ENGINEERING MECHANICS
ME 2401
STATICS OF
PARTICLES
CONTENT
Introduction
Sample Problem 2.3
Resultant of Two Forces
Equilibrium of a Particle
Vectors
Free-Body Diagrams
Addition of Vectors
Sample Problem 2.4
Resultant of Several Concurrent Forces
Problem 2.5
Sample Problem 2.1
Rectangular Components in Space
Sample Problem 2.2
Problem 2.6
Rectangular Components of a Force:
Unit Vectors
Sample Problem 2.7
Addition of Forces by Summing
Components
Sample Problem 2.9
Problem 2.8
Problem 2.10
Problem 2.11
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Introduction
• The objective for the current chapter is to investigate the effects
of forces on particles:
-
replacing multiple forces acting on a particle
with a single equivalent or resultant force,
-
relations between forces acting on a
particle that is in a state of equilibrium.
• The focus on particles does not imply a restriction to miniscule
bodies. Rather, the study is restricted to analyses in which the size
and shape of the bodies is not significant so that all forces may be
assumed to be applied at a single point.
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Resultant of Two Forces
• Force: action of one body on another;
characterized by its point of application,
magnitude, line of action, and sense.
• Experimental evidence shows that the
combined effect of two forces may be
represented by a single resultant force.
• The resultant is equivalent to the diagonal of a
parallelogram which contains the two forces in
adjacent legs.
• Force is a vector quantity.
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Vector
• Vector: parameters possessing magnitude and direction
which add according to the parallelogram law.
Examples: displacements, velocities, accelerations.
• Scalar: parameters possessing magnitude but not
direction. Examples: mass, volume, temperature
• Vector classifications:
- Fixed or bound vectors have well defined points of application that
cannot be changed without affecting an analysis.
- Free vectors may be freely moved in space without changing their
effect on an analysis.
- Sliding vectors may be applied anywhere along their line of action
without affecting an analysis.
• Equal vectors have the same magnitude and direction.
• Negative vector of a given vector has the same magnitude
and the opposite direction.
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Addition of Vectors
• Trapezoid rule for vector addition
• Triangle rule for vector addition
• Law of cosines,
C
B
C
• Law of sines,
B
• Vector addition is commutative,
• Vector subtraction
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Addition of Vectors
• Addition of three or more vectors through
repeated application of the triangle rule
• The polygon rule for the addition of three or
more vectors.
• Vector addition is associative,
• Multiplication of a vector by a scalar
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Resultant of Several Concurrent Forces
• Concurrent forces: set of forces which all
pass through the same point.
A set of concurrent forces applied to a
particle may be replaced by a single
resultant force which is the vector sum of the
applied forces.
• Vector force components: two or more force
vectors which, together, have the same effect
as a single force vector.
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Sample Problem 2.1
SOLUTION:
The two forces act on a bolt at A.
Determine their resultant.
• Graphical solution - construct a
parallelogram with sides in the same
direction as P and Q and lengths in
proportion. Graphically evaluate the
resultant which is equivalent in direction
and proportional in magnitude to the the
diagonal.
• Trigonometric solution - use the triangle
rule for vector addition in conjunction
with the law of cosines and law of sines
to find the resultant.
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Sample Problem 2.1
• Graphical solution - A parallelogram with
sides equal to P and Q is drawn to scale. The
magnitude and direction of the resultant or of
the diagonal to the parallelogram are measured,
• Graphical solution - A triangle is drawn with
P and Q head-to-tail and to scale. The
magnitude and direction of the resultant or of
the third side of the triangle are measured,
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Sample Problem 2.1
• Trigonometric solution - Apply the triangle rule.
From the Law of Cosines,
From the Law of Sines,
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Sample Problem 2.2
SOLUTION:
• Find a graphical solution by applying the
Parallelogram Rule for vector addition. The
parallelogram has sides in the directions of
the two ropes and a diagonal in the direction
of the barge axis and length proportional to
5000 N.
A barge is pulled by two tugboats. If
• Find a trigonometric solution by applying
the resultant of the forces exerted by
the Triangle Rule for vector addition. With
the tugboats is 5000 N directed along
the magnitude and direction of the resultant
the axis of the barge, determine
known and the directions of the other two
sides parallel to the ropes given, apply the
a) the tension in each of the ropes for
Law of Sines to find the rope tensions.
a = 45o,
• The angle for minimum tension in rope 2 is
b) the value of a for which the
determined by applying the Triangle Rule
tension in rope 2 is a minimum.
and observing the effect of variations in a.
