Example 2.7 Find the magnitude of the resultant force on this vertical wall of a tank which has oil, of relative density 0.8, floating on water.

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Example 2.7 Find the magnitude of the resultant force on this vertical wall of a tank which has oil, of relative density 0.8, floating on water.

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Example 2.7 Find the magnitude of the resultant force on this vertical wall of a tank which has oil, of relative density 0.8, floating on water as shown

Draw the pressure diagram as shown to the right. Solution Draw the pressure diagram as shown to the right. The resultant force (per unit length) is simply the area of the pressure diagram. It is convenient to split the area into three, and sum the answers. F1 = A1 = (0.8 x 103) x 9.81 x 0.8 x 0.8x0.5 = 2511.36 F2 = A2 = (0.8 x 103) x 9.81 x 0.8 x 1.2 = 7534.08 F3 = A3 = (103) x 9.81 x 1.2 x 1.2 x 0.5 = 7063.20 R = F1 + F2 + F3 = 17108.64 N/m R acts horizontally through the centroid of the pressure diagram.

Example 2.8 Find the magnitude and direction of the resultant force (in N not N/m) of water on a quadrant gate as shown below

At an angle Horizontal force, RH = Force on projection of curved surface on to a vertical plane = 0.5 x rgh2 x width = 0.5 x 1000 x 9.81 x 12 x 3 = 14715 N Vertical force, RV = weight of fluid above surface = rg x Area covered = 1000 x 9.81 x(pr2/4) x 3 = 23114 N Resultant force At an angle

Example 2.8

Solution ht

A 1. 5m long cylinder, radius 1m, lies as shown in the figure A 1.5m long cylinder, radius 1m, lies as shown in the figure. It holds back oil of relative density 0.8. If the cylinder has a mass of 2250 kg find the reaction at A the reaction at B

Horizontal force RH = projection of vertical plane Reaction at A = -RH = pressure at centroid x area = 0.5rgH x (H Length) = 0.5 rgH2 x Length = 0.5 x 0.8 x 103 x 9.81 x 22 x 1.5 = 23544 N, to the left Vertical force = Reaction at B RV = Force due to weight of fluid, DCE, above (down) + Force due to fluid below BD (upward) + Force due to weight of cylinder Force of fluid above = area of sector DCE x length x rg Force from below = area of real or imaginary fluid above BD = area of BDEC x length x rg Taking downward as positive RV = 0.8 x 103x9.81 x(1x1 - pr2/4) x 1.5 - 0.8 x103 x9.81 x( 1x1 + pr2/4) ) x 1.5 + 22509.81 = 2526 - = 3580 N Reaction at B = 3580 N, vertically up

The resultant and angle of application are given by: