sorry i only have very limited schooling and need final answer as total wieght thanks so muchMarkFL said:First, we need to find the volume of the water. The volume of a cuboid is simply the product of its length, width, and height:
\(\displaystyle V=\left(180\text{ ft}\right)\left(40\text{ ft}\right)\left(2\text{ in}\cdot\frac{1\text{ ft}}{12\text{ in}}\right)=1200\text{ ft}^3\)
Now, weight density $\rho$ is defined as follows (where $w$ is weight):
\(\displaystyle \rho\equiv\frac{w}{V}\implies w=\rho V\)
For water, we have:
\(\displaystyle \rho\approx 62.3\frac{\text{lb.}}{\text{ft}^3}\)
Can you proceed?
xerxies said:sorry i only have very limited schooling and need final answer as total wieght thanks so much
The weight of water on a roof can be calculated by multiplying the area of the roof (180ft x 40ft = 7,200 ft2) by the weight of water per square foot, which is 5.2 pounds. This results in a total weight of 37,440 pounds or 18.72 tons.
No, the weight of water on a roof is not evenly distributed. The majority of the weight is concentrated in the center of the roof, with the weight gradually decreasing towards the edges.
Yes, the pitch of the roof can affect the weight of water. A steeper pitch will allow the water to run off more easily, reducing the weight on the roof. However, a flatter pitch may cause the water to accumulate and increase the weight on the roof.
The weight of water on a roof can put additional stress on the structure, especially in areas where the roof is weak or damaged. This can potentially lead to structural damage or collapse if the weight exceeds the load-bearing capacity of the roof.
To prevent damage from the weight of water on a roof, it is important to regularly inspect and maintain the roof to ensure it is in good condition. Additionally, it is important to have proper drainage systems in place to prevent water from accumulating on the roof. In areas with heavy rainfall, it may also be necessary to reinforce the roof to withstand the weight of water.