Principles
of
Engineering
Mechanics
Second Edition
H.
R.
Harrison
BS~,
PhD, MRAeS
Formerly, Department
of
Mechanical Engineering
and Aeronautics,
The City University, London
T.
Nettleton
MSc, MlMechE
Department
of
Mechanical Engineering and Aeronautics,
The City University, London
Edward
Arnold
A
member
of
the Hodder Headline
Group
LONDON
MELBOURNE

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for
any errors
or
omissions that may be made.
Typeset in 10/11 Times by Wearset, Boldon, Tyne and Wear.
Printed and bound in Great Britain for Edward Arnold, a division

21
Introduction. Newton’s laws
of
motion. Units.
Types
of
force. Gravitation. Frames
of
reference.
Systems of particles. Centre of mass. Free-body
diagrams. Simple harmonic motion. Impulse and
momentum. Work and kinetic energy. Power.
Discussion examples. Problems.
4
Force systems and equilibrium,
37
Addition
of
forces. Moment of a force. Vector
product
of
two vectors. Moments
of
components
of
a force. Couple. Distributed forces. Equivalent
force system in three dimensions. Equilibrium.
Co-planar force system. Equilibrium in three
dimensions. Triple scalar product. Internal
forces. Fluid statics. Buoyancy. Stability

particles. Kinetic energy
of
a rigid body. Potential
energy. Non-conservative systems. The general
energy principle. Summary
of
the energy method.
The power equation. Virtual work. D’Alembert’s
principle. Discussion examples. Problems.
8
Momentum and impulse,
11
1
Linear momentum. Moment of momentum.
Conservation of momentum. Impact of rigid
bodies. Deflection of fluid streams. The rocket in
free space. Illustrative example. Equations
of
motion for a fixed region
of
space. Discussion
examples. Problems.
9
Vibration,
126
Section
A.
One-degree-of-freedom
systems
Introduction. Free vibration of undamped sys-

vi
Contents
11
Dynamics
of
a body in three-dimensional
Introduction. Finite rotation. Angular velocity.
Differentiation
of
a vector when expressed in
terms
of
a moving set
of
axes. Dynamics
of
a
particle in three-dimensional motion. Motion
relative to translating axes. Motion relative to
rotating axes. Kinematics
of
mechanisms. Kine-
tics of a rigid body. Moment
of
force and rate of
change
of
moment
of
momentum. Rotation about

li’s equation.
Section
B.
Two-
and three-dimensional continua
Introduction. Poisson’s ratio. Pure shear. Plane
strain. Plane stress. Rotation
of
reference axes.
Principal strain. Principal stress. The elastic
constants. Strain energy.
Section
C.
Applications to
bars
and beams
Introduction. Compound column. Torsion
of
circular cross-section shafts. Shear force and
bending moment in beams. Stress and strain
distribution within the beam. Deflection
of
beams. Area moment method. Discussion exam-
ples. Problems.
Appendices
1
Vector algebra, 247
2 Units, 249
3
Approximate integration, 251

3
Engineering
Mechanics syllabuses
of
degree courses in
engineering. The emphasis of the book is on the
principles
of
mechanics and examples are drawn
from a wide range
of
engineering applications.
The order
of
presentation has been chosen to
correspond with that which we have found to be
the most easily assimilated by students. Thus,
although in some cases we proceed from the
general to the particular, the gentler approach is
adopted in discussing first two-dimensional and
then three-dimensional problems.
The early part
of
the book deals with the
dynamics
of
particles and of rigid bodies in
two-dimensional motion. Both two- and three-
dimensional statics problems are discussed.
Vector notation is used initially as a label, in

