Explanation Given information: The width of the gate is 4 ft. The height of the gate is 3 ft. The water level in the freshwater marsh is h = 6 ft . Calculation: Consider the density of the freshwater marsh is γ = 62.4 lb/ft 3 . Show the pressure distribution diagram as in Figure 1. Consider the pressure is effective up to the height of gate. Since, the gate needs to be open. Find the intensity of the force as follows; w = width of gate × depth × density =4 ( depth ) γ Refer to the intensity of force calculated; w 1 = 4 γ ( h − 3 ) w 2 = 4 γ h w ′ 1 = 4 γ ′ ( d − 3 ) w ′ 2 = 4 γ ′ d Here, the density of the freshwater marsh is γ , the density of the water in ocean is γ ′ , and the height of the water level in ocean is d . Consider the pressure distribution in the gate portion as triangles for easy calculation. Refer to the figure 1; Calculate the resultant pressure in triangle 1 as follows; P ′ 1 − P 1 = 1 2 × 3 × w ′ 1 − 1 2 × 3 × w 1 Substitute 4 γ ′ ( d − 3 ) for w ′ 1 and 4 γ ( h − 3 ) for w 1 . P ′ 1 − P 1 = 1 2 × 3 × 4 γ ′ ( d − 3 ) − 1 2 × 3 × 4 γ ( h − 3 ) Calculate the resultant pressure in triangle 2 as follows; P ′ 2 − P 2 = 1 2 × 3 × w ′ 2 − 1 2 × 3 × w 2 Substitute 4 γ ′ d for w ′ 2 and 4 γ h for w 2 . P ′ 2 − P 2 = 1 2 × 3 × 4 γ ′ d − 1 2 × 3 × 4 γ h Show the free-body diagram of the gate AB as in Figure 2. Find the horizontal reaction at point B by taking moment about point A

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Author: BEER

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Chapter 5.3, Problem 5.85P

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Find the ocean level d for which the gate will open.

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Chapter 5.3, Problem 5.84P

Chapter 5.3, Problem 5.86P

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