441_PartUniversity Physics Solution

441_PartUniversity Physics Solution - Fluid Mechanics...

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Fluid Mechanics 14-5 (c) 58 92(1.013 10 Pa) (1.72 m)(11.5 m) 5.79 10 N Fp A π == × = × EVALUATE: Most of the force in part (a) is due to the 92 atm of air pressure above the surface of the benzene and the net force on the bottom of the tank is much less than the inward and outward forces. 14.23. IDENTIFY: The gauge pressure at the top of the oil column must produce a force on the disk that is equal to its weight. SET UP: The area of the bottom of the disk is 22 2 (0.150 m) 0.0707 m Ar ππ = . EXECUTE: (a) 0 2 45.0 N 636 Pa 0.0707 m w pp A −== = . (b) The increase in pressure produces a force on the disk equal to the increase in weight. By Pascal’s law the increase in pressure is transmitted to all points in the oil. (i) 2 83.0 N 1170 Pa 0.0707 m p Δ= = . (ii) 1170 Pa EVALUATE: The absolute pressure at the top of the oil produces an upward force on the disk but this force is partially balanced by the force due to the air pressure at the top of the disk. 14.24. IDENTIFY: 2 21 1 A FF A = . 2 F must equal the weight wm g = of the car. SET UP: 2 /4 AD = . 1 D is the diameter of the vessel at the piston where 1 F is applied and 2 D of the diameter at the car. EXECUTE: 2 2 1 2 1 D mg F D = . 2 2 11 (1520 kg)(9.80 m/s ) 10.9 125 N Dm g DF = EVALUATE: The diameter is smaller where the force is smaller, so the pressure will be the same at both pistons. 14.25. IDENTIFY: Apply y y Fm a = to the piston, with y + upward. A = . SET UP: 5 1 atm 1.013 10 Pa . The force diagram for the piston is given in Figure 14.25. p is the absolute pressure of the hydraulic fluid. EXECUTE: atm 0 pA w p A −− = and 2 5 atm gauge (1200 kg)(9.80 m/s ) 1.7 10 Pa 1.7 atm (0.15 m) g p −= = = × = EVALUATE: The larger the diameter of the piston, the smaller the gauge pressure required to lift the car. Figure 14.25 14.26. IDENTIFY: Apply Newton&s 2nd law to the woman plus slab. The buoyancy force exerted by the water is upward and given by water displ , B Vg ρ = where displ V is the volume of water displaced. SET UP: The floating object is the slab of ice plus the woman; the buoyant force must support both. The volume of water displaced equals the volume ice V of the ice. The free-body diagram is given in Figure 14.26. EXECUTE: y y a = tot 0 Bmg = water ice ice (45.0 kg ) m g =+ But / mV = so ice ice ice mp V = Figure 14.26 3 ice 33 water ice 45.0 kg 45.0 kg 0.562 m . 1000 kg/m 920 kg/m V ρρ = EVALUATE: The mass of ice is ice ice ice 517 kg.
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14-6 Chapter 14 14.27. IDENTIFY: Apply y y Fm a = to the sample, with y + upward. water obj B Vg ρ = . SET UP: 17.50 N wm g == and 1.79 kg m = . EXECUTE: 0 TBm g +− = . 17.50 N 11.20 N 6.30 N Bm gT =− = = . 43 obj 33 2 water 6.30 N 6.43 10 m (1.00 10 kg/m )(9.80 m/s ) B V g = × × . 1.79 kg 2.78 10 kg/m m V = × × . EVALUATE: The density of the sample is greater than that of water and it doesn&t float. 14.28. IDENTIFY: The upward buoyant force B exerted by the liquid equals the weight of the fluid displaced by the object. Since the object floats the buoyant force equals its weight. SET UP: Glycerin has density gly 1.26 10 kg/m and seawater has density sw 1.03 10 kg/m . Let obj V be the volume of the apparatus. 2 E 9.80 m/s g = ; 2 C 4.15 m/s g = . Let sub V be the volume submerged on Caasi.
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441_PartUniversity Physics Solution - Fluid Mechanics...

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