Converting Between the SI and USCS
3.5 D IMENSIONAL C ONSISTENCY
When you apply equations of mathematics, science, or engineering, the calculations must be dimensionally consistent, or they are wrong. Dimensional consistency means that the units associated with the numerical values on each side of an equality sign match. Likewise, if two terms are combined in an equation by summation, or if they are subtracted from one another, the two quantities must have the same dimensions. This principle is a straightforward means to double-check your algebraic and numerical work.
In paper-and-pencil calculations, keep the units adjacent to each numerical quantity in an equation so that they can be combined or canceled at each step in the solution. You can manipulate dimensions just as you would any other algebraic quantity. By using the principle of dimensional consistency, you can double- check your calculation and develop greater confi dence in its accuracy. Of course, the result could be incorrect for a reason other than dimensions. Nevertheless, performing a double-check on the units in an equation is always a good idea.
The principle of dimensional consistency can be particularly useful when you perform calculations involving mass and force in the USCS. The defi nitions of the slug and pound-mass in terms of different reference accelerations is often a point of confusion when mass quantities are converted between the USCS and SI. In those cases, the principle of dimensional consistency can be applied to confi rm that the units in the calculation are correct. Dimensional consistency can be illustrated by as simple a calculation as fi nding the weights of two objects, the fi rst having a mass of 1 slug and the second having a mass of 1 lbm. In the fi rst case, the weight of a 1-slug object is
w (1 slug)
(
32.174 ft __s2
)
32.174 slug _______
fts2 32.174 lb
In the fi nal step of this calculation, we used the defi nition of the slug from Table 3.5. This object, having a mass of one slug, weighs 32.174 lb. On the
99
other hand, for the object having a mass of 1 lbm, direct substitution in the equation w 5 mg would give the dimensions of lbm · ft/s2, which is neither the same as a pound nor a conventional unit for force in the USCS. For the calculation to be dimensionally consistent, an intermediate step is necessary to convert m to the units of slug using Equation (3.5):
m 5 (1 lbm)
(
3.1081 3 1022 slugs_____
lbm
)
3.1081 3 1022 slugs
In the second case, the weight of the 1-lbm object becomes w (3.1081 3 1022 slugs)
(
32.174 ft __s2
)
1 slug ? ft
_______
s2 1 lb
The principle of dimensional consistency will help you to make the proper choice of mass units in the USCS. Generally speaking, the slug is the preferred unit for calculations involving Newton’s second law ( f 5 ma), kinetic energy ( 1_ 2 mv2), momentum (mv), gravitational potential energy (mgh), and other mechanical quantities. The process of verifying the dimensional consistency of an equation by keeping track of units is illustrated in the following examples.
The KC-10 Extender tanker aircraft of the United States Air Force is used to refuel other planes in fl ight. The Extender can carry 365,000 lb of jet fuel, which can be transferred to another aircraft through a boom that temporarily connects the two planes. (a) Express the mass of the fuel in the units of slugs and lbm. (b) Express the mass and weight of the fuel in the SI.
Approach
We will calculate the mass m in terms of the fuel’s weight w and the gravita- tional acceleration g 5 32.2 ft/s2. With w expressed in pounds and g having the units of ft/s2, the expression w 5 mg is dimensionally consistent when mass has the units of slugs. We will then convert from slugs to lbm using the conversion factor 1 slug 5 32.174 lbm from Equation (3.5). In part (b), we convert the fuel’s mass in the USCS to the SI using the conversion factor 1 slug 5 14.59 kg from Table 3.6.
Solution
(a) We fi rst determine the mass of the fuel in the units of slugs:
m 5 3.65 ____________ 3 105 lb
32.2 ft/s2 ←
3
m 5 w __ g4
5 1.134 3 104 lb _____ ? s2 ft 5 1.134 3 104 slugs Example 3.4 Aerial Refueling
3.5 Dimensional Consistency
In the last step, we used the defi nition 1 slug 5 1 (lb · s2)/ft. Since an object that weighs 1 lb has a mass of 1 lbm, the mass of the fuel can alternatively be expressed as 365,000 lbm.
(b) We convert the mass quantity 1.134 3 104 slugs into the units of kg in the SI:
m 5 (1.134 3 104 slugs)
(
14.59 ____ slugkg)
5 1.655 3 105 (slugs)
(
____ slugkg)
5 1.655 3 105 kg
Since the numerical value for mass has a large power-of-ten exponent, an SI prefi x from Table 3.3 should be applied. We fi rst write m 5 165.5 3 103 kg so that the exponent is a multiple of 3. Since the “kilo” prefi x already implies a factor of 103 g, m 5 165.5 3 106 g or 165.5 Mg, where
“M” denotes the prefi x “mega.” The fuel’s weight in the SI is w 5 (1.655 3 105 kg)
(
9.81 m __ s2)
← 3 w 5 mg 45 1.62 3 106 kg ? m
______
s2
5 1.62 3 106 N
Because this quantity also has a large power-of-ten exponent, we use the SI prefi x “M” to condense the factor of 1 million. The fuel weighs 1.62 MN.
Discussion
To double-check the calculation of weight in the SI, we note that we can convert the fuel’s 365,000 lb weight directly to the dimensions of newtons.
By using the conversion factor 1 lb 5 4.448 N from Table 3.6, w becomes w 5 (3.65 3 105 lb)
(
4.448 N __ lb)
5 1.62 3 106 (lb)
(
N __ lb)
5 1.62 3 106 N
or 1.62 MN, confi rming our previous answer.
m 5 1.134 3 104 slugs m 5 365,000 lbm m 5 165.5 Mg w 5 1.62 MN Example 3.4 continued
101
The International Space Station has hundreds of shields made of aluminum and bulletproof composite materials that are intended to offer protection against impact with debris present in low Earth orbit (Figure 3.5). With suf- fi cient advance warning, the station’s orbit can even be adjusted slightly to avoid close approaches of larger objects. Over 13,000 pieces of debris have been identifi ed by the United States Space Command, including paint chips, spent booster casings, and even an astronaut’s glove. (a) Calculate the kinetic energy Uk= 1_ 2 mv2 of an m = 1 g particle of debris traveling at v = 8 km/s, which is a typical velocity in low Earth orbit. (b) How fast would a 0.31-lb baseball have to be thrown to have the same kinetic energy?
Approach
We fi rst convert the debris particle’s mass and velocity to the dimensionally con- sistent units of kg and m/s, respectively, using the defi nition of the “kilo” prefi x (Table 3.3). The conventional SI unit for energy in Table 3.2 is the joule, defi ned as 1 N ? m. In part (b), we will convert the kinetic energy to the USCS using the factor 1 J = 0.7376 ft ? lb from Table 3.6. Since the baseball’s weight is specifi ed in the problem statement, we will make an intermediate calculation for its mass.
Solution
(a) With m 5 0.001 kg and v 5 8000 m/s, the kinetic energy of the debris particle is
Uk 1 __
2 (0.001 kg)
(
8000 m __ s)
2 ←3
Uk 1 __ 2 mv24
32,000 (kg)
(
m___ s22)
32,000
(
kg · m______ s2)
(m)Example 3.5 Orbital Debris Collision
3.5 Dimensional Consistency
Figure 3.5
The International Space Station.
Courtesy of NASA.
32,000 N ? m 32,000 J
Applying an SI prefi x to suppress the trailing zeroes, the kinetic energy of the particle is 32 kJ.
(b) Expressed in the USCS, the particle’s kinetic energy is Uk (32,000 J)
(
0.7376 ft _____ ? lbJ
)
23,603 ( J)
(
ft · lb _____ J)
23,603 ft · lb
For a dimensionally consistent calculation of kinetic energy in the USCS, we will determine the baseball’s mass in the units of slugs:
m ________0.31 lb
32.2 ft/s2 ←
3
m w __ g4
9.627 103 lb _____ ? s2 ft 9.627 103 slugs
since 1 slug 5 1 (lb ? s2)/ft in Equation (3.2). To have the same kinetic energy as the debris particle, the baseball must be thrown with velocity
v
冑
________________ 2(23,603 ft ? lb)9.627 103 slugs ←
3
v冑
____ 2Um k4
2214
冑
ft _____ slug? lb2214
冑
ft ? (slug ? ft/s2)_____________
slug 2214 ft __ s
Discussion
Even though orbiting dust and debris particles may be small in size, they can convey large amounts of kinetic energy because their velocities are so great.
