# Fluid mechanics

### 1 個解答

• 天同
Lv 7
6 年前
最愛解答

Let w be the width of the gate. Area of the vertical portion of the gate = wh Area of the horizontal portion of the gate = 1.5w Pressure at the vertical portion = 1000 x 9.81x (h/2) = 4905h where 1000 kg /m^3 is the density of water 9.81 m/s^2 is the acceleration due to gravityPressure force at the vertical portion of the gate = 4905h x (hw) = 4905wh^2 Area moment of inertia of the vertical portion of the gate = wh^3/12 Distance of the action point of pressure force from the centroid of the gate = (wh^3/12)/[(h/2).(wh)] = h/6 Hence, distance of action point of pressure force from point O = h/2 - h/6 = h/3 Consider the horizontal portion of the gate. Pressure acting on the gate = 1000 x 9.81 x h = 9810h Pressure force on the gate = 9810h x 1.5w = 14715wh Because the gate is horizontal, pressure is the same along the gate. Hence, the action point of pressure force is at the mid-point of the gate, i.e. (1.5/2) m = 0.75 m from O. Taking moment about O, (4905wh^2) x (h/3) = (14715wh) x (0.75) i.e. 1635h^2 = 11036 h = 2.6 m

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