a study
of
one-dimensional and plane stress and
strain leading to stresses and deflection
of
beams
and shafts. Also included in this chapter are the
basic elements
of
fluid dynamics, the purpose
of
this material is to show the similarities and the
differences in the methods
of
setting up the
equations for solid and fluid continua. It is not
intended that this should replace a text in fluid
dynamics but to develop the basics in parallel with
solid mechanics. Most students study the two
fields independently,
so
it is hoped that seeing
both Lagrangian and Eulerian co-ordinate sys-
tems in use in the same chapter will assist in the
understanding
of
both disciplines.
There is also a discussion of axial wave
propagation in rods (12.9), this is a topic not
usually covered at this level and may well be

advanced study in any specialisation. The
applications are wide-ranging and chosen to show
as many facets
of
engineering mechanics as is
practical in a book of this size.
We are grateful to The City University for
permission to use examination questions as a
basis for a large number
of
the problems. Thanks
are also due to our fellow teachers
of
Engineering
Mechanics who contributed many
of
the ques-
tions.
July 1993
H.R.H.
T.N.
1
Co-ordinate systems and position vectors
1.1
Introduction
Dynamics is a study of the motion of material
bodies and of the associated forces.
The study of motion is called kinematics and
involves the use of geometry and the concept of
time, whereas the study of the forces associated

path. This is the system used in road maps, where
place
B
(Fig. 1.1) is said to be 10 km (say) from
A
along road
R.
Unless
A
happens to be the end of
road
R,
we must specify the direction which is to
be regarded as positive. This system is often
referred to as a path co-ordinate system.
Two-dimensional systems
If
a point lies on a surface
-
such as that of a
plane, a cylinder or a sphere
-
then two numbers
are required to specify the position
of
the point.
For a plane surface, two systems of co-ordinates
a)
Cartesian co-ordinates.
In this system an

1.1
2
Co-ordinate systems and
positicn
~lnrterr
c)
Spherical co-ordinates.
In this system the
position is specified by the distance of a point
from the origin, and the direction is given by two
angles as shown in Fig. 1.6(a) or (b).
U
Figure
1.3
Three-dimensional systems
Three systems are in common
use:
a)
Cartesian co-ordinates.
This is a simple
extension of the two-dimensional case where a
third axis, the z-axis, has been added. The sense is
not arbitrary but is drawn according to the
right-hand screw convention, as shown in
Fig.
1.4.
This set
of
axes is known as a normal
right-handed triad.

0
of the axes; however, this would involve
an unnecessary complication.
1.3
Vector representation
The position vector
A
line drawn from the origin
0
to the point
P
always completely specifies the position of
P
and
is independent of any co-ordinate system. It
follows that some other line drawn to a
convenient scale can also be used to re resent the
In Fig.
1.7(b),
both
vectors represent the
position of
P
relative to
0,
which is shown in
1.7(a), as both give the magnitude and the
direction of
P
relative to

plus the position
of P relative to Q, as shown in Fig. 1.8(a).
The position
of
P could also be considered as
the position of Q’ relative to
0
plus that of P
relative to Q’.
If
Q’ is chosen such that OQ’PQ is
a parallelogram, i.e. OQ’
=
QP and OQ
=
Q’P,
then the corresponding vector diagram will also
be a parallelogram. Now, since the position magnitude and
is
in the required direction. Hence
vector represented by
oq’,
Fig. 1.8(b), is identical
to that represented by
qp,
and
oq
is identical to
q‘p,
it follows that the sum of two vectors is

vector. Usually the components are taken to be
orthogonal, as shown in Fig. 1.10.
Figure
1.8
rotations depends on the order of addition (see
Chapter 10).
The law of addition may be written symbolic-
ally as
s=g+ep=ep+s
(1.1)
Vector notation
As vector algebra will be used extensively later,
formal vector notation will now be introduced. It
is convenient to represent a vector by a single
symbol and it is conventional to use bold-face
type in printed work or
to
underline a symbol in
manuscript. For position we shall use
S=r
The fact that addition is commutative is
demonstrated in Fig.
1.9:
r=rl+r*=r2+rl
(1.2)
Unit vector
It is often convenient to separate the magnitude
of
a vector from its direction. This is done by
introducing a unit vector

