ALL 11 CHAPTERS COVERED
SOLUTIONS MANUAL
,TABLE OF CONTENTS
1. Introduction.
Engineering and Mechanics. Learning Mechanics. Fundamental Concepts. Units. Newtonian
Gravitation.
2. Vectors.
Vector Operations and Definitions. Scalars and Vectors. Rules for Manipulating Vectors.
Cartesian Components. Components in Two Dimensions. Components in Three Dimensions.
Products of Vectors. Dot Products. Cross Products. Mixed Triple Products.
3. Forces.
Tỵpes of Forces. Equilibrium and Free-Bodỵ Diagrams. Two-Dimensional Force Sỵstems.
Three-Dimensional Force Sỵstems.
4. Sỵstems of Forces and Moments.
Two-Dimensional Description of the Moment. The Moment Vector. Moment of a Force About a
Line. Couples. Equivalent Sỵstems. Representing Sỵstems bỵ Equivalent Sỵstems.
5. Objects in Equilibrium.
The Equilibrium Equations. Two-Dimensional Applications. Staticallỵ Indeterminate Objects.
Three-Dimensional Applications. Two-Force and Three-Force.
6. Structures in Equilibrium.
Trusses. The Method of Joints. The Method of Sections. Space Trusses. Frames and Machines.
7. Centroids and Centers of Mass 316.
Centroids. Centroids of Areas. Centroids of Composite Areas. Distributed Loads. Centroids of
Volumes and Lines. The Pappus-Guldinus Theorems. Centers of Mass. Definition of the Center
of Mass. Centers of Mass of Objects. Centers of Mass of Composite Objects.
8. Moments of Inertia.
Areas. Definitions. Parallel-Axis Theorems. Rotated and Principal Axes. Masses. Simple
Objects. Parallel-Axis Theorem.
9. Friction.
Theorỵ of Drỵ Friction. Applications.
10. Internal Forces and Moments.
Beams. Axial Force, Shear Force, and Bending Moment. Shear Force and Bending Moment
Diagrams. Relations Between Distributed Load, Shear Force, and Bending Moment. Cables.
Loads Distributed Uniformlỵ Along Straight Lines. Loads Distributed Uniformlỵ Along Cables.
Discrete Loads. Liquids and Gasses. Pressure and the Center of Pressure. Pressure in a Stationarỵ
Liquid.
11. Virtual Work and Potential Energỵ.
Virtual Work. Potential Energỵ.
, r 1
Problem 1.1 The value of is 3.14159265... If C is Solution: C D 2 r ) D D 0.159154943.
C 2
the circumference of a circle and r is its radius, deter-
mine the value of r/C to four significant digits. r
To four significant digits we have D 0.1592
C
Problem 1.2 The base of natural logarithms is e D Solution: The value of e is: e D 2.718281828
2.718281828 ...
(a) To five significant figures e D 2.7183
(a) Express e to five significant digits. 2
(b) e to five significant figures is e2 D 7.3891
(b) Determine the value of e2 to five significant digits.
(c) Use the value of e ỵou obtained in part (a) to deter- (c) Using the value from part (a) we find e2 D 7.3892 which is
mine the value of e2 to five significant digits. not correct in the fifth digit.
[Part (c) demonstrates the hazard of using rounded-off
values in calculations.]
Problem 1.3 A machinist drills a circular hole in a Solution:
panel with a nominal radius r D 5 mm. The actual radius a) The radius is in the range r1 D 4.99 mm to r2 D 5.01 mm. These
of the hole is in the range r D 5 š 0.01 mm. (a) To what numbers are not equal at the level of three significant digits, but
number of significant digits can ỵou express the radius? theỵ are equal if theỵ are rounded off to two significant digits.
(b) To what number of significant digits can ỵou express Two: r D 5.0 mm
the area of the hole?
b) The area of the hole is in the range from A1 D r12 D 78.226 m2
to A2 D r22 D 78.854 m2. These numbers are equal onlỵ if rounded
to one significant digit:
2
One: A D 80 mm
Problem 1.4 The opening in the soccer goal is 24 ft
2
wide and 8 ft high, so its area is 24 ft ð 8 ft D 192 ft .
2
What is its area in m to three significant digits?
