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Calculating Weight Using Volume
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Skills Application: Calculating Weight Using Volume
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According to the Guinness World records, the heaviest object ever directly weighed was the Rotating Service Structure (RSS) of launch pad 39B at NASA's Kennedy Space Center. On January 23, 2004 the structure was hydraulically lifted up on 21 jacking points which measured the mass of the RSS as 2,423 tonnes (5,342,000 lbs).

In many instances we need to know the weight of large objects and hydraulic scales like this are not readily available. Thankfully we have math on our side; some of these heavy questions can be calculated with greater precision than a scale could ever provide.

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Calculating the weight of pre-fabricated items based on volume is an often used skill. Weighing items on a scale is plain easier but before you lift that load onto the scale, just how many people do you need for a safe lift? If slings are to be used, how strong? Which one? Determining pre-fabricating sizes to stay within load capacity of transport vehicles is also where these math skills are applied.

This weight calculation combines the volume of cylinder with the density of steel. Of course, other combinations are also possible such as:

• Weight of steel plate calculation uses the volume of a rectangular prism and the density of steel
• Weight of a pipe calculation: volume of a hollow cylinder and the density of PVC
• Weight of a scaffold frame: volume of all hollow cylinders and the density of aluminum
• Weight of a pre-cast manhole cover: volume of a cylinder and density of cast iron or concrete
• Weight of a footing calculation: volume of forms and density of concrete
• Weight of a roof rafter: volume of an oblique rectangular prism and the density of wood
• Weight of a bronze spacer: volume of a cone frustum and the density of bronze

The shapes and materials in this list are found everywhere in construction and manufacturing.

Get a \$20-or-so dual-display scientific calculator, even if you cant take it with you into the exam hall. A calculator is also a learning tool, Technology Use is one of the 9 Essential Skills you cant do without.

Check out the ADVANCED tabs. The exercises will further your understanding and help with building a bigger mental picture. It builds links between numbers and algebra. It is important to sketch the shapes, label the shapes with the given numbers, tracking changes to units of measure, visualizing to cross-link the concepts of volume, weight and density. We use the numbers from this rebar example. ### A Short Review

To calculate weight, the volume of an object and the density of its material needs to be multiplied together.

W = V · ρ

The symbol for Density is the Greek letter “rho”, it looks like this: ρ It is not a lowercase “p”.

Density is often measured in kg/m3 or lb/in3.

Density of materials needs to be found from tables, published by science communities. For example, the density of gold is 19.32 g/cm3, concrete is 150 lb/ft3, air is 1.29 kg/m3.

Temperature further affects these density figures.

Volume can be calculated if the object is a geometric solid, but complex shapes are often submersed in a tank of water or other non-corrosive liquid to determine their volume. ### Worksheet: Level 1

The operations used are clearly specified. Only one type of mathematical operation is used in a task.

### Worksheet: Level 1 Sample Questions

Example:

Density is a property of matter that describes how compacted matter is of a given size.

From the table below find the densities of:

1. Naval Brass (lb/ft³): _____

Further examples:

What is the density of steel? Silver? Crushed rock? Find metric and Imperial figures. ### Worksheet: Level 1 Answer Key

1. Naval Brass  =  546.25 lb/ft³ ### Worksheet: Level 2

Tasks involve one or two types of mathematical operation. Few steps of calculation are required.

### Worksheet: Level 2 Sample Questions

Example:

Calculate the weight of the following objects. Answer in decimals.

1. Calculate a steel cylinder‘s weight if its volume = 10 cm³, steel density is 7.85 g/cm³

Further example:

What is the mass of a steel exit staircase, V = 0.25 m³, density = 7850 kg/m³?

### Worksheet: Level 2 Sample Answer Key

Calculate the weight of the following objects. Answer in decimals.

1. Calculate a steel cylinder‘s weight if its volume = 10 cm³, steel density is 7.85 g/cm³

V × ? = m

volume x density = mass/weight

10 × 7.85 = 78.5 g ### Worksheet: Level 3

Tasks require a combination of operations. Several steps of calculation are required.

