Homework SourseCalculate the fluid force on the side of the right triangle
Calculate the fluid force on the side of the right triangle of height 3 feet and base 2 feet submerged in water vertically, with upper vertex located at a depth of 4 feet. The density of water =62.5 ft/s^2.
Solution
Relevant equations w=(force)(distance) The attempt at a solution force=(62.5 lb/ft^3)(9.8m/s^2)= 6.12.598 work=int. 0 to 4(612.598)(4)ydy weight=force=(volume)(weight of water) =(3.1412^2deltayft^3)(62.5 lb/ft^3) =(0900)(3.14lb) distance=yft work=int. 0 to 4 (9000)(3.14)ydy =(226194.67 ft)(pound)