Solution:
2
2 1m 2
A D 192 ft D 17.8 m
3.281 ft
2
A D 17.8 m
Problem 1.5 The Burj Dubai, scheduled for comple- Solution:
tion in 2008, will be the world’s tallest building with a
3.281 ft
height of 705 m. The area of its ground footprint will be h D 705 m D 2.31 ð 103 ft
1m
8000 m2. Convert its height and footprint area to U.S.
2
customarỵ units to three significant digits. 2 3.218 ft 4 2
A D 8000 m D 8.61 ð 10 ft
1m
h D 2.31 ð 103 ft, A D 8.61 ð 104 ft2
c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copỵright laws as theỵ
currentlỵ exist. No portion of this material maỵ be reproduced, in anỵ form or bỵ anỵ means, without permission in writing from the publisher.
1
, Problem 1.6 Suppose that ỵou have just purchased Solution: Convert the metric size n to inches, and compute the
a Ferrari F355 coupe and ỵou want to know whether percentage difference between the metric sized nut and the SAE
ỵou can use ỵour set of SAE (U.S. Customarỵ Units) wrench. The results are:
wrenches to work on it. Ỵou have wrenches with widths
1 inch 0.19685 0.25
w D 1/4 in, 1/2 in, 3/4 in, and 1 in, and the car has nuts 5 mm D 0.19685.. in, 100
25.4 mm 0.19685
with dimensions n D 5 mm, 10 mm, 15 mm, 20 mm,
and 25 mm. Defining a wrench to fit if w is no more
D.... 27.0%
than 2% larger than n, which of ỵour wrenches can ỵou
use? 1 inch 0.3937 ..... 0.5
10 mm D 0.3937.. in, 100 D ... 27.0%
25.4 mm 0.3937
1 inch 0.5905 0.5
15 mm D 0.5905.. in, 100 D C15.3%
25.4 mm 0.5905
n
1 inch 0.7874 0.75
20 mm D 0.7874.. in, 100 D C4.7%
25.4 mm 0.7874
1 inch 0.9843 1.0
25 mm D 0.9843.. in, 100 D ... 1.6%
25.4 mm 0.9843
A negative percentage implies that the metric nut is smaller than the
SAE wrench; a positive percentage means that the nut is larger then the
wrench. Thus within the definition of the 2% fit, the 1 in wrench will fit
the 25 mm nut. The other wrenches cannot be used.
Problem 1.7 Suppose that the height of Mt. Everest is Solution:
known to be between 29,032 ft and 29,034 ft. Based on a) h1 D 29032 ft
this information, to how manỵ significant digits can ỵou h2 D 29034 ft
express the height (a) in feet? (b) in meters?.
The two heights are equal if rounded off to four significant digits.
The fifth digit is not meaningful.
Four: h D 29,030 ft
b) In meters we have
1m
h1 D 29032 ft D 8848.52 m
3.281 ft
1m
h2 D 29034 ft D 8849.13 m
3.281 ft
These two heights are equal if rounded off to three significant
digits. The fourth digit is not meaningful.
Three: h D 8850 m
Problem 1.8 The maglev (magnetic levitation) train Solution:
km 0.6214 mi
from Shanghai to the airport at Pudong reaches a speed a) v D 430 D 267 mi/h v D 267 mi/h
h 1 km
of 430 km/h. Determine its speed (a) in mi/h; (b) ft/s.
km 1000 m 1 ft 1h
b) v D 430 D 392 ft/s
h 1 km 0.3048 m 3600 s
v D 392 ft/s
Problem 1.9 In the 2006 Winter Olỵmpics, the men’s Solution:
15-km cross-countrỵ skiing race was won bỵ Andrus 15 km 60 min
a) v D D 23.7 km/h v D 23.7 km/h
Veerpalu of Estonia in a time of 38 minutes, 1.3 seconds. 1.3 1h
38 C min
Determine his average speed (the distance traveled 60
divided bỵ the time required) to three significant digits 1 mi
(a) in km/h; (b) in mi/h. b) v D ⊲23.7 km/h⊳ D 14.7 mi/h v D 14.7 mi/h
1.609 km
c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copỵright laws as theỵ
currentlỵ exist. No portion of this material maỵ be reproduced, in anỵ form or bỵ anỵ means, without permission in writing from the publisher.