### Worksheet: Level 3 Sample Questions

Example:

1. Calculate a steel cylinder‘s weight if its radius is 4 in, height is 5 in, steel density is 490 lb/ft³.

Further example:

Calculate the mass of a granite cone, d = 5 cm, h = 8 cm, ρ = 2.7 g/cm³.

### Worksheet: Level 3 Sample Answer Key

r²? ⋅ h = V

radius² × ? × height  =  volume of cylinder

4² × ? × 5 = 251.327 in³  ÷ 1728 = 0.14544 ft3

volume x density = mass (weight)

0.14544 × 490 = 71.26 lb

Note: Mass and weight are different in physics, but not for doing applied math with them in trades. A pound is both a unit of force and unit of mass in non-metric calculations. In the Metric System Kg is a unit of mass, N is a unit of force. Tasks involve multiple steps of calculation. Advanced mathematical techniques may be required.

Another Way to Get the Answer

Calculate the weight of a steel reinforcing bar (long cylinder) if length is 50 feet, the cross-sectional area is 0.785 in2 (Bar diameter is 1 inch.) The density of mild steel is either 0.2836 lb/in3 (= 490 lb/ft3).

One way to get the answer, using the formulas V = A · L and Weight = density x V

Layout your calculations neatly, so you can review, track changes, correct or learn from them. One way of layout can look like this: Calculate the weight of a steel reinforcing bar (long cylinder) if length is 50 feet, the cross-sectional area is 0.785 in2 (Bar diameter is 1 inch.) The density of mild steel is either 0.2836 lb/in3 (= 490 lb/ft3). Another way to get the answer, using the formulas V = A · L and Weight = density x V

1. Identify that the weight of a steel cylinder is to be calculated.

2. Make a sketch of the cylinder: 3. Label the sides of the cylinder:
Use either full words or can abbreviate. The idea is to visualize the problem before calculation. For now, dont sketch or label the 1 inch diameter dimension mentioned in the brackets, its just there to help visualize the steel bar. The density amount is impossible to sketch, just leave it for now.

Look at the 2 formulas, there is one to calculate volume and one to calculate weight. The first formula is for volume. We need an answer for volume to be filled into the second formula. Volume of the cylinder must be calculated first:

4. Write the formula: A · L = V

5. Re-write the formula: change all of the letters into words, according to their meaning: Area of cross section x length = volume of cylinder

6. Look at the units of measure for area and length. Length is feet, and area is based on inches.
Recognize that all measurements have to be in the same unit. At this point, the area of 0.785 in2 will be converted to ft2.

7. Recognize that the conversion factor to convert from in2 to ft2 is 144. It comes from 12 inches times 12 inches making 1 square foot, which 12 x 12 equals 144.

8. Recognize that when converting from smaller unit to bigger unit, that is from in2 to ft2, the conversion factor will be a divisor because many of the smaller units (in2) are to be shared equally by groups of the bigger unit (ft2).

9. Set up the problem:

in2 ÷ conversion factor = ft2

10. Calculate: 0.785 ÷ 144 = 0.005451388

11. Take a pencil and write down “A = 0.00545” or “rebar cross-section = 0.00545”

12. Write ft2 after this number.

13. Decide that the converted cross sectional area of a single rebar is 0.00545 ft2
Now the volume can be calculated:

14. Re-write the formula: change all of the words for actual given numbers: Leave out units of measure words. (eg. feet, cm)

0.00545 x 50 = volume of cylinder

15. Calculate. Use mental math, paper & pencil, or electronic calculator: 0.00545 x 50 = 0.2725

16. Take a pencil and write down “V = 0.2725” or “rebar volume = 0.2725”

17. Look at the units of measure. ft2 was multiplied with ft. The unit for volume is ft3. Take a pencil and write ft3 after the 0.2725

18. Decide that the correct way to write the answer is 0.2725 ft3
Now the weight can be calculated:

19. Write the formula: V · density = Weight

20. Re-write the formula: change all of the letters into words, according to their meaning:

volume x density of steel = weight of steel bar

21. Re-write the formula: change all of the words for actual given numbers:

Leave out units of measure words. (eg. feet, cm):