The equivalent speed of a baseball is some 1500 mph, or about fi fteen times the speed of a major league fastball pitch.
Uk 32 kJ v 2214 ft __ s Example 3.5 continued
103
This example illustrates the full problem-solving process from Sec- tion 3.2, which incorporates the principles of dimensional analysis from Sections 3.3–3.5.
A drill press holds sharpened bits in a rotating chuck and is used to bore holes in a workpiece. The steel drill bit has a diameter d 5 6 mm and length L 5 65 mm. The bit is accidentally bent as the workpiece shifts during a drill- ing operation, and it is subjected to the side force of F 5 50 N. As derived in mechanical engineering courses on stress analysis, the sideways defl ection of the tip is calculated by using the equation
Dx 5 64 FL______ 3
3 Ed4 where the terms have the following units:
Δx (length) the defl ection of the tip
F (force) the magnitude of the force applied at the tip L (length) the drill bit’s length
E (force/length2) a property of the drill bit’s material, called the elastic modulus
d (length) the drill bit’s diameter
By using the numerical value E 5 200 3 109 Pa for steel, calculate the amount Δx that the tip defl ects. (See Figure 3.6.)
Approach
We are tasked to solve for the defl ection at the tip of the steel drill bit given the applied force. We fi rst make a number of assumptions about the system:
Example 3.6 Bending of a Drill Bit
3.5 Dimensional Consistency
Figure 3.6
65 mm
6 mm
50 N F
x L
d
• The curved fl utes on the bit are small and can be neglected in the analysis
• The force is perpendicular to the primary bending axis of the bit
• The channels that spiral along the bit have minimal impact on the bending and can be ignored
We will fi rst combine the units of each quantity in the given equation according to the rules of algebra and verify that the units appearing on each side of the equation are identical. Then we will insert the known quantities, including the bit length, diameter, elastic modulus, and applied force, to solve for the defl ection.
Solution
The quantity 64/3 is a dimensionless scalar, and it therefore has no units to infl uence dimensional consistency. The units of each quantity in the given equation are cancelled:
(length) 5 (force)(length)3
_____________________
(force/(length)2)(length)4 5 (length)
The equation is indeed dimensionally consistent. The tip moves sideways by the amount
Dx 5 64(50 N)(0.065 m)3
___________________________
3(200 3 109 Pa)(6 3 103 m)4 ←
3
Dx 5 64FL______ 3 Ed344
Next, we combine the numerical values and dimensions Dx 5 3.6 3 104 N · m_______ 3
Pa · m4
and then expand the derived unit pascal according to its defi nition in Table 3.2:
Dx 5 3.6 3 104 N __________ ? m3 (N/m2)(m4)
Finally, we cancel units in the numerator and denominator to obtain Δx 5 3.6 × 104 m
Discussion
First, we evaluate the order of magnitude of the solution. For a steel bit of this length, a large defl ection is not expected. Therefore the solution’s order of magnitude is reasonable. Second, we revisit our assumptions to make sure they are reasonable. While the curved fl utes and channels on the bit may slightly infl uence the bending mechanics, we must assume that their impact Example 3.6 continued
105
is negligible for this application. Also, the force may not remain perfectly perpendicular to the bit, but it is reasonable to assume this at the moment of defl ection. Third, we draw conclusions from the solution and explain its physical meaning. Drill bits undergo many forces during operation, and it makes sense that most bits would be made of steel to minimize defl ection.
Because the numerical value has a large negative exponent, we convert it to standard form by using the SI prefi x “milli” to represent a factor of 10−3. The tip moves by the amount Δx 5 0.36 mm, just over one-third of a millimeter.
