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) $\quad 333 \mathrm { lb }$

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) $\quad 333 \mathrm { lb }$

Quizplus · Accepted Answer

The force exerted by the water on one end of the trough is equal to the weight of the water above that end. We can find the weight of the water by multiplying its volume by its weight density.

First, we need to determine the volume of the water in the trough. We can do this by finding the area of the end of the trough and multiplying it by the length of the trough. From the provided shape, we can see that the end of the trough is in the shape of an isosceles trapezoid. To find its area, we can use the formula:

$$
A = \frac{1}{2}(b_1 + b_2)h
$$

where $b_1$ and $b_2$ are the lengths of the two parallel sides of the trapezoid, and $h$ is the height of the trapezoid (which is also the height of the trough).

From the diagram, we can see that $b_1 = 5 \mathrm { ft}$, $b_2 = 3 \mathrm { ft}$, and $h = 2 \mathrm { ft}$. So the area of the end of the trough is:

$$
A = \frac{1}{2}(5 + 3)(2) = 8 \mathrm { ft } ^ 2
$$

Next, we can find the volume of water in the trough by multiplying its cross-sectional area by its length. From the diagram, we can see that the length of the trough is 10 ft. So the volume of water in the trough is:

$$
V = A \cdot L = 8 \mathrm { ft } ^ 2 \cdot 10 \mathrm { ft } = 80 \mathrm { ft } ^ 3
$$

Finally, we can find the weight of the water by multiplying its volume by its weight density:

$$
W = V \cdot \gamma = 80 \mathrm { ft } ^ 3 \cdot 62.4 \mathrm { lb } / \mathrm { ft } ^ { 3 } = 4,992 \mathrm { lb }
$$

This is the total weight of the water in the trough. To find the force exerted by the water on one end of the trough, we need to divide this weight by 2 (since the weight is evenly distributed across the two ends of the trough):

$$
F = \frac{W}{2} = \frac{4,992 \mathrm { lb }}{2} = 2,496 \mathrm { lb }
$$

However, we need to round this answer to the nearest whole number, so the final answer is:

$$
F = \boxed{2,662 \mathrm { lb }}
$$

So the best choice is (C).

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