A vertical gate in a dam has the shape of an isosceles trapezoid 8 feet across the top and 6 feet across the bottom, with a height of 5 feet, as shown below. What is the liquid pressure on the gate if the top of the gate is 4 feet below the surface of the water?
Hint: Density of water [tex]$=62.4 lb / ft ^3$[/tex].
Round your answer to the nearest pound.
Answer (2)
Determine the width of the trapezoid as a function of height: w ( y ) = 6 + 5 2 y .
Calculate the depth of the strip at height y : 9 − y .
Set up the integral for the total force: F = ∫ 0 5 62.4 ( 9 − y ) ( 6 + 5 2 y ) d y .
Evaluate the integral to find the total force: 13936 lb.
Explanation
Problem Setup We are given a vertical gate in a dam shaped like an isosceles trapezoid. The top width is 8 feet, the bottom width is 6 feet, and the height is 5 feet. The top of the gate is 4 feet below the surface of the water. We need to find the liquid pressure on the gate, given the density of water is 62.4 l b / f t 3 .
Finding the Width Function The width of the trapezoid at a height y from the bottom can be expressed as a linear function. Let w ( y ) be the width at height y . We know that w ( 0 ) = 6 and w ( 5 ) = 8 . The slope of the width function is 5 − 0 8 − 6 = 5 2 . Thus, w ( y ) = 6 + 5 2 y .
Setting up the Force Integral The depth of the strip at height y is 4 + ( 5 − y ) = 9 − y feet. The area of a horizontal strip of thickness d y is w ( y ) d y = ( 6 + 5 2 y ) d y . The pressure at depth 9 − y is given by P = ρ ( 9 − y ) , where ρ = 62.4 l b / f t 3 is the density of water. The force on the strip is d F = P ⋅ d A = ρ ( 9 − y ) ( 6 + 5 2 y ) d y = 62.4 ( 9 − y ) ( 6 + 5 2 y ) d y .
Calculating the Total Force To find the total force, we integrate the force over the height of the gate (from y = 0 to y = 5 ): F = ∫ 0 5 62.4 ( 9 − y ) ( 6 + 5 2 y ) d y
Evaluating the Integral Expanding the integrand, we get: 62.4 ( 9 − y ) ( 6 + 5 2 y ) = 62.4 ( 54 + 5 18 y − 6 y − 5 2 y 2 ) = 62.4 ( 54 − 5 12 y − 5 2 y 2 ) .
Now, we integrate from 0 to 5: F = 62.4 ∫ 0 5 ( 54 − 5 12 y − 5 2 y 2 ) d y = 62.4 [ 54 y − 5 6 y 2 − 15 2 y 3 ] 0 5 F = 62.4 [ 54 ( 5 ) − 5 6 ( 5 2 ) − 15 2 ( 5 3 )] = 62.4 [ 270 − 30 − 15 2 ( 125 )] = 62.4 [ 240 − 15 250 ] = 62.4 [ 240 − 3 50 ] = 62.4 [ 3 720 − 50 ] = 62.4 [ 3 670 ] F = 62.4 × 3 670 = 3 41808 = 13936 Therefore, the total force is 13936 lb.
Final Answer The liquid pressure on the gate is equal to the total force F . Therefore, the liquid pressure on the gate is 13936 lb.
Examples
Understanding liquid pressure on submerged surfaces is crucial in designing dams, submarines, and other underwater structures. Engineers use these calculations to ensure the structural integrity of these designs, preventing failures and ensuring safety. For example, calculating the force on a dam gate helps determine the necessary strength of the gate and its supports.
The total liquid pressure on the trapezoidal gate of the dam is approximately 13936 pounds. This was calculated by determining the width of the trapezoid as a function of height, integrating the pressure distribution across the height of the gate, and evaluating the integral. Finally, we found that the force on the gate is significant, necessitating careful engineering considerations.
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