A trough has vertical ends that are equilateral triangles with sides of length $7 \mathrm { ft }$. If the trough is filled with water to a depth of $4 \mathrm { ft }$, find the force exerted by the water on one end of the trough. Round to one decimal place. (The weight density of water is $62.4 \mathrm { lb } / \mathrm { ft } ^ { 3 }$.) Select the correct answer.
A) $\quad 384.3 \mathrm { lb }$
B) $1,337.7 \mathrm { lb }$
C) $768.6 \mathrm { lb }$
D) $2,675.4 \mathrm { lb }$

Question

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. To find the weight of the water, we first need to find the volume of the water using the formula for the volume of a triangular prism:

$$V = \frac{1}{2}bh\times l$$

where $b$ is the length of the base of the triangle (which is also 7 ft), $h$ is the height of the triangle (which is also 7 ft), and $l$ is the length of the trough (which we will call $L$).

Thus,

$$V = \frac{1}{2}\times7\times7\times L = \frac{49}{2}L$$

The weight of the water is given by:

$$W = mg$$

where $m$ is the mass of the water and $g$ is the acceleration due to gravity. The mass of the water is equal to its density times its volume, so we have:

$$m = \rho V$$

where $\rho$ is the density of water. Substituting known values, we have:

$$m = (62.4\mathrm{lb/ft^3})(\frac{49}{2}L) = 3067.2L\mathrm{lb}$$

Therefore,

$$W = mg = (3067.2L\mathrm{lb})(32\mathrm{ft/s^2}) = 98230.4L\mathrm{lb}$$

The weight of the water is distributed evenly over the end of the trough, which is an equilateral triangle with area:

$$A = \frac{\sqrt{3}}{4}s^2$$

where $s$ is the length of a side of the triangle. Substituting known values, we have:

$$A = \frac{\sqrt{3} \times 7^2}{4} = \frac{49\sqrt{3}}{4}$$

Therefore, the force exerted by the water on one end of the trough is:

$$F = \frac{W}{2A} = \frac{98230.4L\mathrm{lb}}{2(\frac{49\sqrt{3}}{4})} = \frac{3991.04L}{\sqrt{3}}\mathrm{lb}$$

We need to find the force when $L=4\mathrm{ft}$. Substituting this value, we get:

$$F = \frac{3991.04(4)}{\sqrt{3}}\mathrm{lb} \approx 768.6\mathrm{lb}$$

Therefore, the answer is C.

A Trough Has Vertical Ends That Are Equilateral Triangles with Sides