## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) $F_A > F_B$ (b) $F_A = F_B$
$P = P_0 + \rho~g~h$ $P$ is the pressure $P_0$ is the atmospheric pressure $\rho$ is the density of the liquid $h$ is the depth below the surface (a) For both tanks, the pressure $P$ at the bottom of the tank is the same because the depth under the surface is the same. We know that $F = P~A$, where $A$ is the area. Since the area of the bottom of Tank A is larger than the area of the bottom of Tank B, $F_A$ is larger than $F_B$ (b) The average pressure from the surface to the bottom of Tank A is equal to the average pressure from the surface to the bottom of Tank B. We know that $F = P~A$, where $A$ is the area. Since the area of the side of Tank A is equal to the area of the side of Tank B, $F_A$ is equal to $F_B$.