The end plates (isosceles triangles) of the trough show below were designed to withstand a fluid force of 5800 lb. How many cubic feet can the tank hold without exceeding this limitation?
Assuming the density is 62.4 lb/ft^3, the maximum volume is ____ ft^3.
View the graph at:
postimg. org/image/4925g8vlp/
I apologize, but as an AI text-based bot, I cannot view or access external images or graphs. However, I can guide you through the steps to calculate the maximum volume of the tank given the provided information.
To calculate the maximum volume, we need to use the concept of fluid force and pressure on the end plates of the trough.
1. Start by calculating the pressure exerted on the end plate. Pressure is defined as force (F) divided by the area (A):
Pressure (P) = F / A
2. We know the force that the end plates can withstand, which is 5800 lb. However, we need to convert this force into units of pressure (lb/ft^2).
3. To determine the area of the end plate, we need to know the dimensions of the isosceles triangle, such as the base and height. Without the graph or specific dimensions, it is not possible to calculate the area accurately.
4. Once you have the area of the end plate, you can calculate the pressure by dividing the force by the area.
5. Now that you have the pressure exerted on the end plate, you can calculate the maximum volume of the tank. This requires knowing the density of the fluid, which is given as 62.4 lb/ft^3.
6. The maximum volume can be determined using the formula:
Volume (V) = Force (F) / (Pressure (P) * Density (D))
By following these steps and knowing the dimensions of the isosceles triangle that represents the end plate, you can calculate the maximum volume of the tank without exceeding the given limitation.
To calculate the maximum volume that the tank can hold without exceeding the limitation of 5800 lb fluid force, we need to use the density and the force to calculate the maximum weight of the fluid that the tank can hold. Then we can convert this weight to volume using the given density.
First, let's calculate the maximum weight of the fluid that the tank can hold:
Weight = Force
Weight = 5800 lb
Next, let's convert this weight to volume using the given density:
Weight = Volume * Density
5800 lb = Volume * 62.4 lb/ft^3
Now, we can solve for the volume:
Volume = 5800 lb / (62.4 lb/ft^3)
Volume ≈ 92.95 ft^3
So, the maximum volume that the tank can hold without exceeding the limitation is approximately 92.95 ft^3.
In my book I have diff answer. Can you double check your work?
Let the bottom of the trough be at (0,0).
The width of the trough at height y is thus
x = 9/14 y
Consider a bunch of horizontal cross-sections of the trough end. Each is of length x and depth 9-y.
The force on the plate when the water is h feet deep, is thus
F = w∫[0,h] (h-y)(9/14 y) dy
where w = 62.f lb/ft^3
= 62.4(3/28 h^3)
So, if we want F=5800, we just solve
62.4(3/28 h^3) = 5800
h = 9.54 ft
Since the tank is 32 ft long, that means it can hold
v = 1/2 (9.54)(9/14 * 9.54)(32) = 936 ft^3