490 x 0. 2725 = weight of steel bar

22. Calculate: 490 x 0.2725 = 133.525

23. Take a pencil and write down 133.5

24. Write lb after the number 133.5 Now youre done with the math.

25. The last step is to check your work: make sure everything was copied and written correctly then determine that the correct way to write the answer is 133.5 lb

We dont think Vaughn can lift that. In addition, long and thin steel bars bend a lot, so even if he lifts in the middle, the ends will still be on the ground, making walking impossible.

Layout your calculations neatly, so you can review, track changes, correct or learn from them. One way a layout can look like is this: This calculation is somewhat lengthy. Here is a way how you can make your very own formula for future reference to speed up calculations:

In the animated calculation, the sequence of calculations were:

We started with 50 This amount was:

1. multiplied by 12,
2. then the result was multiplied by 0.785,
3. then the result was multiplied by 0. 2836 and we got 133.6 lb

It is possible to enter everything in the calculator in one line (on a calculator with a multi-line display):

50 x 12 x 0.785 x 0.2836 = 133.5756 lb

Now, we can make a formula by replacing the numbers we can with words, according to their meaning:

length (feet) x 12 x area (in2) x density (lb/in3) = weight (lbs)

Now we need to replace the words with letter symbols:

L x 12 x A x D = weight

Multiplications are interchangeable. We can re-arrange the letters a bit to make it easier to memorize:

12 · L · A · D = weight

Enter the amounts in your calculator in the order shown in this formula. Do you still get 133.5756?

Meticulously compare this home-made formula with the next calculations formula. Note similarities and differences. Both formulas determine the weight of cylindrical steel objects in pounds. Why are they different?

Similarities:

• Results are the same. Thats reassuring.
• Both calculations use length, area, density to end up with weight.
• Both calculations take the same number of steps.

Differences:

• One calculation is based on in and has an in2 to ft2 conversion in it.
The other is based on feet and has a ft to in conversion in it.
• The calculations use 2 different figures for density. This is why the formulas are different.
• One has a division, the other is all multiplications.
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ADVANCED 2 (based on the second version)

This calculation is somewhat lengthy. Here is a way how you can make your very own formula for future reference to speed up calculations:

In the animated calculation, the sequence of calculations were: We started with 0.785 This amount was:

1. divided by 144,
2. then the result was multiplied by 50,
3. then the result was multiplied by 490
and we got 133.5 lb

It is possible to enter everything in the calculator in one line (on a calculator with a multi-line display):

0.785 ÷ 144 x 50 x 490 = 133.5590278

Now, we can make a formula by replacing the numbers we can with words, according to their meaning:

area (in2) ÷ 144 x length (feet) x density (lb/ft3) = weight (lbs)

Now we need to replace the words with letter symbols:

A ÷ 144 x L x D = weight

Multiplications are interchangeable. We can re-arrange the letters a bit to make it easier to memorize:

L · A · D ÷ 144 = weight

Enter the amounts in your calculator in the order shown in this formula. Do you still get 133.5590278?

Meticulously compare this home-made formula with the previous calculations formula. Both formulas determine the weight of cylindrical steel objects in pounds. Note similarities and differences. Why are they different?

Similarities:

• Results are the same. Thats reassuring.
• Both calculations use length, area, density to end up with weight.
• Both calculations take the same number of steps.

Differences:

• One calculation is based on in and has an in2 to ft2 conversion in it.
The other is based on feet and has a ft to in conversion in it.
• The calculations use 2 different figures for density. This is why the formulas are different.
• One has a division, the other is all multiplications.

Print out and compare the volume of cylinder calculations in episodes 27 & 29.

Volume of a cylinder was calculated by “V = A · h” in the animated version of Episode 27 and “V = A · L” in both versions of Episode 29. One includes “height” the other does not. One includes “length” the other does not. The formulas look different. But they are really not.

