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Chapter 1
Mechanics I: Principles of Statics
Principles of Statics is a subset of Mechanics that deals with bodies at rest despite being under the action of forces. The bodies are considered to be at equilibrium when all the forces sum to zero. To study Statics, we ﬁrst need to understand the nature of forces, quantify them, and then understand how forces can be applied to bodies. The relevant fundamental concepts and mathematical tools related to forces are discussed in Chapters 2–4. The main objective of studying Statics is to conduct correct force analysis for mechanical or integrated systems for design, performance improvement, or failure prevention. The standard method for force analysis, called freebody diagram analysis, is presented in Chapters 5 and 6. In Chapter 7, we demonstrate how proper force analysis can be conducted for various practical applications. However, often, a system might be too complicated to be analyzed by the free-body diagram analysis. Chapter 8 addresses this situation. In Chapter 9, we discuss the stability of equilibrium. In Chapter 10, we present a case study to demonstrate how to conduct proper force analysis to design a sensor for monitoring machining forces. Finally, in Chapter 11, we provide detailed solutions to 31 diﬃcult problems in Statics.
As stated, the main objective of studying Statics is to conduct correct force analysis for mechanical or integrated systems. To this end, we, as engineers, must approach the analysis quantitatively, with proper knowledge about precision and uncertainties. We will discuss key concepts and practices to ensure conﬁdence in the calculations.
1

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Figure 1.1: A Cartesian coordinate with metrics deﬁned along each axis.
Finally, in Chapter 1, we will suggest a general problem-solving protocol to promote a systematic and creative thought process. Again, the idea is that we can do a little thinking to solve new problems, not limited to the problems we have studied before.
1.1 Space and Time
Space and time (or spacetime) are fundamental metrics we need for solving problems. To quantify a space, typically, we will deﬁne a reference coordinate. Typically, in Mechanics, we deal with a 1D, 2D, or 3D space, deﬁned commonly by a suitable coordinate such as the O − x − y − z coordinate, shown in Fig. 1.1. Each axis of the O − x − y − z coordinate is typically ﬁxed in direction in Statics.
We need to deﬁne metrics along each axis. In other words, how do we deﬁne length, for example, the length of one meter, in the SI system. In the old days, there was a “National Prototype Metre Bar No. 27, made in 1889 by the international Bureau of Weights and Measures (BIPM) and it was given to the United States to serve as the standard for deﬁning all units of length in the US from 1893 to 1960.”1 “The bars were to be made of a special alloy, 90% platinum and 10% iridium, which is signiﬁcantly harder
1https://en.wikipedia.org/wiki/History of the metre#International prototype metre.

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Mechanics I: Principles of Statics

3

than pure platinum, and have a special X-shaped cross section (a ‘Tresca section’, named after French engineer Henri Tresca) to minimize the eﬀects of torsional strain during length comparisons.”2
Today, the standard of length is no longer deﬁned by a meter bar, but by time, based on Einstein’s theory of special relativity, which assumes that the speed of light in vacuum is constant. The oﬃcial deﬁnition is3
“The metre, symbol m, is the SI unit of length. It is deﬁned by taking the ﬁxed numerical value of the speed of light in vacuum c to be 299792458 when expressed in the unit m/s, where the second is deﬁned in terms of the caesium frequency ΔνCs.”
Therefore, length is based on time. The question now is how do we achieve precise time-keeping? According to NIST,4 “since 1967, the International System of Units (SI) has deﬁned a second as the period equal to 9,192,631,770 cycles of the radiation, which corresponds to the transition between two energy levels of the ground state of the Cesium-133 atom. This deﬁnition makes the cesium oscillator (sometimes referred to generically as an atomic clock) the primary standard for time and frequency measurements.” In other words, we rely on counting to establish time and thus the standard of length.

1.2 Concept of Precision
Being quantitative is one important requirement for engineers. We need to handle numbers properly.
In Section 1.1, we talked about using counting for time-keeping, precisely, at least down to 1/9, 192, 631, 770 seconds, or 0.10873 nanoseconds based on the deﬁnition of 1 second.
As we count, we could make mistakes or simply reach diﬀerent counts because the event has changed. In other words, there are errors and uncertainties. Errors are due to mistakes. The uncertainties are due to factors out of our control, no matter how carefully we try to avoid mistakes. The subject of precision deals with errors and uncertainties. There are
2Nelson, R.A. (1981). “Foundations of the international system of units (SI)”. The Physics Teacher. 19(9): 596–613. 3BIPM. (2019). SI Brochure, 9th edition, p. 131. 4https://www.nist.gov/pml/time-and-frequency-division/timekeeping-and-clocks-faqs.