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Sample Problem 2.2
• Graphical solution - Parallelogram Rule
with known resultant direction and
magnitude, known directions for sides.
• Trigonometric solution - Triangle Rule
with Law of Sines
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Sample Problem 2.2
• The angle for minimum tension in rope 2 is
determined by applying the Triangle Rule
and observing the effect of variations in a.
• The minimum tension in rope 2 occurs when
T1 and T2 are perpendicular.
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Rectangular Components of a Force: Unit Vectors
• May resolve a force vector into perpendicular
components so that the resulting parallelogram is a
rectangle.
are referred to as rectangular
vector components and
• Define perpendicular unit vectors
parallel to the x and y axes.
which are
• Vector components may be expressed as products of
the unit vectors with the scalar magnitudes of the
vector components.
Fx and Fy are referred to as the scalar components of
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Addition of Forces by Summing Components
• Wish to find the resultant of 3 or more
concurrent forces,
• Resolve each force into rectangular components
• The scalar components of the resultant are
equal to the sum of the corresponding scalar
components of the given forces.
• To find the resultant magnitude and direction,
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Sample Problem 2.3
Four forces act on bolt A as shown. Determine the resultant of the force on the bolt.
SOLUTION:
• Resolve each force into rectangular
components.
• Determine the components of the
resultant by adding the corresponding
force components.
• Calculate the magnitude and direction
of the resultant.
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Sample Problem 2.3
SOLUTION:
• Resolve each force into rectangular components.
• Determine the components of the resultant by
adding the corresponding force components.
• Calculate the magnitude and direction.
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Equilibrium of a Particle
• When the resultant of all forces acting on a particle is zero, the particle is
in equilibrium.
• Newton’s First Law: If the resultant force on a particle is zero, the particle will
remain at rest or will continue at constant speed in a straight line.
• Particle acted upon by
two forces:
- equal magnitude
- same line of action
- opposite sense
• Particle acted upon by three or more forces:
- graphical solution yields a closed polygon
- algebraic solution
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Free-Body Diagrams
Space Diagram: A sketch showing
the physical conditions of the
problem.
Free-Body Diagram: A sketch showing
only the forces on the selected particle.
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Sample Problem 2.4
In a ship-unloading operation, a 3500-N automobile is supported by a cable. A rope is
tied to the cable and pulled to center the automobile over its intended position. What is
the tension in the rope?
SOLUTION:
• Construct a free-body diagram for the
particle at the junction of the rope and
cable.
• Apply the conditions for equilibrium
by creating a closed polygon from the
forces applied to the particle.
• Apply trigonometric relations to
determine the unknown force
magnitudes.
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Sample Problem 2.4
SOLUTION:
• Construct a free-body diagram for the
particle at A.
• Apply the conditions for equilibrium.
• Solve for the unknown force magnitudes.
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Problem 2.5
The direction of the 75-lb forces may vary, but the angle between the forces is always
50o. Determine the value of a for which the resultant of the forces acting at A is
directed horizontally to the left.
580 lb
A
30o
a
75 lb
50o
75 lb
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Problem 2.5
1. Determine the resultant R of two or more forces.
2. Draw a parallelogram with the applied forces as two adjacent sides and the
resultant as the included diagonal.
3. Set the resultant, or sum of the forces, directed horizontally.
580 lb
A
30o
a
75 lb
50o
75 lb
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Problem 2.5 Solution
580 lb
o
30
a
75 lb
50o
Determine the resultant R of two or more forces.
A
We first Replace the two 75-lb forces by
their resultant R1, using the triangle rule.
