Question

A long solid cylinder of radius 0.8 m hinged
at point A is used as an automatic gate, as
shown in Fig. beside. 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

Answer 1

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:

$𝐹_𝐻 = 𝐹_𝑥 = 𝑃_{𝑎𝑣𝑒}𝐴 = 9810\times(4.2 + \frac{0.8}{2}) (0.8\times1) = 36101 N$

Vertical force on horizontal surface (upward):

$𝐹_𝑦 = 𝑃_{𝑎𝑣𝑒}𝐴 = 9810\times5\times(0.8\times1) = 39240 N$

Weight of fluid block per m length (downward):

$𝑊 = 𝑚𝑔 = ρgV = ρg(𝑅^2 − \frac{π𝑅^2}{4})\times(1)$

$𝑊 = 9810 (0.8^2 − 0.8^2\times\frac{π}{4})\times(1) = 1347 N$

The net upward vertical force is

$𝐹_𝑉 = 𝐹𝑦 − 𝑊 = 39240 − 1347 = 37893 N$

Then the magnitude and direction of the hydrostatic force acting on the cylindrical surface become

$𝐹_𝑅 = \sqrt{(𝐹_𝐻)^2 + (𝐹_𝑉)^2}$

$𝐹_𝑅 = \sqrt{(36101)^2 + (37893)^2} =52337 N$

The tangent of the angle it makes with the horizontal is

$\theta = arctan(\frac{𝐹_𝑉}{𝐹_𝐻}) = arctan(\frac{37893}{36101}) = 46.4°$

(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

$F_RRsin\theta – W_{cyl}R = 0 → 𝑊_{𝑐𝑦𝑙} = F_RRsin\theta = 52337\times𝑠𝑖𝑛46.4 = 37900 N$