Question

An inclined rectangular sluice gate AB 1.2 m by 5 m size as shown in Fig. Q3 is installed to control the discharge of water. The end A is hinged. Determine the force normal to the gate applied at B to open it.

Answer 1

Answer:

138.68 kN

Explanation:

I assume the figure is the one included in my answer.

Let's say r is the distance from the hinge A.  For a narrow section of the gate at this position, the length is dr, the width is w, and the area is dA.

dA = w dr

The pressure is:

P = ρgh

Using geometry, we can write h in terms of r.

(OA + r)² = h² + h²

(5√2 − 1.2 + r)² = 2h²

5√2 − 1.2 + r = √2 h

h = (5√2 − 1.2 + r) / √2

So the pressure at position r is:

P = ρg (5√2 − 1.2 + r) / √2

The force at position r is:

dF = P dA

dF = ρgw (5√2 − 1.2 + r) / √2 dr

The moment about hinge A caused by this force is:

dM = dF r

dM = (ρgw/√2) ((5√2 − 1.2) r + r²) dr

The total torque caused by the pressure is is:

M = ∫ dM

M = (ρgw/√2) ∫ ((5√2 − 1.2) r + r²) dr

M = (ρgw/√2) (½ (5√2 − 1.2) r² + ⅓ r³) [from r=0 to r=1.2]

M = (ρgw/√2) (½ (5√2 − 1.2) (1.2)² + ⅓ (1.2)³)

Sum of the moments on the gate:

∑τ = Iα

F(1.2) − M = 0

F = M / 1.2

F = (ρgw/√2) (½ (5√2 − 1.2) (1.2) + ⅓ (1.2)²)

Given that ρ = 1000 kg/m³, g = 9.81 m/s², and w = 5 m:

F = (1000 kg/m³ × 9.8 m/s² × 5 m /√2) (½ (5√2 − 1.2) (1.2) + ⅓ (1.2)²)

F = 138.68 kN

Round as needed.