75 lb
a
25o
50o
25o
R1
R1 = 2(75 lb) cos25o = 135.95 lb
R1 = 135.95 lb
a +25o
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Problem 2.5 Solution
Draw a parallelogram with the applied forces as two adjacent sides and the resultant as
the included diagonal. Set the resultant, or sum of the forces, directed horizontally.
R2
o
30
580 lb
a+25
o
R1 = 135.95 lb
Consider the resultant R2
of R1 and the 580-lb force
and recall that R2 must be
horizontal and directed to
the left.
Law of sines:
sin(30o)
sin(a+25o)
=
580 lb
135.95 lb
o
(580
lb)
sin(30
)
o
sin(a+25 ) =
135.95 lb
a + 25o = 61.97o
= 0.88270
a = 37.0o
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Rectangular Components in Space
• With the angles between
•
and the axes,
is a unit vector along the line of action of
and
are the direction
cosines for
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Rectangular Components in Space
Since the magnitude of 𝜆 is equal to unity,
We must have
(cos 𝜃𝑥 )2 +(cos 𝜃𝑦 )2 +(cos 𝜃𝑧 )2 =1
• When the rectangular component 𝐹𝑥 , 𝐹𝑦
and 𝐹𝑧 of a force F are given, the
magnitude F of the force
F = 𝐹𝑥2 + 𝐹𝑦2 + 𝐹𝑧2
• And the direction cosines of F are
𝐹𝑦
𝐹
𝐹
cos 𝜃𝑥 = 𝑥
cos 𝜃𝑦 =
and cos 𝜃𝑧 = 𝑧
𝐹
𝐹
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Rectangular Components in Space
Direction of the force is defined by the location of two points,
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Problem 2.6
Find the magnitude and direction of the resultant of the two forces shown knowing that
P = 600 N and Q = 450 N.
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Problem 2.6
Find the magnitude and direction of the resultant of the two forces shown knowing that
P = 600 N and Q = 450 N.
SOLUTION:
• Based on the relative locations of the
Force P and Q, determine the resultant
force R
• Apply the unit vector to determine the
components of the force P and Q.
• Noting that the components of the unit
vector are the direction cosines for the
vector, calculate the corresponding
angles.
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Problem 2.6
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Sample Problem 2.7
The tension in the guy wire is 2500 N. Determine:
a) components Fx, Fy, Fz of the force acting on the bolt at A,
b) the angles qx, qy, qz defining the direction of the force
SOLUTION:
• Based on the relative locations of the
points A and B, determine the unit
vector pointing from A towards B.
• Apply the unit vector to determine the
components of the force acting on A.
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Sample Problem 2.9
A 200-kg cylinder is hung by means of two cables AB and AC, which are attached to
the top of a vertical wall. A horizontal force P perpendicular to the wall holds the
cylinder in the position shown. Determine the magnitude of P and the tension in each
cable.
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Sample Problem 2.9
SOLUTION:
A ≡ 1.2, 2, 0
B ≡ 0, 12, 8
C ≡ 0, 12, -10
Free-Body Diagram. Point A is chosen as a free body; this point is subjected to
four forces, three of which are of unknown magnitude. Introducing the unit
vectors i, j, k, we resolve each force into rectangular components.
P =Pi
W = -mgj = - 2(200 kg)(9.81 m/𝑠 2 )j = -(1962 N)j
(1)
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Sample Problem 2.9
A ≡ 1.2, 2, 0
B ≡ 0, 12, 8
C ≡ 0, 12, -10
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Sample Problem 2.9
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Problem 2.8
A container of weight W = 1165 N is supported by three cables as shown. Determine
the tension in each cable.
y
360 mm
C
D
450 mm
O
500 mm
B
z
A
320 mm
600 mm
x
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y
360 mm
Problem 2.8
C
D
450 mm
O
500 mm
B
z
A
320 mm
600 mm
x
A container of weight
W = 1165 N is supported
by three cables as shown.
Determine the tension in
each cable.
1. Draw a free-body diagram of the particle. This diagram shows
the particle and all the forces acting on it.
2. Resolve each of the forces into rectangular components.
Follow the method outlined in the text.
