An aquarium 6 ft long, 4 ft wide, and 3 ft deep is full of water. a) Find the hydrostatic pressure on the bottom of the aquarium. b) Find the hydrostatic force on the bottom of the aquarium. c) Fin

a) To find the hydrostatic pressure on the bottom of the aquarium, we use the formula: {eq}P = rho gh {/eq} where {eq}P {/eq} is the pressure, {eq}rho {/eq} is the density of the fluid, {eq}g {/eq} is the acceleration due to gravity, and {eq}h {/eq} is the depth of the fluid. In this case, we know that the aquarium is full of water, so {eq}rho {/eq} is the density of water, which is \approx imately {eq}1000,\text{ kg/m}^3 {/eq}. We also know that {eq}g {/eq} is \approx imately {eq}9.81,\text{ m/s}^2 {/eq}, the acceleration due to gravity on Earth. The depth of the water is {eq}3,\text{ ft} {/eq}, which is equivalent to {eq}0.9144,\text{ m} {/eq}. Therefore: {eq}P = (1000,\text{ kg/m}^3)(9.81,\text{ m/s}^2)(0.9144,\text{ m}) \approx 8984,\text{ Pa} {/eq} So the hydrostatic pressure on the bottom of the aquarium is \approx imately {eq}8984,\text{ Pa} {/eq}. b) To find the hydrostatic force on the bottom of the aquarium, we use the formula: {eq}F = PA {/eq} where {eq}F {/eq} is the force, {eq}P {/eq} is the pressure, and {eq}A {/eq} is the area over which the pressure acts. In this case, we can assume that the bottom of the aquarium is flat, so the area over which the pressure acts is simply the area of the bottom, which is: {eq}A = (6,\text{ ft})(4,\text{ ft}) = 24,\text{ ft}^2 {/eq} To find the pressure, we can use the value we found in part (a), which is \approx imately {eq}8984,\text{ Pa} {/eq}. Therefore, the hydrostatic force on the bottom of the aquarium is: {eq}F = (8984,\text{ Pa})(24,\text{ ft}^2) \approx 20,885,\text{ lbf} {/eq} So the hydrostatic force on the bottom of the aquarium is \approx imately {eq}20,885,\text{ lbf} {/eq}. c) To find the hydrostatic force on one end of the aquarium, we need to know the area of one end and the pressure acting on it. Assuming that the end is a rectangle and that the depth of the water is the same as before, the area of one end is: {eq}A = (4,\text{ ft})(3,\text{ ft}) = 12,\text{ ft}^2 {/eq} To find the pressure, we need to know the depth of the water at the end. Assuming that the end is perpendicular to the ground, the depth will be half of the depth of the aquarium, which is {eq}1.5,\text{ ft} {/eq} or \approx imately {eq}0.4572,\text{ m} {/eq}. Therefore, the pressure at one end of the aquarium is: {eq}P = (1000,\text{ kg/m}^3)(9.81,\text{ m/s}^2)(0.4572,\text{ m}) \approx 4474,\text{ Pa} {/eq} Using this pressure and the area, we can find the hydrostatic force on one end of the aquarium: {eq}F = (4474,\text{ Pa})(12,\text{ ft}^2) \approx 64,488,\text{ N} {/eq} So the hydrostatic force on one end of the aquarium is \approx imately {eq}64,488,\text{ N} {/eq}.