components
of
A
with respect to the
x-,
y-,
z-axes.
It follows that, if
B
=
B,i+B,j+
B,k,
then
A
+
B
=
(A,
+
B,)i
+
(A,
+
By)j
+(Az+Bz)k
(1.5)
4
Co-ordinate systems and position vectors
It is also easily shown that
Direction cosines

is defined as the cosine
of the angle between the vector and the positive
x-axis, i.e. from Fig. 1.13.
Notice that
because
A
and
B
are free vectors.
Scalarproduct
of
two
vectors
The scalar product of two vectors
A
and
B
(sometimes referred to as the
dot product)
is
formally defined as
IA
1
IB
1
cos0, Fig. 1.12, where
0
is the smallest angle between the two vectors.
The scalar product is denoted by a dot placed
between the two vector symbols:

I
(1.10b)
n
=
cos(LP0N)
=
A,/IA
I
(1.10~)
From equations 1.3 and 1.10,
AA,AA
e=-
=-i++j+Ik
IAl IAl IAI IAl
=
li+rnj+nk
that is the direction cosines are the components of
the unit vector; hence
12+m2+n2
=
1
(1.11)
Figure
1.12
Discussion
examples
From Fig. 1.12 it is seen that
[A
I
cos0 is the

of
e.
Example
1.1
See Fig. 1.14.
A
surveying instrument at C can
measure distance and angle.
Relative to the fixed
x-, y-,
z-axes at C, point
A
is at an elevation of 9.2" above the horizontal
(xy)
plane. The body of the instrument has to be
rotated about the vertical axis through 41" from
the
x
direction in order to be aligned with
A.
The.
distance from C to
A
is
5005
m. Corresponding
values for point
B
are 1.3", 73.4" and 7037
m.

See Fig. 1.15. For point A,
r
=
5005
m,
e
=
410,
Q,
=
9.2".
z
=
rsinQ,
=
5005sin9.2"
=
800.2 m
R
=
rcos4
=
5005~0~9.2"
=
4941.0 m
x
=
Rcose
=
4941~0~41"

or
AB=CB-CA
=
(2010i+6742j+ 159.78)
-
(3729i+ 3241j+ 800.2k)
=
(-1719i+3501j-640.5k) m
The distance from
A
to B is given by
131
=
d[(-1791)2+ (3501)2+
(-640.5)2]
=
3952 m
and the component
of
AB
in the xy-plane is
d[(
-
1719)2
+
(3501)2]
=
3900 m
Example
1.2

=
(3j+ 2k)
+
(-2i+
4k)
=
-2i+3j+6k
Hence point
C
is located at (-2,3, 6) m.
Similarly
Z
=
3
+
2
so
that
Z=Z-G
=
(-2i
+
3j+ 6k)
-
(3i
+
4j+
5k)
=
(-5i-

component, otherwise known as the
scalar
component.
The vector
a
is determined from the
relationship
thus
s=OA-OB
+
++
OB+BA
=
Z?
-+
=
(2i
+
2j
-
4k)
-
(5i
+
7j
-
1
k
)
=

=
(3i+4j+5k)
therefore cos (LCOD)
*
(-3i
-
5j
-
3k)/d43
=
-
(3
x
3
+
4
x
5
+
5
x
3)/d43
=
-6.17
m (OC)(OD)
- -
2.3
The minus sign indicates that the component of
OP
(taking the direction from

A.
-
If we wish to represent the component of
OP
in
-
d21
d6
d126
the specified direction as a vector, we multiply the
scalar component by the unit vector for the
specified direction. Thus and LCOD
=
69.12”
=
0.3563
m
-(3i
+
5j
+
3k)( -6.17)/d43
As
a check, we can determine LCOD from the
=
(2.82+4.70j+2.82k)
m
cosine rule:
OC2
+