Δx 5 0.36 mm Example 3.6 continued
This example also illustrates the full problem-solving process from Section 3.2, which incorporates the principles of dimensional analysis from Sections 3.3–3.5.
A person with a mass of 70 kg is standing on a scale in an elevator that reads 140 lb at a given instant. Determine which way the elevator is mov- ing and whether it is accelerating. From Newton’s second law, if a body is accelerating, then the sum of the forces equals the body’s mass m times its acceleration a through the equation
兺F 5 ma
If the sum of the forces in any direction is equal to zero, then the body is not accelerating in that direction. (See Figure 3.7.)
Example 3.7 Elevator acceleration
3.5 Dimensional Consistency
Figure 3.7
Fweight Fnormal Fweight Fnormal 70 kg
Approach
We are tasked to determine which direction the elevator is moving and whether the elevator is accelerating. We fi rst make a number of assumptions about the system:
• The person and elevator are moving in tandem; so we only have to analyze the forces on the person
• The only movement is in the vertical or y-direction
• Our analysis is done on Earth, and therefore gravity is 9.81 m/s2 or 32.2 ft/s2
• The scale does not move relative to the elevator fl oor or the person We will fi rst convert the mass of the person in kilograms to the equiva- lent weight in newtons. Then, we will convert their weight in newtons to pounds. We will compare the person’s weight to the reading on the scale to determine which direction the elevator is moving. Then we will use the difference in weight to determine the acceleration of the elevator. If there is no difference in weight, then we know that the elevator is not accelerating.
Solution
The weight W of the person can be found as follows:
W 5 (mass)(gravity) 5 (70 kg)(9.81 m/s2) 5 687 N
Then, the weight can be converted into pounds using the factor from Table 3.6 as follows,
W 5 687 N
(
0.22481 lb __N
)
5 154 N ? lb __
N 5 154 lb
This is the downward force of the person, represented by Fweight in Figure 3.7.
The scale reading represents the upward force exerted on the person by the scale, or the normal force, Fnormal. Since the weight of the person is larger than the scale reading, the elevator is accelerating downward which decreases the scale reading. This is illustrated in the equation
兺F 5 Fnormal 2 Fweight 5 140 lb 2 154 lb 5 214 lb
Finally, we solve for the acceleration, noting that the mass of the person must be converted to slugs.
a 5 ____ 兺F
m 5
(
________ 154 lbm214 lb)
(
32.2 lbm ________1 slug
)
(
slug ? ft __s2
_______
lb
)
5 22.9 ft __s2 Example 3.7 continued
107
Discussion
First, we evaluate the order of magnitude of the solution. The acceleration is not large, which is not expected since the scale reading is not signifi cantly dif- ferent from the person’s weight. Second, we revisit our assumptions to make sure they are reasonable. All the assumptions are very logical. The person, scale, and elevator may undergo some motion relative to one other in reality, but the impact on the analysis would be minimal. Third, we draw conclusions from the solution and explain its physical meaning. The acceleration is nega- tive, indicating a downward acceleration, which aligns with the scale reading.
When an elevator begins to accelerate downward, the passengers temporarily feel lighter. Their mass does not change since gravity did not change. However, their perceived weight has changed, which is what the scale measures.
Note that the same analysis can be done using the SI. First, we convert the scale reading to newtons
W 5 (140 lb)
(
4.45 N _______ 1 lb)
5 623 lb ? N __
lb 5 623 N
Since this is less than the person’s actual weight of 687 N, we conclude that the elevator is accelerating downward. Solving for acceleration gives
a 5 ____ 兺F
m 5
(
623 N ______________ 2 687 N70 kg
)
5 2_______ 64 N
70 kg 5 20.91
(
kg ? m ______ kg__ s2)
5 20.91 m __s2
Using the conversion 1 ft 5 0.3048 m from Table 3.6, this solution can be converted to 2.9 ft/s2, which matches our previous analysis.
a 5 22.9 ft __
s2 Example 3.7 continued