1. a large vertical rectangular plate of glass is to be inserted in the wall of an aquarium underwater so visitors can see into the tank fo fish. the glass is 10 feet high, 25 feet long and the top fo the glass is 3 feet below the top of the water. The pressure exerted by water at depth b feet below the surface is .433b pounds/square inch. Your task is to set up the integral to compute the total pressure of the water (in pounds)against the glass.
2. What fraction of the pressure if exerted against the bottom half of the glass?
Once again, whoever wrote this question does not seem to know the difference begtween force and pressure, and probably should not be teaching the subject.
I already answered part 1, in a separate post. For the second question, compare the average depth of the top half of the glass (3 + 2.5 = 5.5 feet) to the average depth of the bottom half (3 + 7.5 = 10.5 feet). The fraction of the total FORCE exterted on the bottom half is 10.5/(10.5 + 5.5) = 10.5/16 = 21/32.
To set up the integral to compute the total pressure of the water against the glass, we need to consider the pressure exerted by each small horizontal strip of water on the glass.
First, let's divide the height of the glass into small segments. Since the glass is 10 feet high and the top of the glass is 3 feet below the water surface, the total height of the water column against the glass is 7 feet.
Next, let's consider a small horizontal strip of water at a height y from the bottom of the glass. The width of this strip would be 25 feet.
The pressure exerted by this small strip of water is given by the formula P = 0.433y, based on the given information that the pressure exerted by water at a depth b feet is 0.433b pounds/square inch.
To calculate the total pressure on the glass, we need to integrate the pressure exerted by each small strip of water over the height of the glass.
The integral to compute the total pressure of the water against the glass is:
∫[from 0 to 7] 0.433y * (25) dy
Explanation of the integral:
- We integrate with respect to y since we are considering the height of each small strip of water.
- The limits of integration are from 0 to 7, representing the entire height of the glass.
- The integrand is 0.433y * (25), which is the pressure exerted by each small strip of water multiplied by its width.
Now, to address the second question:
To find the fraction of the pressure exerted against the bottom half of the glass, we need to consider the height of the glass and compare it to the height of the bottom half.
The bottom half of the glass is 5 feet high, given that the glass is 10 feet high in total.
To calculate the pressure exerted against the bottom half of the glass, we need to integrate the pressure exerted by each small strip of water that lies within the height of the bottom half.
The integral to compute the pressure exerted against the bottom half of the glass is:
∫[from 0 to 5] 0.433y * (25) dy
Explanation of the integral:
- We integrate with respect to y since we are considering the height of each small strip of water within the bottom half of the glass.
- The limits of integration are from 0 to 5, representing the height of the bottom half of the glass.
- The integrand is 0.433y * (25), which is the pressure exerted by each small strip of water within the bottom half multiplied by its width.
To find the fraction of the pressure exerted against the bottom half, we can divide the integral for the pressure exerted against the bottom half by the total pressure (which is the integral calculated in the first question). So the fraction would be:
∫[from 0 to 5] 0.433y * (25) dy / ∫[from 0 to 7] 0.433y * (25) dy