A
position vector is given by OP
=
(3i+2j+ lk)
m. Determine its unit vector.
1.2
A
line PQ has a length
of
6
m and a direction
given by the unit vector
gi+G++k.
Write PQ as a
vector.
1.3
Point
A
is at
(1,2,3)
m
and the position vector of
point
B,
relative to
A,
is
(6i+3k)
m. Determine the
position

m and point
B
at
(2,2,6)
m. Determine the position vector (a) from
A
to
B
and (b) from
B
to
A.
1.6
P is located at point
(0,
3, 2)
m and Q at point
(3,2,1).
Determine the position vector from P to Q
1.7
A
is at the point
(1,
1,2)
m. The position
of
point
B
relative to
A

at
(5,
53
O)
m-
=
(2i-
lj+
lk)
-
(li+ 2j+ 4k)
=
(li-3j-3k)
m
=
4.36
m
and
lal
=
d[12+(-3)2+(-3)2]
=
d19
and its unit vector.
The scalar or dot product involves the angle
Problems
7
relative to B is
(-3i-2j+2k)
rn. Determine the

shown in Fig.
1.18.
A cable is suspended from the point
P
in the ceiling and a lamp
L
at the end of the cable is
1.2 rn vertically below
P.
Figure
1.20
1.11
What are the angles between the line joining the
origin
0
and a point at (2,
-5,6)
m and the positive x-,
y-,
z-axes?
1.12
In problem 1.7, determine the angle ABC.
Determine the Cartesian and cylindrical co-ordinates
of
the lamp
L
relative to the x-, y-, z-axes and also find
1.13
A vector is given by
(2i+ 3j+

Points A, B and C are located at (1,2, 1) m,
(5,6,7)
rn and (-2,
-5, 6)
rn respectively. Determine
(a) the perpendicular distance from B to the line AC
and
(b)
the angle BAC.
1.9
Show that the relationship between Cartesian and
cylindrical co-ordinates is governed by the following
equations (see Fig. 1.19):
x
=
RcosO, y
=
Rsin
0,
R
=
(x2+y2)"',
i
=
cos
BeR
-
sin
Beo,
eR

Kinematics
of
a particle in plane motion
2.1
Displacement, velocity and
acceleration
of
a particle
A
particle may be defined as a material object
whose dimensions are
of
no
consequence to the
SO
v
=
limA,o
-e,
=
describing the kinematics
of
such an object, the
motion
may
be
taken
as
being
that

The
tem
&Idt
is
the
rate
of
change
of
distance
along the path and is a scalar quantity usually
Acceleration
of
a particle
The
acceleration
of
a
particle
is
defined (see
Fig.
2.2)
as
Av
dv d2r
dt dt2
a
=
limAr+,,( )

interval At is defined to be
If the time difference At
=
t2
-
tl is small, then
Ar
-_
-
At
Vaverage
This is a vector quantity in the direction
of
Ar.
The instantaneous velocity
is
defined as
v
=
limA,+o
-
-
-
(:)
-
:
Figure
2.2
tangential to the path unless the path is straight.
Having defined velocity and acceleration in a

=
ii+yj
then
v
+
Av
=
(i
+
Ai
)
i
+
(
y
+
Ay
)
j
giving
Av
=
Aii
+
Ayj.
Av
Ai
Ay
Figure
2.3

(2.6)
that the vector diagram (Fig.
2.3)
is an isosceles
=
ii+yj
(2.7)
and
lal
=
d(n2
+j2)
the motion in Cartesian co-ordinates.
i)
Motion in a straight line with constant
acceleration
Choosing the x-axis to coincide with the path
of
motion, we have
Let
us
consider two simple cases and describe
x=Q
Intregration with respect to time gives
Jidt
=
J(dv/dt) dt
=
v
=

-
Ar
=-1+-J*
Ax. Ay
At At At
where
C2
is another constant depending on the
value
of
x
at
t
=
0.
ii)
Motion with constant speed along a
For the circular path shown in Fig.
2.6,
(2.4)
circularpath
v
=
limk+o(z)
Ar
=
zi+zj
dx
dy
From Fig.