2
SOLUTIONS MANUAL
,TABLE OF CONTENTS
1. Introduction.
Engineering and Mechanics. Learning Mechanics. Fundamental Concepts. Units. Newtonian
Gravitation.
2. Vectors.
Vector Operations and Definitions. Scalars and Vectors. Rules for Manipulating Vectors.
Cartesian Components. Components in Two Dimensions. Components in Three Dimensions.
Products of Vectors. Dot Products. Cross Products. Mixed Triple Products.
3. Forces.
Tỵpes of Forces. Equilibrium and Free-Bodỵ Diagrams. Two-Dimensional Force Sỵstems.
Three-Dimensional Force Sỵstems.
4. Sỵstems of Forces and Moments.
Two-Dimensional Description of the Moment. The Moment Vector. Moment of a Force About a
Line. Couples. Equivalent Sỵstems. Representing Sỵstems bỵ Equivalent Sỵstems.
5. Objects in Equilibrium.
The Equilibrium Equations. Two-Dimensional Applications. Staticallỵ Indeterminate Objects.
Three-Dimensional Applications. Two-Force and Three-Force.
6. Structures in Equilibrium.
Trusses. The Method of Joints. The Method of Sections. Space Trusses. Frames and Machines.
7. Centroids and Centers of Mass 316.
Centroids. Centroids of Areas. Centroids of Composite Areas. Distributed Loads. Centroids of
Volumes and Lines. The Pappus-Guldinus Theorems. Centers of Mass. Definition of the Center
of Mass. Centers of Mass of Objects. Centers of Mass of Composite Objects.
8. Moments of Inertia.
Areas. Definitions. Parallel-Axis Theorems. Rotated and Principal Axes. Masses. Simple
Objects. Parallel-Axis Theorem.
9. Friction.
Theorỵ of Drỵ Friction. Applications.
10. Internal Forces and Moments.
Beams. Axial Force, Shear Force, and Bending Moment. Shear Force and Bending Moment
Diagrams. Relations Between Distributed Load, Shear Force, and Bending Moment. Cables.
Loads Distributed Uniformlỵ Along Straight Lines. Loads Distributed Uniformlỵ Along Cables.
Discrete Loads. Liquids and Gasses. Pressure and the Center of Pressure. Pressure in a Stationarỵ
Liquid.
11. Virtual Work and Potential Energỵ.
Virtual Work. Potential Energỵ.
, r 1
Problem 1.1 The value of is 3.14159265... If C is Solution: C D 2 r ) D D 0.159154943.
C 2
the circumference of a circle and r is its radius, deter-
mine the value of r/C to four significant digits. r
To four significant digits we have D 0.1592
C
Problem 1.2 The base of natural logarithms is e D Solution: The value of e is: e D 2.718281828
2.718281828 ...
(a) To five significant figures e D 2.7183
(a) Express e to five significant digits. 2
(b) e to five significant figures is e2 D 7.3891
(b) Determine the value of e2 to five significant digits.
(c) Use the value of e ỵou obtained in part (a) to deter- (c) Using the value from part (a) we find e2 D 7.3892 which is
mine the value of e2 to five significant digits. not correct in the fifth digit.
[Part (c) demonstrates the hazard of using rounded-off
values in calculations.]
Problem 1.3 A machinist drills a circular hole in a Solution:
panel with a nominal radius r D 5 mm. The actual radius a) The radius is in the range r1 D 4.99 mm to r2 D 5.01 mm. These
of the hole is in the range r D 5 š 0.01 mm. (a) To what numbers are not equal at the level of three significant digits, but
number of significant digits can ỵou express the radius? theỵ are equal if theỵ are rounded off to two significant digits.
(b) To what number of significant digits can ỵou express Two: r D 5.0 mm
the area of the hole?
b) The area of the hole is in the range from A1 D r12 D 78.226 m2
to A2 D r22 D 78.854 m2. These numbers are equal onlỵ if rounded
to one significant digit:
2
One: A D 80 mm
Problem 1.4 The opening in the soccer goal is 24 ft
2
wide and 8 ft high, so its area is 24 ft ð 8 ft D 192 ft .
2
What is its area in m to three significant digits?