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three main concepts in precision, which are accuracy, repeatability, and resolution.5
For example, if we use a ruler to measure the length of a rod, the accuracy is the deviation of the measurement from the true value of the length. The problem is that we usually do not know the true value. When we take measurement with a ruler, the ruler has its limit of resolution, which is, for example, 1 mm, for a simple oﬃce ruler. We can get a 0.01 mm resolution using a caliper.
With a simple ruler, we align one end of the rod to the zero mark of the ruler and check where the other end ends. If the other end lands between the marks of 11 and 12 mm, we can only guess that the true length is, judging where it lands between the marks, 11.4 mm, for example. In fact, we are not sure if it is 11.3, 11.4, or 11.5 mm, and diﬀerent people might guess it diﬀerently. As a result, we have an uncertainty, which is at least 0.2 mm, in this case. This uncertainty aﬀects how repeatable the measurement could be.
One practical way of achieving a higher precision is to make multiple measurements. For example, if we want to estimate the value of an antique piece, we can ask many experts to estimate its value. Because we cannot trust one expert more than others, the best practice is to calculate the average of all the estimates. The average, thus, is the best estimate to incorporate all the inputs from all the experts. To determine the accuracy, we should use the best estimate (the average) and calculate its deviation from the true value (if we know it). The variation among the estimates from diﬀerent experts represents repeatability. The resolution is related to the smallest money unit used in the estimates for this case.

1.2.1 Signiﬁcance of ﬁgures
When we make a measurement or present a number, we often do not explicitly state its precision in terms of accuracy, repeatability, and resolution. Instead, we use the concept of signiﬁcant ﬁgures to indicate the precision of a measurement. This is the subject of arithmetic precision. “A signiﬁcant ﬁgure is any one of the digits 1 to 9; zero is a signiﬁcant ﬁgure except when it is used to ﬁx the decimal point or to ﬁll the places of unknown or discarded digits.”6 “Thus in the number 0.000532, the

5Slocum, A. (1992). Precision Machine Design. Society of Manufacturing Engineers,
Dearborn, Michigan. ISBN-13: 978-0872634923. 6Doebelin, E.O. (1990). Measurement Systems Application and Design. McGraw-Hill
Publishing Company, New York, NY. ISBN 0-07-017338-9.

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Mechanics I: Principles of Statics

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signiﬁcant ﬁgures are 5, 3, and 2, while in the number 2,076, all the digits, including the zero, are signiﬁcant. For a number such as 2,300, the zeros may or may not be signiﬁcant. To convey which ﬁgures are signiﬁcant, we write this as 2.3 × 103 if two signiﬁcant ﬁgures are intended, 2.30 × 103 if three, 2.300 × 103 if four, and so forth (see footnote 6).”
We often have to compute by using many measurements with diﬀerent signiﬁcant ﬁgures. When we are doing hand calculation, it is preferred to round the number with more signiﬁcant ﬁgures to be the same as the one with less signiﬁcant ﬁgures. There are practical reasons for this rounding practice such as (1) the ﬁnal results cannot have a higher number of signiﬁcant ﬁgures and (2) handling a long string of digits can incur more mistakes. However, in today’s computing practice using computers, the second concern is no longer an issue. We should still round the number of the ﬁnal result to recognize its limited signiﬁcant ﬁgures and to enable easy reading.
A widely used rounding practice is as follows (see footnote 6):

“To round a number to n signiﬁcant ﬁgures, discard all digits to the right of the nth place. If the discarded number is less than onehalf a unit in the nth place, leave the nth digit unchanged. If the discarded number is greater than one-half a unit in the nth place, increase the nth digit by 1. If the discarded number is exactly onehalf a unit in the nth place, leave the nth digit unchanged if it is an even number and add 1 to it if it is odd.”

Therefore, if we are given a number, 12.35, we should know that the actual number could range from 12.346 to 12.355, or 12.35+−00..000054. The uncertainty range could be larger than 0.009 because most people may not follow the rounding rule consistently.
In some applications, we need to emphasize the number of decimal places (the number of signiﬁcant ﬁgures following the decimal point). For example, in your bank statement, the balance of your checking account is rounded to two places after the decimal point. This is of course for practical reasons because we do not have a coin smaller than one cent.
In general, when one is performing addition and/or subtraction, it is better to round the result to the same number of decimal places as the number with the lowest number of decimal places. When the computation involves multiplication and/or division, it is better to round the result to be the same as the one with the lowest signiﬁcant ﬁgures.