and
2xi+2i2+2yy+2y2
=
0
2xx+2yy
=
0
Since
2i2
+
2y2
=
2v2,
xx+yy
=
-v2
(2.11)
We see that, when
y
=
0
and
x
=
R,
Figure
2.8
x
=
-v2/R

over a time interval
At
is
shown in Fig.
2.8,
where
AY
is the elemental path
length. Referring to Fig.
2.9,
the direction
of
the
path has changed by an angle
A0
and the speed
has increased by
Av.
Noting that the magnitude of
v(t+At)
is
(v+Av),
the change in velocity
resolved along the original normal is
f
v
+
Av
)
sin

v-
dt
=
YXIR
-
XylR
Differentiating
x2
+y2
=
v2
with respect to
time, we have
and is directed towards the centre of curvature,
i.e. in the direction of
e,.
2ix
+
2yy
=
0
hence
If
p
is the radius of curvature, then
y/x
=
-x/y
ds
=

a,
=
0.
This analysis should be contrasted with the
more direct approach in terms
of
path and polar
The change in velocity resolved tangentially to
the
path
is
co-ordinates shown later
in
this chapter.
(V
+
AV)CosAO
-
v
2.4
Polar co-ordinates
11
=
-
e,+r
-
e,
=
ie,+rbee (2.17)
Resolving the components of Av along the e,

-e,+v-en
dt dt
d2s v2
dt2
t
p
cases.
i)
Straight-line motion with constant
acceleration
x
(b+A8)sinA8-i]er
or d2sldt2
=
a (2.15)
replaced by
s.
ii)
Motion in a circle at constant speed
2.4
Polar co-ordinates
Polar co-ordinates are a special case of cylindrical
co-ordinates with
z=
0, or of spherical co-
ordinates with
4
=
0.
Figure

lim*,o
(E)
=
(z
-
reZ) e,
de de dr.
+
i-+r-+-O ee
The
so1ution
is
the Same as before, with
x
For small angles,
sinA8+
A,g
and
COSA~+
1;
thus
Ai
di .de
a
=
(v2/p)en (v and
p
are constant) (2.16)
(2.18)
An alternative approach to deriving equations

which are rotating at an angular rate
o
=
8
as
shown in Fig. 2.12. The derivative with respect to
time
of
e, is
Aer
er
=
limii-0
(E)
where Aer is the change in e, which occurs in the
time interval At. During this interval e, and ee
12
Kinematics
of
a particle in plane motion
r
=
constant for all time
v
=
roe,
Because
r
and
ZJ

we see that there is a constant component
of
vectors
e’,
and
ere.
The difference between
e’,
acceleration,
2ib,
at right angles to the spoke,
and
e,
is
be,
=
e’,
-
e,.
The magnitude
of
Ae,
for independent
of
r.
This component is often called
small
AB
is
1xAO

(2.19)
rB/A
=
position of B relative to
A
iB/A
=
velocity
of
B relative to
A,
etc.
Similarly it can be shown that
eo
=
-Be, (2.20)
From Fig.
2.14,
The velocity
v
is the derivative with respect to
time
of
the position vector r
=
re,.
From the chain
rule for differentiation we obtain
rB/O
=

fer
+
roee
from equation
2.19,
which is the result previously
obtained in equation
(2.17).
The acceleration
a
can also be found from the
chain rule, thus
d
dt
Figure 2.14
a
=
C
=
-
(ie,
+
roe,)
=
re,
+
ie,
+
ihe,
+

reference
axes.l
Consider now the case
of
a wheel radius
r,
centre
A,
moving
so
that
A
has rectilinear motion
in the x-direction and the wheel is rotating at
angular speed
w
=
h
(Fig.
2.15).
The path traced
out by a point B on the rim
of
the wheel is
complex, but the velocity and acceleration
of
B
may be easily obtained by use
of
equations

ii+
(roeo)
=ii-i+(-rwsinei+rwcosej)
(2.26)
Similarly,
FB/~
=
xi
+
(
-rw2e,
+
rko
)
We may write
=
xi
-
rw2
(cos
~i
+
sin
ej)
dv
ds
dv dv
dt dt
ds ds
+

