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10 December, 19:21

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.

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  1. 10 December, 19:25
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    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.
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