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http://www.cramster.com/answers-mar-12/mechanical-engineering/fluid-static-3-wall-reservoir-approximately-75m-tall_2297209.aspx?rec=0

http://www.cramster.com/answers-mar-12/mechanical-engineering/fluid-static-2-length-342-cross-section-shown_2297195.aspx?rec=0

http://www.cramster.com/answers-mar-12/mechanical-engineering/fluid-static-1-square-cross-section-16cm-side-diagram-wrong_2297193.aspx?rec=0

http://www.cramster.com/answers-mar-12/mechanical-engineering/hw-needed-follow-link-questionhttp-ima_2297111.aspx?rec=0


please!!!!

Explanation / Answer

This solution is for the last question.

(a) Depth of pin from the free surface of water on right side = H - (LSin).
Pressure on the gate from water on right side of the gate will vary linearly from g (H - LSin) at the pin to gH at the bottom. Hence average pressure on the gate from right side = [g (H - LSin) + gH]/2 = g (2H - LSin)/2 = 26893.6 lb/ft-s^2.

This average pressure will act at the C.G of the trapezium whose parallel sides are g (H - LSin) and gH and the distance between them is L and its direction will be normal to the gate.

Location of this Centre of Pressure from the pin is given by

x = [(g (H - LSin)*L)*(L/2) + (0.5*L*( gH - g (H - LSin))*(2L/3) ] / [g (H - LSin)*L + (0.5*L*( gH - g (H - LSin)))]

which can be simplified to

x = (H - (LSin/3))*L / (2H - LSin)

By putting values, we get x = 2.6 ft.

Similarly, average pressure on the gate by the water on left side = g (2h - LSin)/2 = 8821.34 lb/ft-s^2.

Location of centre of pressure due to water on left side of gate from the pin y = (h - (LSin/3))*L / (2h - LSin).

By putting values, y = 2.805 ft. This pressure will also act normal to the gate but towards right side.

For equilibrium, we take moments of these forces about the pin and equate it to the moment due to the weight of the gate.

Therefore, (26893.6*(5*3)*2.6) - (8821.34*(5*3)*2.805) = W*(L/2*Cos)

By putting L = 5 ft, it gives,

Weight of the gate, W = 353865 lb-ft/s^2.

Hence, mass of the gate = W/g = 10996 lb.

(b) For Reactions at the pin joint, we need to balance the forces in horizontal and vertical direction.

Net force in horizontal direction will be (26893.6*(5*3))(Sin40) - (8821.34*(5*3))(Sin40) = 174249 lb-ft/s^2 = 5414.8 lbf. This force will be acting towards left. Hence net reaction force at the pin in horizontal direction will be 5414.8 lbf acting towards right.

Similarly, net force on the gate in vertical direction will be 26893.6*(5*3)(Cos40) - 8821.34*(5*3)(Cos40) - 353865 = -146202 lb-ft/s^2 = -4543.3 lbf. Negative sign indicates that this force will be acting downwards.

Hence reaction at the pin in vertical direction will be 4543.3 lbf in upward direction.

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