F=Fl=
F
(dx i + dy j + dz k)
d
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Problem 2.8
y
360 mm
C
D
450 mm
O
500 mm
B
z
A
320 mm
600 mm
A container of weight
W = 1165 N is supported
by three cables as shown.
Determine the tension in
each cable.
x
3. Set the resultant, or sum, of the forces exerted on the particle
equal to zero. You will obtain a vectorial equation consisting
of terms containing the unit vectors i, j, and k. Three scalar
equations result, which can be solved for the unknowns.
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y
360 mm
C
D
450 mm
SF=0
TAB + TAC + TAD + W = 0
O
TAC
500 mm
z
Problem 2.8 Solution
Draw a free-body diagram of the particle.
B
TAD
A
320 mm
TAB
600 mm
x
W = _ (1165 N) j
AB = (450 mm)i + (600 mm)j
AB = 750 mm
AC = (600 mm)j _ (320mm)k
AC = 680 mm
AD = (_500 mm)i + (600 mm)j + (360 mm)k
AD = 860 mm
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y
360 mm
D
500 mm
z
O
TAC
TAD
A
Problem 2.8 Solution
Resolve each of the forces
450 mm
C
into rectangular components.
AB
TAB = TAB lAB = TAB
AB
320 mm
B
600
TAB
450
j TAB
i +
=
x
600 mm
750
750
(
)
= (0.6 i + 0.8 j) TAB
W = _ (1165 N) j
600 _ 320
AC
15 j _ 8 k
TAC = TAC lAC = TAC
= 680 j
TAC
k TAC =
17
AC
17
680
AD
500
600
360
=
k TAD
i+
j+
TAD = TAD lAD = TAD
860
860
860
AD
25
30
18
=
k TAD
i+
j+
43
43
43
(
) (
(
(
)
)
)
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Problem 2.8 Solution
y
360 mm
C
D
O
TAC
500 mm
z
Set the resultant, or sum, of
the forces exerted on the
particle equal to zero.
450 mm
B
TAD
A
320 mm
TAB
600 mm
x
Substitution into S F = 0,
factor i, j, k and set their
coefficients to zero:
W = _ (1165 N) j
TAB = 0.9690 TAD
0.6 TAB _ 25 TAD = 0
43
15
0.8 TAB + TAC + 30 TAD _ 1165 N = 0
17
43
_ 8
TAC + 18 TAD = 0
TAC = 0.8895 TAD
43
17
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(1)
(2)
(3)
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Problem 2.8 Solution
y
360 mm
C
D
O
TAC
500 mm
z
TAD
A
Substitution for TAB and
TAC from (1) and (3)
into (2):
450 mm
B
320 mm
TAB
600 mm
x
W = _ (1165 N) j
( 0.8 x 0.9690 + 15 x 0.8895 + 30 )TAD _ 1165 N = 0
43
17
TAD = 516 N
_
2.2578 TAD 1165 N = 0
From (1): TAB = 0.9690 (516 N)
T = 500 N
AB
From (3): TAC = 0.8895 (516 N)
TAC = 459 N
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Problem 2.10
y
Cable AB is 65 ft long, and the
tension in that cable is 3900 lb.
Determine
(a) the x, y, and z components of
the force exerted by the cable on
the anchor B,
(b) the angles qx, qy, and qz
defining the direction of that force.
A
56 ft
D
a
O
B
20o
50o
z
C
x
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y
Problem 2.10
A
Cable AB is 65 ft long, and
the tension in that cable is
3900 lb. Determine
(a) The x, y, and z components of
the force exerted by the cable on the
anchor B,
(b) the angles qx, qy, and qz
defining the direction of that force.
56 ft
D
a
O
B
20o
o
50
z
C
x
1. Determine the rectangular components of a force defined by its
magnitude and direction. If the direction of the force F is defined by
the angles qy and f, projections of F through these angles or their
components will yield the components of F.
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Problem 2.10
y
A
Cable AB is 65 ft long, and
the tension in that cable is
3900 lb. Determine (a) the
x, y, and z components of
the force exerted by the
cable on the anchor B, (b)
the angles qx, qy, and qz defining
the direction of that force.