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Number of Occurrences

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Diﬃcult Engineering Concepts Better Explained

In this book, armed with modern computing power, all numbers of ﬁnal results will be rounded to two decimal places unless a higher precision is required. If a number is very small, such as 0.0017345, it would be rounded to 0.00173, to contain three signiﬁcant ﬁgures, unless a higher precision is required.

1.2.2 Statistical nature of uncertainty
In the above discussion, we speciﬁed the range of uncertainty of a measurement using the plus and minus expression. If a number is expressed as 12.35+−00..000054, we only specify that all the measurements would be between 12.346 and 12.355. To understand this uncertainty more, we should also deﬁne the probability distribution within the range of uncertainty. Essentially, there are no fundamental laws to help us decide how often a number would appear within the range of uncertainty.
To answer this question, a basic discussion on statistics is in order. If we measure the blood pressure of a person several times, it is highly unlikely we will get the same reading. Let us say that we take the measurement 100 times, consecutively (hypothetically, not advised to do so), and obtained the measurements of the systolic pressure as shown in Fig. 1.2 (this ﬁgure is denoted as a histogram).
The measurements are grouped into 26 bins and each bin covers 2 mmHg. From Fig. 1.2, we found that, for example, there are 15 readings between 106 and 108 mmHg, only 2 readings over 130 mmHg, 10 readings between 120 and 130 mmHg, and 5 readings below 90 mmHg. Therefore, is the blood pressure of the person normal? There are 83 readings within the
Histogram of Systolic Pressure Measurements 15
Fitted Normal Distribution Curve 10

5

0

70

80

90

100

110

120

130

140

Systolic Pressure, mmHg

Figure 1.2: Histogram of 100 measurements of systolic pressure.

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Mechanics I: Principles of Statics

7

normal range between 90 and 120 mmHg, but 12 times when the pressure is in the pre–high blood pressure range. What is causing all these variations? The actual blood pressure could ﬂuctuate, and the measurement device could have measurement errors. The question is still “Is it normal?” The same scenario can be related to the strength of a structure or the quality of a part after we obtain multiple measurements.
One common statistical practice is to assume that the probability distribution of the reading is a normal distribution. We can ﬁt a normal distribution curve to the histogram, as shown in Fig. 1.2. This assumption is typically valid due to the Central Limit Theorem, or we just assume that it is for convenience. There are, of course, many exceptions. Based on the ﬁtted normal distribution, we assume that if we repeat the measurement an almost inﬁnite number of times, the histogram will approach a normal distribution curve (not necessary the same as the ﬁtted one in Fig. 1.2). Based on the ﬁtted curve, we estimate the average of the reading, m, is 108.0 mmHg, and the standard deviation, σ, is 10.2 mmHg. From the normal distribution curve, we know that 68.27% of the readings will fall in the range of 108.0 ± 10.2 mmHg (97.8−118.2 mmHg), 95.45% in the range of 108.0 ± 2 × 10.2 mmHg (87.6−128.4 mmHg), and 99.73% in the range of 108.0 ± 3 × 10.2 mmHg (77.4−138.6 mmHg).
From these values, we are pretty conﬁdent that the person is not suﬀering from high blood pressure because there is only 0.14% chance for it to be so.
The average of 108.0 mmHg provides us with the best guess of the true blood pressure, while we are 99.73% conﬁdent that the true blood pressure is between 77.4 and 138.6 mmHg, which represents the repeatability or the range of the uncertainty. Unfortunately, we can never know what the true blood pressure is; we can only give our best guess and then decide based on how conﬁdent we are with our guess.
When we have additional information about a measurement, the implied uncertainty of using the signiﬁcant ﬁgures and number of decimal places is no longer adequate. We should represent the measurement in the format of plus and minus, such as 108.0 ± 30.6 mmHg. If not speciﬁcally stated, we will assume that this is deﬁned with 99.73% conﬁdence. In many cases, a 95.45% conﬁdence is suﬃcient; therefore, the measurement should be expressed as 108.0 ± 20.4 mmHg with 95.45% or simply 95% conﬁdence. The conﬁdence should be explicitly stated if it is not 99.73%.

