ball just barely floats in the water. Calculate (a) the outer radius of the ball and (b)the radius of the cavity. (ρAl = 2.7 g/cc)

mg = FB ρAlVg = ρH20Vg (“g” cancels)

2.7*4/3*π(R3- r3) = 1 (4/3*πR3)1.7 (R)3 = 2.7 (r3)R3 = 1.588r3

m = ρAlV1260gram = 2.7*4/3*π(R3- r3)1260 = 2.7*4/3*π(1.588r3- r3)r = 5.74 cmR3 = 1.588r3

R = 6.70 cm #43At a faucet the diameter of the stream of water is 0.960 cm. The stream fills a 125 cccontainer in 16.3 seconds. Find the reduced diameter of stream 13.0 cm below the opening of the

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faucet? ans 0.247 cmUse conservation of energy to solve for velocitymgh = ½mv2 v = 159.6 cm/sthen use Same formula as in #41

πr2 (v) = V2 / tr2 = (V2 / t) / π (v)r2 = (125cc/16.3s) / (π*159.6)r = 0.1237d = 0.246 cm

#51A siphon is used to drain water from a tank. The siphon hasa uniform diameter. Assume steady flow without friction.(a) If the distance h = 1.00 m, find the speed of outflow atthe end of the siphon. (b) What If? What is the limitationon the height of the top of the siphon above the watersurface? (For the flow of the liquid to be continuous, the pressure must not dropbelow the vapor pressure of the liquid.)

(a)Po + ½rv2+rgh = Po + ½rv2+rghPo + 0 +rgh = Po + ½rv2 + 0rgh = ½rv2

v = (2gh)1/2 v = (2g*1)1/2 v = 4.43 m/s

(b)P2 + ½rv2

2 + rgh2 = P3 + ½rv32+rgh3

P2 + rgh2 = Po + rg(0)P2 = Po - rgh2h2 = (P2 - Po) / -r gh2 = (0 - Po) / r gh2 = 1.013 x 105 / (1000*9.8)h2 = 10.3 meters

v2 = v3 v1 << (v2 or v3) P3 = Po (open) h3 = 0 (at the bottom) vtop ≈ 0 m/s… soP2 = A2v2 ≈ 0 m/s

Instructor solutions incorrect…it states that since P2 is > 0, we can drop P2 out of the equation. If the studentincludes this…then give the student only ½ credit since the student is just copying directly from the solutionsmanual without thinking.