56 ft
D
a
O
B
20o
50o
z
C
x
2. Determine the direction cosines of the line of action of a force.
The direction cosines of the line of action of a force F are
determined by dividing the components of the force by F.
Fy
Fz
Fx
cos qy= F
cos qz= F
cos qx= F
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A
y
65 ft
qy
56 ft
Problem 2.10 Solution
Determine the direction cosines of
the line of action of a force.
From triangle AOB:
Fy
F
cos qy = 56 ft = 0.86154
65 ft
qy = 30.51o
Fx
B
O
o
20
z
Fz
(a) Fx = _ F sin qy cos 20o
= _ (3900 lb) sin 30.51o cos 20o
Fx = _1861 lb
x
Fy = + F cos qy = (3900 lb)(0.86154)
Fy = + 3360 lb
Fz = + (3900 lb) sin 30.51o sin 20o
Fz = + 677 lb
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y
Problem 2.10 Solution
A
65 ft
qy
56 ft
Determine the direction cosines
of the line of action of a force.
F
Fy
Fx
B
O
_
Fx
1861 lb
=
(b) cos qx =
F
3900 lb
cos qx = _ 0.4771
qx = 118.5o
20o Fz
From above (a):
qy = 30.5o
677 lb
= +
= + 0.1736
3900 lb
qz = 80.0o
x
Fz
cos qz =
F
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Problem 2.11
Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless
rods. If a 60-lb force Q is applied to collar B as shown, Determine (a) the tension in the
wire when x = 9 in., (b) the corresponding magnitude of the force P required to
maintain the equilibrium of the system.
y
x
P
A
O
20 in
Q
z
z
B
x
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y
Problem 2.11
x
P
A
O
20 in
Q
z
z
B
x
Collars A and B are connected
by a 25-in.-long wire and can
slide freely on frictionless rods.
If a 60-lb force Q is applied to
collar B as shown, Determine
(a) the tension in the wire when
x = 9 in., (b) the corresponding
magnitude of the force P required
to maintain the equilibrium of
the system.
1. Draw a free-body diagram of the particle. This diagram shows
the particle and all the forces acting on it.
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y
Problem 2.11
x
P
A
O
20 in
Q
z
z
B
x
Collars A and B are connected
by a 25-in.-long wire and can
slide freely on frictionless rods.
If a 60-lb force Q is applied to
collar B as shown, Determine
(a) the tension in the wire when
x = 9 in., (b) the corresponding
magnitude of the force P required
to maintain the equilibrium of
the system.
2. Set the resultant, or sum, of the forces exerted on the particle equal
to zero. You will obtain a vectorial equation consisting of terms
containing the unit vectors i, j, and k. Three scalar equations result,
which can be solved for the unknowns.
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y
Problem 2.11 Solution
x
P
lAB = AB
AB =
A
O
20 in
_
x i _ (20 in) j + z k
25 in
Draw a free-body diagram of the particle.
Q
z
B
z
x
Ny j
A
Nz k
TAB lAB
Free Body: Collar A
S F = 0: P i + Ny j + Nz k + TAB lAB = 0
Substitute for lAB and set coefficients of i
P i equal to zero:
P
_
TAB x
= 0
25
(1)
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Problem 2.11 Solution
_
TAB lAB
Free Body: Collar B
Ny j
B
𝚺F = 0: (60 lb) k + Nx i + Ny j _ TAB lAB = 0
Substitute for lAB and set coefficients
of k equal to zero:
Nx i
60
Q = (60 lb) k
_
TAB z
25
= 0
(2)
(a) Since x = 9 in.: (9 in)2 + (20 in) 2 + z 2 = (25 in) 2
From eq. (2):
60
(b) From eq. (1): P =
_
TAB(12)
25
= 0
(125.0 lb)(9 in)
25 in
z = 12 in
TAB = 125.0 lb
P = 45.0 lb
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Acknowledgement
In this PowerPoint presentation, most of the slide has been made from the lecture
note of Ferdinand P. Beer. Russell Johnston, Jr. Special thanks to J. Walt Oler
(Texas Tech University ) for his valuable information about the Statics of Particles.
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