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1.2.3 Computation with uncertainties
We will use an excellent example and discussion from Doeblin (1990)7 to demonstrate how to handle uncertainties in computation. Let a variable y = f (x1, x2, x3, . . . , xn), where xi’s denote the measured variables with the uncertainties Δx1, Δx2, Δx3, . . . , Δxn. What will be the uncertainty Δy?
Using the Taylor series, we expand function f as,

y = y0 + Δy = f (x1 + Δx1, x2 + Δx2, x3 + Δx3, . . . , xn + Δxn)

=

f

(x10,

x20 ,

x30, . . . ,

xn0)

+

Δx1

∂f ∂x1

+

Δx2

∂f ∂x2

+

Δx3

∂f ∂x3

+...

+

Δxn

∂f ∂xn

+ H.O.T.

(1.1)

where all the partial derivatives are to be evaluated at the known values of

xi0’s and the values of Δxi’s could be positive or negative. The variable y0 is the nominal value of f with respect to xi0’s. We can ignore the Higher Order Terms (H.O.T.) if the values of Δxi’s are small.
To determine Δy, we usually consider the worst-case and the best-case

scenarios. The real case most likely falls somewhere in between.

For the worst-case scenario, we deﬁne the range of Δxi as RΔxi;

therefore,

−

1 2

RΔxi

<

Δxi

<

1 2

RΔxi

and

the

range

of

Δy

becomes,

RΔy =

RΔx1

∂f ∂x1

+

RΔx2

∂f ∂x2

+

RΔx3

∂f ∂x3

+...+

RΔxn

∂f ∂xn

(1.2)

We should express y as

y

=

y0

±

1 RΔy 2

(1.3)

7Doebelin, E.O. (1990). Measurement Systems Application and Design. McGraw-Hill Publishing Company, New York, NY. pp. 58–67., ISBN 0-07-017338-9.

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Mechanics I: Principles of Statics

9

If we consider Δxi’s as random variables with normal distributions, then we can compute the standard deviation of Δy as

σΔy

=

s„ σΔx1

∂f ∂x1

«2

+

„ σΔx2

∂f ∂x2

«2

+

„ σΔx3

∂f «2 ∂x3

+

...

+

„ σΔxn

∂f «2 ∂xn

(1.4)

From Equation (1.4), if we present the uncertainty of y with 99.73% conﬁdence, we have

y = yo + Δy = yo ± 3σΔy

(1.5)

Equation (1.5) represents the best-case scenario. What if the uncertainties are not in normal distribution? In that case,
there is no easy analytical way to determine Δy. However, with today’s computing power, we can use the Monte Carlo simulation to determine the probability distribution of Δy and plot a histogram similar to that in Fig. 1.2.
Let us consider an example from Doeblin (1990) regarding the measurement obtained from a dynamometer. The output power of a dynamometer can be written as

P

=

2πRF

(

L 12

)

550 t

=

2π 550 × 12

RF L t

(1.6)

where P is the power in hp, R is the revolution of the shaft per second, F is the measured force in lbf at the end of the torque arm, L is the length of the torque arm in inches, and t is time in second, s.
For a speciﬁc run, if the data are

F = 10.12 ± 0.040 lbf R = 1202 ± 1.0 rev/s L = 15.63 ± 0.050 in t = 60.0 ± 0.50 s

(1.7) (1.8) (1.9) (1.10)

the uncertainties of these measurements are determined from the sensors’ calibration. Note that all the uncertainty terms are expressed with one extra decimal place than the nominal value.

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The nominal value of P is computed based on Equation (1.1). We then have

P0 =

2π 550 × 12

1202.0

×

10.12 60.0

×

15.63

=

3.01668

∼=

3.02

(1.11)

We round the above value to two decimal places. Now, we compute the partial derivatives in Equation (1.1) to three
signiﬁcant ﬁgures as

∂f ∂x1

|0

=

∂P ∂F

|0

=

2π 550 × 12

LR t

|0

=

2π 550 × 12

15.63 × 1202 60.0

=

0.298 hp/lbf

(1.12)

∂f ∂x2

|0

=

∂P ∂R

|0

=

2π 550 ×

12

FL t

|0

=

0.00251 hp/r

∂f ∂x3

|0

=

∂P ∂L

|0

=

2π 550 ×

12

FR t

|0

=

0.193 hp/in

∂f ∂x4

|0

=

∂P ∂t

|0

=

−2π 550 × 12

F

LR t2

|0

=

−0.0500 hp/lbf

(1.13) (1.14) (1.15)