(2.28)
Most problems in one-dimensional kinematics
involve converting data given in one set of
variables to other data. As an example: given the
way in which a component
of
acceleration varies
with displacement, determine the variation
of
speed with time. In such problems the sketching
of appropriate graphs is a useful aid to the
2.7
Graphical methods
Speed-time graph
(Fig.
2.17)
dv d2s dv
'-
dt
dt2
-
'd,i
a =
iB/o
=
(i+rw)i+Oj= 0
then
X=
-rw
Also,

linear motion is one-dimensional,
not all one-
dimensional motion is linear.
We have one-dimensional motion in path
co-ordinates if we consider only displacement
along the path; in polar co-ordinates we can
consider only variations in angle, regarding the
radius as constant. Let us consider a problem in
path co-ordinates, Fig.
2.16,
the location
of
P
being determined by
s
measured along the path
from some origin
0.
(This path could,
of
course,
be a straight line.)
Speed is defined as
v
=
dsldt, and dvldt
=
rate
Figure 2.17
Slope

=
change
of
distance
If
a,
is constant, then the graph is a straight line
14 Kinematics
of
a particle in plane motion
area
=
i(v1+v2)(t2-tl)
=s2-s1
(2.31)
and slope
=
a,
Distance-time graph
(Fig.
2.18)
Figure 2.21
The advantages
of
sketching the graphs are many
-
even for cases
of
constant acceleration (see
examples 2.2 and 2.3).

and v
=
vo. Show
that
(3
v= vo+atot, (b)
s
=
vot+Ba,ot2,
(c)
v2
=
vo
+
2atos and (d)
s
=
+(v
+
vo)t.
Solution
Rate-of-change-of-speed-time graph
(Fig.
2.19)
Figure 2.19
a) Since
a,
=
dv/dt,
dv

atot
0
It,
dt
Q
If
a,
is constant, then
area
=
at(t2-tl)
=
v2-vl
(2.34)
(9
or v
=
vo
+
a,ot
Rate-of-change-of-speed-displacement graph
Fig.
2.20).
Here we make use
of
equation
b) Since v
=
ds/dt,
f2.28).

vo)/t and substituting

s2
dv (2.35) for
ato
in (ii) gives
v-
ds
=
iv2
-
;vl2
Area
=
s
=
B(vo+v)t (iv)
I,,
ds
[As these equations for constant acceleration are
(2.36) often introduced before the case
of
variable
acceleration has been discussed, it is a common
mistake to try to apply them to problems dealing
with variable acceleration. For such problems,
however, the methods of section 2.7 should
always be used (cf. example 2.3).]
If
a,

of a vehicle is given in Fig.
2.22.
At distance, find for
s
=
40 m (a) the speed and (b)
time
t
=
0
the speed is zero. Determine the speed the time taken.
when
t
=
t3.
Solution
a) We are given
a,
in terms
of
s
and require to
find
v,
therefore we must use an expression
relating these three parameters. The constant-
acceleration formulae are of course not relevant
here. The basic definition
a,
=

Fig. 2.23.
Letting
s1
=
0
and
s2
=
40
m, the area is found
to be 32.0
(m/s)’.
This area can be determined by
counting the squares under the graph, by the
trapezium rule, by Simpson’s rule, etc., depend-
ing on the order of accuracy required. (The
trapezium rule and Simpson’s rule are given in
Appendix 3
.)
L2
‘3
Figure
2.22
Solution
Each portion
of
the graph represents
constant acceleration and
so
we can use the

