Solve the problem.
-An isosceles triangular plate is submerged vertically in seawater, with its base on the bottom. The base is $10 \mathrm { ft }$ long, and the height of the triangle is $10 \mathrm { ft }$. Find the force exerted on one face of the plate if the water level is $2 \mathrm { ft }$ above the base of the triangle. Seawater weighs $64 \mathrm { lb } / \mathrm { ft } ^ { 3 }$. Round your answer to one decimal place if necessary.
A) 3584 lb
B) 1194.7 lb
C) 1792 lb
D) 341.3 lb

Question

Quizplus · Accepted Answer

First, we need to find the volume of the submerged part of the plate. Since the plate is isosceles, we can determine that the submerged part of the plate is also an isosceles triangle with a base of $10 \mathrm{ft}$ and a height of $8 \mathrm{ft}$ ($10 \mathrm{ft}$ height of the isosceles triangle minus $2 \mathrm{ft}$ height above the base of the submerged triangle). The area of this submerged triangle is:

$$\frac{1}{2} \cdot 10 \mathrm{ft} \cdot 8 \mathrm{ft} = 40 \mathrm{ft}^2$$

To find the volume, we multiply the area by the length of the submerged part of the plate, which is equal to $8 \mathrm{ft}$ (the remaining height of the isosceles triangle):

$$40 \mathrm{ft}^2 \cdot 8 \mathrm{ft} = 320 \mathrm{ft}^3$$

Now we can find the weight of the water that the plate displaces. Since seawater weighs $64 \mathrm{lb} / \mathrm{ft}^3$, the weight of the displaced water is:

$$320 \mathrm{ft}^3 \cdot 64 \mathrm{lb} / \mathrm{ft}^3 = 20480 \mathrm{lb}$$

This is the upward force exerted on the plate by the water. By Newton's third law, the plate exerts an equal and opposite force on the water, which is also equal to the weight of the displaced water. Therefore, the force exerted on one face of the plate is also 20480 lb.

Rounding this value to one decimal place, we get:

$$20480 \mathrm{lb} \approx \boxed{1194.7 \mathrm{lb}}$$

Solve the Problem 10ft10 \mathrm { ft }10ft Long, and the Height of the Triangle Is

Solve the Problem $10 \mathrm { ft }$ Long, and the Height of the Triangle Is