For the worst-case scenario, the ranges of uncertainties are based on Equations (1.7–1.10)

RΔF = 0.080 lbf RΔR = 2.0 rev/s RΔL = 0.100 in RΔt = 1.00 s

(1.16) (1.17) (1.18) (1.19)

From Equation (1.2), we have

RΔy =

RΔF

∂ ∂

P F

+

RΔR

∂P ∂R

+

RΔL

∂P ∂L

+

RΔt

∂P ∂t

= 0.098 hp (1.20)

From Equation (1.3), we have

P

=

P0

±

1 RΔy 2

=

3.02 ± 0.049

(hp)

∼=

3.02 ±

0.05(hp)

(1.21)

For the best-case scenario, we ﬁrst assume that the expressions of Equations (1.7–1.10) are based on 99.73% conﬁdence level of normal

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Mechanics I: Principles of Statics

11

distributions. Therefore, we have

σΔF

=

1 (0.040)lbf 3

σΔR

=

1 (1.0)rev/s 3

σΔL

=

1 (0.050)in 3

σΔt

=

1 (0.50)s 3

From Equation (1.4), the standard deviation of ΔP is

(1.22) (1.23) (1.24) (1.25)

σΔP =

σΔF

∂P ∂F

2
+

σΔR

∂ ∂

P R

2
+

σΔL

∂P ∂L

2
+

σΔt

∂P ∂t

2

= 0.00977 hp

(1.26)

Finally,
P = P0 ± 3σΔP = 3.017 ± 3 (0.00977) ∼= 3.017 ± 0.029 (hp) (1.27)
In other words, we are conﬁdent (99.73% sure) that the actual value of P will lie between 2.988 and 3.046 hp. With a lower conﬁdence of 95.45%, the true value of P will lie between 2.997 and 3.036 hp. There is only a 4.28% chance that the true value lies between 3.036 and 3.046 hp or between 2.988 and 2.997 hp. We use three decimal places for the best-case scenario.

1.2.4 Precision consideration in diﬀerential quantities

As shown in Equation (1.1), the higher order terms can be neglected if the values of Δxi’s are small. This is the case when we discuss the diﬀerential terms. For example, the Taylor expansion of sinθ is

sinθ = θ − θ3 + θ5 − . . . 3! 5!
For a diﬀerential angle dθ, it becomes

(1.28)

sindθ = dθ − dθ3 + dθ5 − . . . ∼= dθ 3! 5!

(1.29)

when the H.O.T are neglected. What is the error if we neglect the high order terms? In this case, the error ratio, ε, deﬁned as the error to the true

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12 (a)

Diﬃcult Engineering Concepts Better Explained (b)

(c)

Figure 1.3: (a) Volume calculation of a straight cone; (b) An inﬁnitesimal conical layer; (c) An inﬁnitesimal cylindrical layer.

value, is

ε

=

|sindθ − dθ| sindθ

=

dθ3 3!

−

dθ5 5!

+

dθ7 7!

sindθ

−...

<

dθ3 3!
sindθ

∼=

dθ3 3!
dθ

dθ2 =
3!

(1.30)

which is inﬁnitesimal. Similarly, we found that cosdθ ∼= 1 and tandθ ∼= dθ. In Statics, there will be cases where integration will need to be done

over diﬀerent geometric shapes to determine the surface area, volume, and

moments of inertia. Often, we have to decide on the inclusion or exclusion of diﬀerential terms. We will present a few cases for illustration.
As shown in Fig. 1.3, we want to determine the volume of a straight cone with a height h and a base radius r. We can slice the cone into inﬁnitesimally thin disks and add up the volume of all the disks. The top surface of the thin disk in Fig. 1.3(b) has a radius x and the bottom surface has a radius x + dx. Let us say that we do not know how to determine the diﬀerential volume, dV , of the thin disk of (b), but we know it will be between those of two cylinders with a radius x and x + dx, respectively, as shown in (c). Therefore, we have

πx2dy < dV < π(x + dx)2dy

(1.31)

If we decide to set dV = πx2dy, the error ratio, ε, will be

ε

<

π(x

+

dx)2dy − πx2dy πx2dy

=

2dx x2

+

dx2 x2

(1.32)

which is still negligible.

