Find the height of each right cylinder

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Homework Help Overview

The discussion revolves around determining the height of concrete right cylinders needed for hurricane protection, given their weight and radius. The problem involves understanding the relationship between weight, volume, and density in the context of concrete construction.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula for volume and how to incorporate the weight of concrete into their calculations. There is a focus on converting measurements from inches to feet and clarifying the role of density in determining volume.

Discussion Status

Some participants have provided guidance on necessary conversions and highlighted the importance of using the correct equations. There are multiple interpretations of how to approach the problem, particularly regarding the relationship between weight and volume.

Contextual Notes

Participants note the need to work in consistent units (feet instead of inches) and question the application of the density of concrete in the calculations. There is an emphasis on understanding how to derive volume from weight and density.

Meadow Delorto
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Homework Statement


I have to construct 8 concrete right cylinders for hurricane protection on windows, but I need to know the height to fit the given criteria. Each has a 4 inch radius and each weighs 1200 pounds. Concrete weighs 150 pounds per cubic foot.

V= volume of right cylinder
r=radius
h= height of right cylinder

Homework Equations


V=pi*r^2*h

The Attempt at a Solution


V=pi*r^2*h
1200 lbs= pi*16*h
h=23.8732 inches (1’11.87”).

I just don’t know where the weight of concrete, 150 pounds per cubic foot, comes into play. Thank you very much.
 
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:welcome:You need to work in feet, not inches, and the equation you are missing is ## W=\delta \, V ##, where ## \delta ## is the density in pounds per cubic foot. When you compute the volume, the radius ## r ## needs to be converted to "x" feet. The answer you get for height ## h ## will be in feet.
 
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In your statement, you set V=1200 pounds. That is not right; ##V=\frac{weight}{density}##. The density is what give is 150 pounds per cubic foot.
 
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Charles Link said:
:welcome:You need to work in feet, not inches, and the equation you are missing is ## W=\delta \, V ##, where ## \delta ## is the density in pounds per cubic foot. When you compute the volume, the radius ## r ## needs to be converted to "x" feet. The answer you get for height ## h ## will be in feet.
Would I be correct if I said it would have to be 22.96 feet?
 
Meadow Delorto said:

Homework Statement


I have to construct 8 concrete right cylinders for hurricane protection on windows, but I need to know the height to fit the given criteria. Each has a 4 inch radius and each weighs 1200 pounds. Concrete weighs 150 pounds per cubic foot.

V= volume of right cylinder
r=radius
h= height of right cylinder

Homework Equations


V=pi*r^2*h

The Attempt at a Solution


V=pi*r^2*h
1200 lbs= pi*16*h
h=23.8732 inches (1’11.87”).

I just don’t know where the weight of concrete, 150 pounds per cubic foot, comes into play. Thank you very much.

If the final object weighs 1200 lb and concrete weighs 150 lb/cu.ft., how many cubic feet of concrete does the object consist of? So, what is the object's volume, in cubic feet?
 
Last edited:
You have the volume formula, and you know how much concrete and you have the density. So?

Convert 1200 pounds into the quantity in cubic feet.
Now use your volume formula to calculate the unknown height.
 

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