```Activity 5.4 Calculating Properties of Solids
Introduction
Have you ever stopped to think why it is that you are able to float in water? The
reason has to do with the concept of buoyancy. The volume of water that your body
displaces has weight. The weight of the displaced water pushes upward on you,
while the weight of your body pushes down. If the weight of the displaced water
pushing upward is greater than your weight, then you will rise out of the water to a
point where equilibrium has been achieved. This means that the weight of your body
that is submerged in the water, let’s say from your shoulders down, is equal to the
weight of the volume of water that your body is displacing. Another way to explain
this is to look at density. If your average weight density, or the average amount of
weight per unit volume of your body, is less than the weight density of the water,
then you will float. The word average is used because skin, bone, hair, and muscle
tissue all have different weight densities. If you were to drop a coin in the water, you
would notice that it sinks. The weight density of metal is higher than the weight
density of water; therefore, the volume of water that is displaced by the coin cannot
weigh as much as the coin itself.
Engineers design vessels to travel on and under the surface of the water. Typically,
these vessels are comprised mostly of metal. How is it that these vessels do not sink
straight to the bottom of the sea floor? The answer again deals with weight, average
density, and volume. If you were to fill a glass with water and place it in a tub full of
water, the glass would sink. Like metal, glass has a weight density that is greater
than water. If you were to place the same glass in a tub full of water, but with no
liquid in the glass, it would float. In this case, the majority of the volume of the space
that is taken up by the glass is the air on the inside of the glass, which is significantly
less dense than water. Therefore, the average weight density of the glass is less
than the weight density of water. This is the reason why ships are able to float. The
majority of the volume of a ship is air. This example is only one engineering
application of physics and the mathematics associated with the properties of solids.
Equipment


Engineering notebook
Pencil
Procedure
In this activity you will learn how to hand calculate the volume, weight, and surface
area of common solids. You will then apply your knowledge by calculating these
properties for each of your puzzle piece solutions.
&copy; 2012 Project Lead The Way, Inc.
Introduction to Engineering Design Activity 5.4 Calculating Properties of Solids – Page 1
1. A cast iron cylinder serves as a counterbalance that is used in a window
manufacturer’s double-hung window design. The cylinder has a height of 5.25 inches
and a diameter of 1.75 inches. The weight density of cast iron is 0.259 lb/in3. Use this
information to answer the following questions.
a. What is the volume of the cylinder? Precision = 0.00
V = r2h
V = 3.14 x (.875 in)2 x 5.25 in
V = 3.14 x .766 in2 x 5.25 in
V = 12.63 in3
a. What is the surface area of the cylinder? Precision = 0.00
SA = (2r)h + 2(r2)
SA = (2 x 3.14 x 0.875 in) x 5.25 in + 2(3.14 x (0.875 in)2)
SA = 28.85 in2 + 4.82 in2
SA = 33.67 in2
b. What is the weight of the cylinder? Precision = 0.00
W = VDW
W = 12.63 in3 x 0.259 lb/in3
W = 3.27 lb
c. If one quart of cleaning solution will clean 7,200 square inches of surface,
how many quarts will be required to clean 1,500 cylinders? Round your
answer to the nearest quart needed to complete the cleaning.
total SA = # of cylinders x SA of a cylinder
total SA = 1,500 cylinders x 33.65 in2
total SA = 50,475 in2
# quarts of cleaning solution needed = total SA  SA cleaned per
quart
# quarts of cleaning solution needed = 50,475 in2  7200 in2 per quart
# quarts of cleaning solution needed = 7.01 quarts, therefore 8 quarts
are needed.
d. What will the total cost be to ship 200 of the cylinders if the shipping rate is
\$4.25 per pound?
&copy; 2012 Project Lead The Way, Inc.
Introduction to Engineering Design Activity 5.4 Calculating Properties of Solids – Page 2
total W of all cylinders = # of cylinders to be shipped x W of one
cylinder
total W of all cylinders = 200 x 3.27 lb
total W of all cylinders = 654 lb
total shipping cost = total W of all cylinders x cost per pound
total shipping cost = 654 lb x \$4.25 per pound
total shipping cost = \$2,779.50
e. If a cylinder for a larger window is cut from the same stock (a rod of the same
diameter) and must have a weight of 4.25 lb in order for the window to
operate properly, how long must the counterweight cylinder be?