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Mechanics I: Principles of Statics

13

Now, integrating dV over y, we have the well-known result,

V
dV =
0

h
πx2dy =
0

h 0

π

r2 h2

y2dy

=

π r2h 3

(1.33)

For the same straight cone, if we would like to determine the exterior cone surface, can we use the exterior surface of the thin cylindrical disk and carry out similar integration?
If we do so, then the exterior cone surface becomes

Sˆ =

Sˆ
dSˆ =
0

h
2πxdy =
0

h 0

2π

r h

y

dy

=

πrh

(1.34)

which is incorrect. So what went wrong? The actual exterior surface of the thin disk in Fig. 1.3(b) should be

dS = 2πxdl = 2πx dx2 + dy2 = 2π r h

r2

+ h2

h2

y

dy

(1.35)

The error ratio between dS and dSˆ is

ε

=

dS − dSˆ dS

=

√ r√2 + h2 − h r2 + h2

=

l

−h l

(1.36)

√ where l = r2 + h2, which is the length along the cone exterior surface.

As shown in Equation (1.36), the error ratio is ﬁnite, not negligible. As a

result, the error will accumulate, leading to mistakes. The correct answer is

S=

S
dS =
0

h
2πxdl =
0

h 0

2π

r h

r2

+ h2 h2

y

dy

=

πrl

(1.37)

Between Sˆ and S, we have the same error ratio as in Equation (1.36). A similar consideration should be applied when calculating the exterior surface of a paraboloid disk, as discussed in Problem 11.19 of Chapter 11.

1.3 Problem-Solving Protocol in Statics
To solve a problem in Statics, we should follow a set of systematic steps as follows:
(a) Problem statements; (b) Force analysis; (c) Governing equations; (d) Solving equations;

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Diﬃcult Engineering Concepts Better Explained

(e) Answer veriﬁcations; (f) Extensions.

In the problem statements, one should make proper sketches and deﬁne proper reference coordinates. List all the given conditions and the unknowns to be found.
For force analysis, we will present a systematic method in Chapters 5 and 6 so that the force analysis can be conducted correctly every time.
As regards governing equations, we typically construct force equilibrium equations based on the force analysis (Chapters 7 and 8). Additional governing equations can be found in Chapters 9 and 10.
As regards solving equations, after we have established relevant governing equations, we should always count the number of the unknowns associated with the governing equations. If the number of unknowns is greater than the number of the governing equations, we are not able to solve for all the unknowns. When this occurs, we should carry out the following steps to overcome this problem. First, we should check if there is an equation involving only one unknown. If so, this particular unknown can be solved. As a result, the number of unknowns is reduced by one, but the number of equations is also reduced by one. We should also check if there is a subset of the governing equations involving the same number of unknowns. If so, we can solve for those unknowns. Once we exhaust all the equations that can be solved, we still need to ﬁnd additional equations.
To ﬁnd additional equations, we should examine the unknowns and consider if they have speciﬁc relationships among them so that we can establish additional equations.
If we cannot ﬁnd additional equations, then we have to make assumptions to deﬁne the values of some unknowns. For example, in the case of a smooth surface the friction can be assumed to be zero.
Finally, as engineers, we should know that we can conduct measurements using sensors to determine the values of some unknowns. Once we reduce the number of unknowns to be the same as that of the governing equations, we can proceed to solve for all the unknowns.
In solving equations, we can use computing aids for help. Today, there are equation solvers useful for solving complicated equations, numerically or symbolically. We provide a few programming examples in this book.
After we have solved the equations and obtained the values of unknowns, we should conduct answer veriﬁcations. Do they look

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Mechanics I: Principles of Statics

15

reasonable? We could have made some mistakes along the way. Do they violate the assumptions? Do they violate basic laws?
Finally, we should think about the implication of the results for extensions.

1.4 Concluding Remarks
Practice makes perfect. It applies to both honing the problem-solving skill and deepening the understanding of the concept. In Chapter 11, we provided detailed solutions to 31 diﬃcult problems based on Appendix A of the classic Statics textbook by Meriam.8 These problems, along with the examples presented in Chapters 1–10 are useful to learn how to think properly to solve diﬃcult Statics problems through proper understanding of diﬃcult concepts in Statics.

8Meriam, J.L. (1975). Statics, 2nd Edition. John Wiley & Sons, Inc., Hoboken, NJ.

