Sm1 42thru1 43

PROBLEM 1.42 A = 0.0018 m² patm = 100 kPa Spring force varies linearly from 900 N when y1 = 0.003 m to zero when y2 = 0.002 m AIR At the initial and final states, static equilibrium exists so effect of friction (between the piston and cylinder) can be ignored. Thus, focusing at the interface between the air and the friction (dashed) we have: PAIR = patm + Fs A Initially, Fspring = 900 N. So, p1 = 100 kPa + Fs = 150 kPa. Finally, Fspring = 0 N. So, p2 = 100 kPa. P2 = 0.999 atm. PROBLEM 1.43 patm = 14.7 lbf/in² 60 Building Storage Tank HIGH g = 62.4 The pressure at the bottom of the storage tank is p = patm + p g l = 14.7 (20 ft )(30 ft/s² ) + 62.4 = 14.7 + 8.6 = 23.3 FORWARD