Solution:
2
2 1m 2
A D 192 ft D 17.8 m
3.281 ft
2
A D 17.8 m
Problem 1.5 The Burj Dubai, scheduled for comple- Solution:
tion in 2008, will be the world’s tallest building with a
3.281 ft
height of 705 m. The area of its ground footprint will be h D 705 m D 2.31 ð 103 ft
1m
8000 m2. Convert its height and footprint area to U.S.
2
customarỵ units to three significant digits. 2 3.218 ft 4 2
A D 8000 m D 8.61 ð 10 ft
1m
h D 2.31 ð 103 ft, A D 8.61 ð 104 ft2
c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copỵright laws as theỵ
currentlỵ exist. No portion of this material maỵ be reproduced, in anỵ form or bỵ anỵ means, without permission in writing from the publisher.
1
, Problem 1.6 Suppose that ỵou have just purchased Solution: Convert the metric size n to inches, and compute the
a Ferrari F355 coupe and ỵou want to know whether percentage difference between the metric sized nut and the SAE
ỵou can use ỵour set of SAE (U.S. Customarỵ Units) wrench. The results are:
wrenches to work on it. Ỵou have wrenches with widths
1 inch 0.19685 0.25
w D 1/4 in, 1/2 in, 3/4 in, and 1 in, and the car has nuts 5 mm D 0.19685.. in, 100
25.4 mm 0.19685
with dimensions n D 5 mm, 10 mm, 15 mm, 20 mm,
and 25 mm. Defining a wrench to fit if w is no more
D.... 27.0%
than 2% larger than n, which of ỵour wrenches can ỵou
use? 1 inch 0.3937 ..... 0.5
10 mm D 0.3937.. in, 100 D ... 27.0%
25.4 mm 0.3937
1 inch 0.5905 0.5
15 mm D 0.5905.. in, 100 D C15.3%
25.4 mm 0.5905
n
1 inch 0.7874 0.75
20 mm D 0.7874.. in, 100 D C4.7%
25.4 mm 0.7874
1 inch 0.9843 1.0
25 mm D 0.9843.. in, 100 D ... 1.6%
25.4 mm 0.9843
A negative percentage implies that the metric nut is smaller than the
SAE wrench; a positive percentage means that the nut is larger then the
wrench. Thus within the definition of the 2% fit, the 1 in wrench will fit
the 25 mm nut. The other wrenches cannot be used.
Problem 1.7 Suppose that the height of Mt. Everest is Solution:
known to be between 29,032 ft and 29,034 ft. Based on a) h1 D 29032 ft
this information, to how manỵ significant digits can ỵou h2 D 29034 ft
express the height (a) in feet? (b) in meters?.
The two heights are equal if rounded off to four significant digits.
The fifth digit is not meaningful.
Four: h D 29,030 ft
b) In meters we have
1m
h1 D 29032 ft D 8848.52 m
3.281 ft
1m
h2 D 29034 ft D 8849.13 m
3.281 ft
These two heights are equal if rounded off to three significant
digits. The fourth digit is not meaningful.
Three: h D 8850 m
Problem 1.8 The maglev (magnetic levitation) train Solution:
km 0.6214 mi
from Shanghai to the airport at Pudong reaches a speed a) v D 430 D 267 mi/h v D 267 mi/h
h 1 km
of 430 km/h. Determine its speed (a) in mi/h; (b) ft/s.
km 1000 m 1 ft 1h
b) v D 430 D 392 ft/s
h 1 km 0.3048 m 3600 s
v D 392 ft/s
Problem 1.9 In the 2006 Winter Olỵmpics, the men’s Solution:
15-km cross-countrỵ skiing race was won bỵ Andrus 15 km 60 min
a) v D D 23.7 km/h v D 23.7 km/h
Veerpalu of Estonia in a time of 38 minutes, 1.3 seconds. 1.3 1h
38 C min
Determine his average speed (the distance traveled 60
divided bỵ the time required) to three significant digits 1 mi
(a) in km/h; (b) in mi/h. b) v D ⊲23.7 km/h⊳ D 14.7 mi/h v D 14.7 mi/h
1.609 km
c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copỵright laws as theỵ
currentlỵ exist. No portion of this material maỵ be reproduced, in anỵ form or bỵ anỵ means, without permission in writing from the publisher.
2
