Question

A water tank is divided into two separate sections with a window between them. Section 1 is filled with water that is H₂feet high and section 2 is filled H₃ feet high. The window is W₁ feet wide, H₁ feet high and C feet thick. The window is H₄ feet above the bottom of the tank. Calculate the magnitude of the force on the window due to section 1. (Note: Neglect the thickness of the window)  The magnitude of the force on the window due to section 1.Please be as detailed as possible :) thank you very much

A water tank is divided into two separate sections with a window between them. Section 1 is filled with water that is Hafeet high and section 2 is filled H3 feet high. The window is W4 feet wide, Hy feet high and C feet thick. The window is H4 feet above the bottom of the tank. Calculate the magnitude of the force on the window due to section 1. (Note: Neglect the thickness of the window) COLLAPSE IMAGES H= 1.2 ft H2 = 8.2 ft H3 = 19.2 ft He=2.4ft W1 = 2.2 The pressure at the top of the window due to section 1. bift ENTER 3 tres remaining points possible The pressure at the bottom of the window due to section 1. Ib/ft ENTER 3 tries remaining 1 points possible The magnitude of the force on the window due to section 1. ENTER

Answer 1

Pressure at top of window due to section 1=[H2 -(H4 + H1)]*62.4 = 4.6*62.4 lb/ft2 = 287.04 lb/ft2 = 287.04*W1 = 287.04*2.2 = 631.488 lb/ft

Pressure at bottom of window due to section 1 = [H2 - H4]*62.4*2.2 = 796.224 lb/ft

Magnitude of force on window = (631.488*H1) +(0.5*(796.224-631.488)*H1)= 856.6272 lb