W = VDw
V=
W 4.25 lb
=
= 16.41 in.3
Dw 0.259 lb3
.in.
V = πr2 h
V
16.41 in.3
h= 2 =
= 6.82 in.
πr
1.75 in. 2
π(
)
2
f. If the original window frame is redesigned such that the space provided for
the counterweight requires that the height of the counterweight be reduced to
4.5 in. with the original weight, what is the diameter of the new counterweight
design?
If the weight remains the same, the volume will remain the same
(assuming the same density). Therefore,
V = 12.63 in3 = r2h
V
r2 = πh
V
12.63 in.3
√
√
𝑟=
=
= 0.945 𝑖𝑛.
πh
(3.14)(4.5 𝑖𝑛. )
𝑑 = 2𝑟 = 2(0.945 𝑖𝑛. ) = 1.89 𝑖𝑛.
2. A wood board is one of a dozen different parts in a homemade robot kit. The width,
depth, and height dimensions of the board are 3.5 x 17 x 1.5 inches, respectively. The
board is made from southern yellow pine, which has an air dry weight density of .021
lb/in3.
&copy; 2012 Project Lead The Way, Inc.
Introduction to Engineering Design Activity 5.4 Calculating Properties of Solids – Page 3
a. What is the volume of the wood board? Precision = 0.00
V = wdh
V = 3.5 in x 17 in x 1.5 in
V = 89.25 in3
b. What is the surface area of the wood board? Precision = 0.00
SA = 2(wd + wh + dh)
SA = 2((3.5 in x 17 in) + (3.5 in x 1.5 in) + (17 in x 1.5 in))
SA = 2(59.5 in2 + 5.25 in2 + 25.5 in2)
SA = 180.5 in2
c. What is the weight of the wood board? Precision = 0.00
W = VDW
W = 89.25 in3 x .021 lb/in3
W = 1.87 lb
d. If one gallon of paint will cover 57,600 square inches, how many gallons
would be needed to give two coats of paint to 25,000 boards? Round your
total SA = # of boards x SA of a board
total SA = 25,000 boards x 180.5 in2
total SA = 4,512,500 in2
# gallons of paint needed for 2 coats = 2 x total SA  SA covered per
gallon
# gallons of paint needed for 2 coats = 2(4,512,500 in2  57,600 in2 per
gallon)
# gallons of paint needed for 2 coats = 157 gallons
e. What will the total cost be to ship the 25,000 boards to a facility for
assembling into the finish kit form if the shipping rate is \$4.25 per pound?
total W of all boards = # of boards to be shipped x W of one board
total W of all boards = 25,000 x 1.87 lb
total W of all boards = 46,750 lb
total shipping cost = total W of all boards x cost per pound
total shipping cost = 46,750 lb x \$4.25 per pound
total shipping cost = \$198,687.50
&copy; 2012 Project Lead The Way, Inc.
Introduction to Engineering Design Activity 5.4 Calculating Properties of Solids – Page 4
f. If a 2 x 6 (which means the actual dimensions of the board are 1.5 in. x 5.5
in.) milled from the same wood (and therefore has the same density) weighs
3.2 lb, how long is the board?
W = VDw
V=
W
3.2 lb
=
= 152.38 in.3
Dw 0.021. lb3
.in.
V = lwd
V
152.38 in.3
l=
=
= 18.5 in.
wd (1.5 in.)(5.5 in.)
3. Calculate the properties of your puzzle cube pieces.
a. Fill in the following table with the appropriate information. Be sure to use
SI units. You may need to re-measure your puzzle pieces using a metric
ruler. Answers will vary. The following is an example only.
Piece
identity
Number of Mass Surface Volume
cubes
(g)
area
(cm3)
(cm2)
Density
(g/cm3)
Red
4
18
65.0
27.4
0.66
Blue
5
23
80.3
34.3
0.67
Purple
6
26
87.6
41.2
0.63
Yellow
6
27
87.6
41.2
0.66
Green
6
27
94.9
41.2
0.66
b. Add Density as the header in the final column of the table. Be sure to
indicate appropriate units. Calculate the density for each piece and record
the value in the table.
c. Create a scatterplot in Excel displaying mass and volume as the
variables.

Which variable (mass or volume) should you use as the independent
variable (which will be plotted on the horizontal axis) in order to best
represent the density? Why?
