Ch 14.1      #3

A 50 kg woman balances on one heel of a pair of high heeled shoes.  If the heel is circular and has a 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

 Ch 14.2     Dam In the example on our class notes, let’s say a hatch 5 meters in width was placed near the bottom of a 30 meter dam.  The hinged hatch is located 20 to 22 meters below the surface.  The hinge is along the top of the hatch at 20 meters.  What is the torque exerted by the water about the hinge? dF = P       dA         dA = w dh dF = ρgh (5 dh)       w = 5 meters F = 5ρg ∫h dh dτ =      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 22 meters deep.

 Ch 14.3     #20 A U-tube of uniform cross sectional area, open to the atmosphere, is partially filled with mercury.  Water is then poured into both arms.  If the equilibrium configuration of the tube is with h2 = 1.00 cm determine the value of h1. Pleft = Pright  (Aleft = Aright) Fleft Aright = Fright Aleft  Fleft = Fright ρ  g       h      = ρ g h  +  ρ g h 1“g”(h+h1+h2) = 1“g”h + 13.6“g”h2 1“g”(h+h1+h2) = 1“g”h + 13.6“g”h2 h1+h2 = 13.6h2 h1 = 12.6h2    h2 = 1.00 cm h1 = 12.6 cm

Ch 14.4     Ball

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

 FNet = Fdown + FW Fdown + mg Fdown + 30 Fdown + 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

Ch 14.5     #41

Water flows through a fire hose of diameter 6.35 cm at a rate of 0.0120 m3/sec.  The fire hose ends in a nozzle of inner diameter (id) of 2.20 cm.  What is the speed with which the water exits the nozzle?

 We know mass/time is constant…so V1 / t           = V2 / t   Initial diameter (and radius) is irrelevant A1 Δx1/t      = V2 / t πr2 (v)         = 0.0120 m3/s v                  = 0.012/ π(0.011)2 v                  = 31.6 m/s

Ch 14.6     #45

Through a pipe 15.0 cm in diameter, water is pumped from the Colorado River up to Grand Canyon Village, located on the rim of the canyon.  The river is at an elevation of 564 m and the village is at an elevation of 2096 m.  (a) What is the minimum pressure at which the water must be pumped if it is to arrive at the village?  (b) If 4500 m3 are pumped per day, what is the speed of the water 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   + ½ρv22  + 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/sec flow rate =   A    v .052083 = π¼d2  v .052083 = π¼ .152  v v = 2.95 m/s (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 Pa P = 1 atm + 4340 Pa + 150 x 105 Pa P = 1.013 x 105 Pa + 4340 Pa + 150 x 105 Pa P = 151.05 x 105 Pa

Ch 14.7     #53

A hypodermic syringe contains water.  The barrel of the syringe has a cross-sectional area of 2.50 x 10-5 m2, and the needle has a cross-sectional area of 10-8 m2.  In the absence of a force on the plunger, the pressure everywhere is 1 atm.  A force of magnitude 2.00 N acts on the plunger, making water squirt horizontally from the needle.  Determine the speed of the medicine as it leaves the needle’s tip.     ans 12.6m/s

 A1v1 = A2v2 2.50 x 10-5 v1 = 10-8 v2 v2 = 2500 v1 So v12 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