A Gravity-Controlled Cylindrical Gate
A long solid cylinder of radius 0.8 m hinged at point A is used as an automatic gate, as shown in Fig. 3–41. When the water level reaches 5 m, the gate opens by turning about the hinge at point A. Determine (a) the hydrostatic force acting on the cylinder and its line of action when the gate opens and (b) the weight of the cylinder per m length of the cylinder.
The height of a water reservoir is controlled by a cylindrical gate hinged to the reservoir. The hydrostatic force on the cylinder and the weight of the cylinder per m length are to be determined.
Assumptions 1 Friction at the hinge is negligible. 2 Atmospheric pressure acts on both sides of the gate, and thus it cancels out.
Properties We take the density of water to be 1000 kg/m³ throughout.
Analysis (a) We consider the free-body diagram of the liquid block enclosed by the circular surface of the cylinder and its vertical and horizontal projections. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are determined as
Horizontal force on vertical surface:
\begin{aligned}F_{H} &=F_{x}=P_{ avg } A=\rho g h_{C} A=\rho g(s+R / 2) A \\&=\left(1000 kg / m ^{3}\right)\left(9.81 m / s ^{2}\right)(4.2+0.8 / 2 m )(0.8 m \times 1 m )\left(\frac{1 kN }{1000 kg \cdot m / s ^{2}}\right) \\&=36.1 kN\end{aligned}
Vertical force on horizontal surface (upward):
\begin{aligned}F_{y} &=P_{ avg } A=\rho g h_{C} A=\rho g h_{ bottom } A \\&=\left(1000 kg / m ^{3}\right)\left(9.81 m / s ^{2}\right)(5 m )(0.8 m \times 1 m )\left(\frac{1 kN }{1000 kg \cdot m / s ^{2}}\right) \\&=39.2 kN\end{aligned}
Weight (downward) of fluid block for one m width into the page:
\begin{aligned}W &=m g=\rho g V=\rho g\left(R^{2}-\pi R^{2} / 4\right)(1 m ) \\&=\left(1000 kg / m ^{3}\right)\left(9.81 m / s ^{2}\right)(0.8 m )^{2}(1-\pi / 4)(1 m )\left(\frac{1 kN }{1000 kg \cdot m / s ^{2}}\right) \\&=1.3 kN\end{aligned}
Therefore, the net upward vertical force is
F_{V}=F_{y}-W=39.2-1.3=37.9 kN
Then the magnitude and direction of the hydrostatic force acting on the cylindrical surface become
\begin{gathered}F_{R}=\sqrt{F_{H}^{2}+F_{V}^{2}}=\sqrt{36.1^{2}+37.9^{2}}=52.3 kN \\\tan \theta=F_{V} / F_{H}=37.9 / 36.1=1.05 \rightarrow \theta=46.4^{\circ}\end{gathered}
Therefore, the magnitude of the hydrostatic force acting on the cylinder is 52.3 kN per m length of the cylinder, and its line of action passes through the center of the cylinder making an angle 46.4° with the horizontal.
(b) When the water level is 5 m high, the gate is about to open and thus the reaction force at the bottom of the cylinder is zero. Then the forces other than those at the hinge acting on the cylinder are its weight, acting through the center, and the hydrostatic force exerted by water. Taking a moment about point A at the location of the hinge and equating it to zero gives
F_{R} R \sin \theta-W_{\text {cyl }} R=0 \rightarrow W_{\text {cyl }}=F_{R} \sin \theta=(52.3 kN ) \sin 46.4^{\circ}=37.9 kN
Discussion The weight of the cylinder per m length is determined to be 37.9 kN. It can be shown that this corresponds to a mass of 3863 kg per m length and to a density of 1921 kg/m³ for the material of the cylinder.