=
Ultl
Time
tl
to
t2:
=
a2(2-t1)+a1t1
v3-v2= as(t3-t2), v3
=
a3(t3-t2)+02
03
=
a3(t3
-
t2)
+
@(t2
-
tl
)
+
Ultl
Time
t2
to
t3:
Alternatively we can dispense with the con-
stant-acceleration formulae and obtain the same
result more rapidly by noting that the speed

v40
=
d[2(32)
+
3’1
=
8.54
m/s
b) Given
a,
as a function
of
s,
time cannot be
found directly. We can, however, make use of the
relationship
v
=
ds/dt in the form
dt
=
(l1v)ds
a,/(m
s-’)
s/m a,/(m
sp2)
s/m provided we can first establish the relationship
1.2
0
-1.3 25 between

s/m area/(m2 s-~)
=
[2(area
+
2)02)]1'2
0-5
8.4
5.08
0-10
20.2 7.03
CL15
32.6 8.61
0-20
39.4 9.37
0-25 36.8 9.09
0-30 31.2 8.45
0-35 29.0 8.91
0-40
32.0 8.54
Since
t2
-
tl
=
(l/v)
ds,
the area under the graph
of
l/v
versus

of
2.5m/s to the
right. The angular velocity
of
the wheel is
constant and equal to 6 rads clockwise. Point
P
is
at the bottom
of
the wheel and is in contact with a
horizontal surface. Points
Q
and
R
are as shown
in the figure.
-
IY.yrv
Figure 2.24
drn
(l/v)/ s/m
(l/v)/
s/m
(s
m-')
(s
m-')
0.333
0

of
5.0
m/s
and a tangential rotating and translating is to determine the
acceleration
of
magnitude 2.0
m/s2.
If
the motion
of
the wheel centre and add on the motion
magnitude
of
the total acceleration is 3.0m/s2,
of
the point relative to the centre.
So
for an
what is the radius of curvature
of
the path being arbitrary point A and centre C we can make use
of
Solution
Choice
of
co-ordinates is not difficult
vA
=
vc

(v2/p)
e,
to C
is
given by
The time taken is found to be approximately
traced out by the point at this instant?
a
=
atet
+
anen
(see equations 2.14)
vplc
=
i-e,
+
r6ee
aQ
=
aC
+
aQ/C
where
r
is the length of the line
CP
and
I3
is the

sense); thus
&
=
-6
rads, and [see Fig. 2.27(a)]
but
vc
is constant, and
so
ac
=
0.
Therefore
UQ
=
-
18j
m/s2
e,
=
sin
30"i
+
cos
307
=
4
i
+
td3j

=
(5.1i- 1.5j)
m/s
The same result can be obtained from a velocity
vector diagram, Fig. 2.28. Here
vc
and
vR/c
are
drawn
to
some appropriate scale in the correct
directions and are added graphically to give
VR
.

Figure
2.27
Figure
2.28
For the acceleration
of
R
relative to
C
we have
vplc
=
veee
=

decelerating at 2
m/s2.
The telescopic arm AB has
a length of
1.5
m which
is
increasing at a constant
rate of 2ds. At the same time, the arm has an
anticlockwise angular velocity
of
3
rads and a
clockwise angular acceleration
of
0.5
rads'.
The velocity of
C
is
v
=
2.5
m/s
and the total
velocity of
P
is
=
(9i+

=
&ee
=
(-3)(-i)
=
3i
m/s
so
that
VQ
=
VC
+
vQ/C
=
2.5i
+
3i
=
5.5
m/~
From equation
2.18,
the acceleration
of
Q
relative to
C
is given by
uQIC

=
[(-18.54)2+ (5.96)2]1/2
=
19.47
m/s2
A
graphical solution
is
again appropriate,
and somewhat quicker. For the velocity vector
diagram we first draw, to scale,
vA,
the velocity of
A,
4
m/s
to the right (Fig.
2.31).
The velocity
of
B
relative to
A,
Z)B/~,
having the components
i.
=
2
and re
=

and
j
(see Fig.
2.30).
e,
=
cos
20"i
+
sin
203
ee
=
-sin
20"i
+
cos
203
From equation
2.17,
Z)BIA=i.e,+rke
wherer= 1.5andi.=2
Figure 2.31
thus
%/A
=
2(0.940i+ 0.3421')
=
O.341i+4.91jm/s2
For the acceleration vector diagram of Fig.