1. Why? Is there still a way that explains how these 2 different formulas are the same, consistent and valid?
Look for clues in episode 09 & 26 ADVANCED.

In Episode 27b the volume of the cylinder was calculated by the formula V = r²? · h, which looks totally different from either “V = A · h” or “V = A · L” Hmmm. Looks messed up. Whats going on?

The volume of cylinders can be reliably and correctly calculated either way.

2. How is the longer cylinder formula “V = r²? · h” the same as “V = A · h” ?

3. Based on clues from Episode 20, area of circles, come up with another formula for calculating the
volume of cylinders thats based on diameter.

4. What if your cylinder is oblique, leaning at a 45° angle to the left? (As with a solid bar, end cut on an angle.) How would you calculate its volume? Write a formula to do that.

1. Length and height are synonyms, interchangeable. Laid horizontally, the height of a cylinder becomes the length of the steel bar.
2. “r²?” is replaced by the letter A
3. 2 examples are here:
$\text{V = }\frac{d^2 \pi}{4}\cdot{h}$
or
$\text{V = }d^2\frac{\pi}{4}\cdot{h}$
…, but any of the 6 circle area formulas in Episode 20 ADVANCED can be either before or after the multiplication by “ h ”. So thats 12 possible volume of cylinder formulas.
4. Same formulas as above work. It doesn’t matter how much the cylinder leans. Height is a perpendicular measurement between the 2 parallel circle bases. Look at the final answers using different calculations:     133.525     133.5756     133.5590278

There are some slight differences in the decimal digits. This is due to rounding from step to step. This rounding introduces a level of uncertainty into the calculations, undermining the degree of accuracy of the final answer.

There are 3 ways to eliminate some of the rounded numbers. The decimals for area (0.785) and density (0.2836) are both already rounded amounts.

1. We can eliminate inaccuracy by using either whole numbers,
2. or unrounded & non-infinite decimals.
3. We can also enter more terms per calculation into the calculator without pressing the = button.

The rebar with the A = 0.785 in2 actually is a 1 inch rebar. Its diameter, 1 inch, is a nice whole number.

• Given the 1 inch diameter rebar size, we can calculate a more accurate cross sectional area:
Diameter = 1 inch, so radius = ½ inch
A = r2 ?, so 0.52? = 0.785398163 in2 This new amount is more accurate than the 0.785 in2

The 0.2836 figure for density has also been rounded. We can calculate more accurate density:

• Mild steel density is 490 lb/ft3, this is a whole number, the weight of mild steel for a cubic foot. Since there are 1728 in3 in a ft3, mild steels density is 490 ÷ 1728 = 0.283564814 lb/in3
The figure 0.283564814 lb/in3 is more accurate than 0.2836 lb/in3.

For the most accurate result:

• replace the 0.758 by entering 0.52? for area in the formula.
also replace 0.2836 by entering 490 ÷ 1728 for density in the formula. length (feet) x 12 x area (in2) x density (lb/in3) = weight (lbs)

Now enter in your calculator:  50  x 12 x  0.52?  x  490 ÷ 1728  = 133.6267709 lb

This level of accuracy is important if you have hundreds or thousands of bars to calculate with.

This is still a mathematical model of a rebar based on a smooth cylinder. Real rebars have ridges, which may add more weight to the calculated figure. Rebar can also be rusty or dirty, adding even more weight.

There is a solution to all this puzzle. Manufacturers actually specify and publish the real weight of manufactured materials per foot length, eg:

 Imperial Bar Size Nominal Diameter (in) "Soft" Metric Size Weight per unit length (lb/ft) Mass per unit length (kg/m) Nominal Diameter (mm) Nominal Area(in2) Nominal Area (mm2) #8 1.000 #25 2.67 3.982 25.4 0.79 509

http://www.harrissupplysolutions.com/8-rebar.html

The manufacturer`s numbers are the best choice to work with. Weight = 50 feet x 2.67 lb/ft = 133.5 lb

Note: "Nominal" means "in name only". Actual measurements vary slightly during shifts, due to environmental or minute physical differences in identical models of machines.