You are given the shape of the vertical ends of a trough that is completely filled with water. Find the force exerted by the water on one end of the trough. (The weight density of water is $62.4 \mathrm { lb } / \mathrm { ft } ^ { 3 }$.)

A) $666 \mathrm { lb }$
B) $1,331 \mathrm { lb }$
C) $2,662 \mathrm { lb }$
D) $333 \mathrm { lb }$

Question

You are given the shape of the vertical ends of a trough that is completely filled with water. Find the force exerted by the water on one end of the trough. (The weight density of water is $62.4 \mathrm { lb } / \mathrm { ft } ^ { 3 }$.)

A) $666 \mathrm { lb }$ 
B) $1,331 \mathrm { lb }$ 
C) $2,662 \mathrm { lb }$ 
D) $333 \mathrm { lb }$

Quizplus · Accepted Answer

The force exerted by the water on one end of the trough depends on the weight of the water and the shape of the end. We can use the formula $F = \gamma A h$, where $\gamma$ is the weight density of water, $A$ is the area of the end, and $h$ is the depth of the water. The weight density of water is given as $62.4 \mathrm{lb/ft^3}$.

From the diagram, we can see that the end of the trough is a trapezoid. We can divide the end into two parts: a rectangle with height 3 ft and width 10 ft, and a trapezoid with height 3 ft, bottom base 10 ft, and top base 6 ft. The area of the rectangle is $A_1 = 3 \mathrm{ft} \times 10 \mathrm{ft} = 30 \mathrm{ft^2}$, and the area of the trapezoid is $A_2 = \frac{1}{2} (6\mathrm{ft}+10\mathrm{ft}) \times 3\mathrm{ft} = 24 \mathrm{ft^2}$. Thus, the total area of the end is $A = A_1 + A_2 = 54 \mathrm{ft^2}$.

The depth of the water is the height of the end, which is 3 ft. Therefore, $h=3 \mathrm{ft}$.

Now we can plug in the numbers to find the force exerted by the water: $F = \gamma A h = (62.4 \mathrm{lb/ft^3}) \times 54 \mathrm{ft^2} \times 3 \mathrm{ft} = 9,547.2 \mathrm{lb} \approx 2,662 \mathrm{lb}$. Therefore, the answer is (C), $2,662 \mathrm { lb }$.

You Are Given the Shape of the Vertical Ends of a Trough