The independent variable should be Volume. Since the slope
of the line is represented by rise/run and density =
mass/volume, the rise (vertical axis) should be mass and the
run (horizontal axis) should be volume.

Since a zero volume would result in zero mass, to what value should
you set the y-intercept?
Zero. That is, the point (0,0) = (volume, mass) is a point on the
line.
&copy; 2012 Project Lead The Way, Inc.
Introduction to Engineering Design Activity 5.4 Calculating Properties of Solids – Page 5

How can you use the trendline of the scatterplot to provide information
about the density of the wood? Explain.
If the data is plotted such that the independent variable is
volume and the dependent variable is mass, the slope of the
line (mass/volume) will represent density.
d. Find a trendline for your mass/volume data. Print a copy of the scatterplot
with the trendline and mathematical model displayed.

Rewrite the equation of the trendline in function notation such that M(v)
= mass and v = volume.
M(v) = 0.6522v

What is the slope of the trendline (include units)?
Answers will vary. For the example above, the slope = 0.6522
g/cm3.

How does this compare to your calculated value of density above?
Explain any difference.
Answers will vary. For the example above, the density as
indictated by the slope of the trendline is very similar to the
densities calculated using the volume and mass of the puzzle
pieces.
e. Convert the density indicated by the equation of your trendline to weight
density in pounds per cubic inch. Show your work including all conversion
factors.
&copy; 2012 Project Lead The Way, Inc.
Introduction to Engineering Design Activity 5.4 Calculating Properties of Solids – Page 6
Answers will vary. For the example,
0.6522
g
cm3
2.54 cm 3 1 kg
2.205 lb
lb
(
) (
)(
) = 0.0236 3
1 in.
1000 g
1 kg
in.
f. What is the mass of your complete puzzle (in grams)?
Answers will vary. For the example,
Total mass = 18 + 23 + 26 + 27 + 27 = 121 g
g. Convert the mass of your complete puzzle to weight in pounds. Show your
work.
1kg
2.205 lb
121 g (
)(
) = 0.266 lb
1000 g
1 kg
h. Convert the mass of your complete puzzle to weight in ounces. Note that
1 lb = 16 oz. Show your work.
1kg
2.205 lb 16 oz
121 g (
)(
)(
) = 4.27 oz
1000 g
1 kg
1 lb
i.
If WeShipIt offers a flat rate of \$7.95 to ship a parcel up to ten pounds (as
long as it fits within a 18 cm x 18 cm x 18 cm box), how many cubes can
be shipped in this box for \$7.95? Be sure to check both the size and the
weight constraints. Ignore the weight of the box.
Check weight: The maximum number of puzzles that will weigh less than
10 lb is (10 lb / 0.266 lb =) 37 full puzzles.
Check size: Each cube puzzle measures 5.7 cm x 5.7 cm x 5.7 cm.
Therefore, (18cm/5.7cm =) 3 puzzles will fit within the 18 cm side length
restriction. So, a total of (3 x 3 x 3 =) 27 puzzles will fit snuggly into the
box. Note that the total weight of 27 puzzles is (343 x 0.266 lb =) 7.2 lb.
Therefore, the size constraint will control the number of puzzles that can
be shipped in the box for \$7.95
j.
If the same wood is used to build a CD storage unit according to the
drawing below, what is the weight of the shelving unit? All dimensions in
the drawing are in inches.
Volume = 2(12 in.)(0.5 in.)(5 in.) + 3(23 in.)(0.5 in)(5 in.) = 232.5 in.3
lb
Weight = VDw = (232.5 in.2 ) (0.0236 3) = 5.49 lb
in.
&copy; 2012 Project Lead The Way, Inc.
Introduction to Engineering Design Activity 5.4 Calculating Properties of Solids – Page 7
Conclusion
1. What is the difference between area and volume?
Area is the measure of a two-dimensional surface. Volume is the measure
of a three-dimensional space.
2. What is weight density?
Weight density is how much an object or substance weighs per unit
volume.
3. What is a “physical” property? Give examples of physical properties.
A physical property is a property that can be observed or measured
without changing the identity of the matter. Examples of physical
properties include length, volume, surface area, mass, density, center of
gravity, centroid and principle axes.
&copy; 2012 Project Lead The Way, Inc.
Introduction to Engineering Design Activity 5.4 Calculating Properties of Solids – Page 8
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