![]() Evaluate the definite integral \( \int^50_0 B(h) dh\).Taking the moments of all forces about A, we get Let the total force F is acting at a depth of h from the free surface of liquid, i.e., from A. Total pressure force on one vertical face of the tank, of the triangle EFC, i.e., at a distance of 0.9 +(2/3) X 0.6 = 1.30 m below A. (iii) Forceğ 3 = Area of triangle EFC x Width of tank of the rectangle BDEF i.e., at a distance of 0.9+ (0.6/2)= 1.2 m (ii) Force F 2 = Area of rectangle BDEF x Width of tank of the triangle ADE, i.e., at a distance of (2/3 ) X 0.9 = 0.6 m below A (i) Force F 1 = Area of triangle ADE X Width of tank The total pressure force consists of the following components: The pressure diagram is split into triangle ADE, rectangle BDEF and triangle EFC. Intensity of pressure at D, Pd = P1g x h = 900 x 9.81 x 0.9 = 7946.1 N/m 2 (a) Total pressure on one vertical side is calculated by drawing pressure diagram, which is shown in Fig.8-1 Given : Cubical tank of sides 1.5 m means the dimensions of the tank are 1.5 mx 1.5 mx 1.5 m.ĭensity of liquid, Pi = 0.9 x 1000 = 900 kg/m 3 (a) total pressure, and (b) position of centre of pressure. Calculate for one vertical side of the tank : The upper remaining part is filled with oil of specific gravity 0.9. It contains water for the lower 0.6 m depth. ![]()
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