Weight of Water on 180ft x 40ft Roof: Help Needed!

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Discussion Overview

The discussion revolves around calculating the weight of water on a roof measuring 180ft by 40ft with a depth of 2 inches. Participants explore the necessary calculations and concepts related to volume and weight density of water.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant, xerxies, requests assistance in calculating the weight of water on the specified roof area.
  • Another participant provides a formula for calculating the volume of water, stating that the volume is 1200 ft³ based on the dimensions given.
  • The same participant explains the relationship between weight density and weight, introducing the formula for weight as the product of volume and density.
  • There is a clarification that the density of water is approximately 62.3 lb/ft³.
  • A later reply emphasizes the importance of understanding the calculation process rather than just receiving the final answer, while also noting a distinction between mass and weight.

Areas of Agreement / Disagreement

Participants generally agree on the method for calculating the weight of water, but there is no consensus on providing a final numerical answer, as one participant expresses a desire for a direct answer while others encourage understanding the process.

Contextual Notes

There is a mention of a potential distinction between mass and weight, indicating that the figures provided relate to mass rather than weight, which may require further clarification.

xerxies
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hopeing some one can help me...180ft x 40 ft x 2 inches of water ...what is weight of water on roof thanks to all who ponder this as i am sadly in need of help...should have stayed in school lol thanks again people...xerxies
 
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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:

$$V=\left(180\text{ ft}\right)\left(40\text{ ft}\right)\left(2\text{ in}\cdot\frac{1\text{ ft}}{12\text{ in}}\right)=1200\text{ ft}^3$$

Now, weight density $\rho$ is defined as follows (where $w$ is weight):

$$\rho\equiv\frac{w}{V}\implies w=\rho V$$

For water, we have:

$$\rho\approx 62.3\frac{\text{lb.}}{\text{ft}^3}$$

Can you proceed?
 
MarkFL 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:

$$V=\left(180\text{ ft}\right)\left(40\text{ ft}\right)\left(2\text{ in}\cdot\frac{1\text{ ft}}{12\text{ in}}\right)=1200\text{ ft}^3$$

Now, weight density $\rho$ is defined as follows (where $w$ is weight):

$$\rho\equiv\frac{w}{V}\implies w=\rho V$$

For water, we have:

$$\rho\approx 62.3\frac{\text{lb.}}{\text{ft}^3}$$

Can you proceed?
sorry i only have very limited schooling and need final answer as total wieght thanks so much
 
xerxies said:
sorry i only have very limited schooling and need final answer as total wieght thanks so much

Hello xerxies, MHB is a board which aims to help you find the answer by yourself. We could give you the answer but it wouldn't help if this came up again in future.

Mark has given you the total volume as 1200 ft³, the density 62.3 lb/ft³ and the formula for weight weight = volume * density

Thus to find the weight multiply 1200 ft³ x 62.3 lb/ft³

edit: Technically your figures are for the mass of water and not its weight although the two are used interchangeably. If you need to find the force of the water then let us know
 
thanks all who helped
 

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