Explanation Calculation: Sketch the trapezoidal section as shown in Figure 1. Find the hydrostatic force using the formula. R = γ y ¯ A Find the location of point of application of resultant force using the formula: y P = I S S ′ y ¯ A Here, R is the resultant of the hydrostatic forces acting on the gate and y P is the depth to the point of application of R . Find the centroid y ¯ A using the relation: Refer to Figure 1. y ¯ A = ∑ y ¯ A = [ h + 0.17 ] ( 1 2 × 1.2 × 0.51 ) + [ h + 0.34 ] ( 1 2 × 0.84 × 0.51 ) = [ h + 0.17 ] ( 0.306 ) + [ h + 0.34 ] ( 0.2142 ) = 0.306 h + 0.05202 + 0.2142 h + 0.0728 = ( 0.5202 h + 0.12484 ) m 3 Find the resultant force ( R ) acting on the gate using the relation: R = γ y ¯ A = p y ( y ¯ A ) (1) Substitute 10 3 kg / m 3 for p , 9.81 m / s 2 for y , and ( 0.5202 h + 0.12484 ) m 3 for y ¯ A in Equation (1). R = 10 3 ( 9.81 ) ( 0.5202 h + 0.12484 ) = 5 , 103.162 ( h + 0.24 ) N Find the moment of inertia ( I S S ′ ) 1 using the relation as shown below: ( I S S ′ ) 1 = I x + A d 2 = b h 3 36 + ( 1 2 b h ) d 2 (2) Here, b is the width of the trapezoidal section, d is vertical distance from the centroid of the segment to the neutral axis, and h is the height of the trapezoidal section. Substitute 1.2 m for b , 0.51 m for h , and ( h + 0.17 ) m for d in Equation (2). ( I S S ′ ) 1 = 1.2 × 0.51 3 36 + ( 1 2 × 1.2 × 0.51 ) ( h + 0.17 ) 2 = 0.0044217 + 0.306 ( h 2 + 0.0289 + 0.33 h ) = 0.306 h 2 + 0.10404 h + 0.013265 = ( 0.306 h 2 + 0.10404 h + 0.013265 ) m 4 Find the moment of inertia ( I S S ′ ) 2 using the relation as shown below: ( I S S ′ ) 2 = I ¯ x 2 + A d 2 = b h 3 36 + ( 1 2 b h ) d 2 (3) Substitute 0.84 m for b , 0.51 m for h , and ( h + 0.34 ) m for d in Equation (3). ( I S S ′ ) 2 = 0.84 × 0.51 3 36 + ( 1 2 × 0

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Chapter 9.2, Problem 9.61P

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Find the depth of of water for which the gate will open.

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Find the depth of of water for which the gate will open.

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