## College Physics (4th Edition)

We can rank the barrels in order of the force on the plug due to the liquid in the barrel, from largest to smallest: $c = e \gt b = d \gt a$
The pressure on the plug from the inside of the barrel is $P_{atm}+ P_{gauge}$ while the pressure on the plug from outside the barrel is $P_{atm}$. Therefore, the net pressure on the plug is $P_{gauge}$. We can calculate the force on the plug for each barrel. (a) $F = P_g~A$ $F = [\rho~g~(h-0.20~m)](\pi~r^2)$ $F = [(1000~kg/m^3)~(9.80~m/s^2)~(1.00~m-0.20~m)](\pi)~(0.010~m)^2$ $F = 2.46~N$ (b) $F = P_g~A$ $F = [\rho~g~(h-0.20~m)](\pi~r^2)$ $F = [(1000~kg/m^3)~(9.80~m/s^2)~(1.20~m-0.20~m)](\pi)~(0.010~m)^2$ $F = 3.08~N$ (c) $F = P_g~A$ $F = [\rho~g~(h-0.20~m)](\pi~r^2)$ $F = [(800~kg/m^3)~(9.80~m/s^2)~(1.20~m-0.20~m)](\pi)~(0.0125~m)^2$ $F = 3.85~N$ (d) $F = P_g~A$ $F = [\rho~g~(h-0.20~m)](\pi~r^2)$ $F = [(800~kg/m^3)~(9.80~m/s^2)~(1.00~m-0.20~m)](\pi)~(0.0125~m)^2$ $F = 3.08~N$ (e) $F = P_g~A$ $F = [\rho~g~(h-0.20~m)](\pi~r^2)$ $F = [(1000~kg/m^3)~(9.80~m/s^2)~(1.45~m-0.20~m)](\pi)~(0.010~m)^2$ $F = 3.85~N$ We can rank the barrels in order of the force on the plug due to the liquid in the barrel, from largest to smallest: $c = e \gt b = d \gt a$