e,
direc-
tion and
r8+2i.h=
1.5(-0.5)+2(2)3
=
11.25
m/s2
in the
eo
direction. The acceleration of
B,
aB
,
can be scaled from the figure.
+
1.5(3)(-0.3423+ 0.940j)
%
=
VA +%/A
%
=
4i
+
(0.34i+ 4.01j)
thus
=
(4.34i+4.91.)
m/s2
The speed of

e,
Figure 2.32
and
i:
=
0
since
i.
is constant.
aB/A
=
-1.5(3)2(0.940i+0.342j)
Example
2.7
+
[(1.5)(-0.5) +2(2)3]
A
racing car
B
is being filmed from a camera
mounted on car
A
which is travelling along a
straight road at a constant speed
of
72
km/h. The
racing car is moving at a constant speed of
144
km/h along the circular track, centre

the camera
so
that the image
of
B
remains centrally positioned in the viewfinder.
Solution
In order to find the required angular
velocity and angular acceleration, we shall first
need to determine the velocity and acceleration
of
B
relative to A in the given polar co-ordinates and
then make use of equations 2.17 and 2.18.
The velocity
of
B
is perpendicular to the line
AB,
so
that
%
=
lM(E)(-eo)
=
-Neo
m/s
The velocity of A is
vA
=

(9
Also, from equation 2.17,
(ii)
Comparing equations (i) and (ii) and noting
from Fig. 2.33 that
r
=
(230/cos30")
-
200
=
65.58 m
we find
i=
10ds
and the angular velocity of the camera is
h
=
-22.5fU65.58
=
-0.346
rads
The acceleration of
B
is most conveniently
found from path co-ordinates (equations 2.14)
and is
Problems
19
4d

(iv)
Comparing equations (iii) and (iv) we see that
0
=
re+2fh
=
65.588+2(10)(-0.346)
hence the angular acceleration
of
the camera is
8
=
20(0.346)/65.58
=
0.106 rads'
Problems
2.1
The position
of
a point, in metres, is given by
r
=
(6t-5t2)i+
(7+8t3)j,
where
t
is the time in
seconds. Determine the position, velocity and the
acceleration
of

+
213')
ds.
Determine the position and velocity at
t
=
1.
2.3
A
point
A
is following
a
curved path and at a
particular instant the radius
of
curvature of the path is
16m.
The
speed of the point
A
is
8ds
and its
component of acceleration tangential to the path is
3
ds2.
Determine the magnitude of the total accelera-
tion.
2.4

A.
(1
knot
=
1
nautical milem
=
6082.66
ft/h
=
0.515
ds.)
2.6
A
telescopic arm
AB
pivots about
A
in a vertical
20
Kinematics
of
a particle in plane motion
plane and is extending at a constant rate of
1
ds,
the
angular velocity of the arm remaining constant at
5
rads anticlockwise, Fig.

3
s
and the distance travelled
between
t
=
0
and
t
=
6
s.
(b) If, at
t=3s,
the magnitude
of
the total
acceleration is
1.0
ds2,
estimate the magnitude
of
the
acceleration normal to the path and also the radius
of
curvature
of
the path.
motion
of

of
C
relative to
B
and the velocity
of
C.
ds2.
Also determine for this cae the speed
of
B
and
2'11
The
forward
(tangential)
acce1eration
at
Of
the
At
s
=
4
m, the forward speed is
4.6
ds.
Estimate
(a) the speed at
s

+
2)
ds2,
where
s
is the distance moved in metres. When
s
=
0
its speed is
zero. Find its speed when
s
=
4
m.