Questions: How much work is required to empty the full tank shown below by pumping water up and over the top edge? Hint: Density of water = 62.4 lb / ft^3. W=[?] ft-lb Round your answer to the nearest whole number

How much work is required to empty the full tank shown below by pumping water up and over the top edge?
Hint: Density of water = 62.4 lb / ft^3.
W=[?] ft-lb

Round your answer to the nearest whole number
Transcript text: How much work is required to empty the full tank shown below by pumping water up and over the top edge? Hint: Density of water $=62.4 \mathrm{lb} / \mathrm{ft}^{3}$. \[ \mathrm{W}=[?] \mathrm{ft}-\mathrm{lb} \] Round your answer to the nearest whole number
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Solution

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Solution Steps

Step 1: Find the cross-sectional area

The triangular cross-section has a base of 2 ft and a height of 2 ft. Therefore, its area is: A = (1/2) * base * height = (1/2) * 2 ft * 2 ft = 2 ft²

Step 2: Set up the integral for work

The work required to lift a slice of water at depth 'y' with thickness 'dy' over the top edge is given by: dW = (density of water) * (cross-sectional area) * (distance lifted) * (thickness) * g dW = 62.4 lb/ft³ * 2 ft² * y * dy * 32.2 ft/s² (Here, g is implicit since we're given lb, not slugs). Since we're taking slices horizontally, y represents the distance the slice has to be lifted. The length of the tank is 6 ft. This gives us: dW = 62.4 lb/ft³ * 2 ft² * y dy * 6 ft dW = 748.8y dy ft-lb

Step 3: Integrate to find the total work

Integrating the above expression from y = 0 to y = 2 (the depth of the tank) gives the total work: W = ∫(from 0 to 2) 748.8y dy W = 748.8 [y²/2](from 0 to 2) W = 748.8 * (4/2 - 0/2) W = 748.8 * 2 W = 1497.6 ft-lb

Final Answer The final answer is $\boxed{1498}$

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