Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
Forces in a plane: 2/1 Forces on particles, Resultant of two forces
Force:
Represent the action of one body on another and is generally characterized by:
i.
Point of application
ii.
Magnitude
iii.
Direction
The direction of a force is defined by the line of action and the sense of the force.
The line of action is the infinite straight line along which the force act; it is
characterized by the angle its forms with some fixed axis (Fig. (2-1)).
Line of action
1
F
Point of application
θ
0
Fig. (2-1)
Note: Force is measured by Newton (N) or (Ib)
KN = 1000 N
Kip = 1000 Ib
2/2 Vectors and Scalar
Vectors: are defined as mathematical expression possessing magnitude and direction,
which add according to parallelogram law. Examples: displacements, velocities,
accelerations.
Scalar: parameters possessing magnitude but not direction.
volume, temperature.
8
Examples:
mass,
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
Two vectors which have the same magnitude and the same direction are said to be
equal, whether or not they also have the same point of application Fig. (2-2a).
Equal vectors may be denoted by the same letter.
The negative vector of a given vector p is defined as a vector having the same
magnitude as p and a direction opposite to that of p Fig. (2-2b). The negative of
vector p is denoted by –p. The vectors p and –p are commonly referred to as equal
and opposite vectors. Cleary, we have
+ − = ……………………………………….. (2.1)
p
p
p
-p
(a)
(b)
Fig. (2-2)
2/3 Addition of Vectors
Vectors add according to the parallelogram law. Thus, the sum of two vectors P and
Q is obtained by attaching the two vectors to the same point A and constructing a
parallelogram, using P and Q as two sides of parallelogram (Fig. (2-3)). The diagonal
that passes through A represents the sum of two vectors P and Q, and this sum is
denoted by P+Q.
P+Q
P
A
Q
Fig. (2-3)
9
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
From parallelogram law, we can derive an alternative method for determining the
sum of two vectors. This method, known as the triangle rule is derived as follows
Fig. (2-4).
Q
α
P
P
β
P+Q
Q
β
α
A
A
P+Q=Q+P
P
Q+P
β
Q
α
A
Fig. (2-4)
The sum of three or more vectors: The sum of three vectors P, Q, and S will by
definition, be obtained by first adding vectors P and Q and then adding the vector S
to the vector P + Q Fig. (2-5). We thus write
P + Q + S = (P + Q) + S = P + (Q + S)……………..(2.2)
Q
Q
S
P+Q
P
P
P+Q+S
P+Q+S
Q
A
S
Q+S
P
P+Q+S
A
Fig. (2-5)
10
A
S
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
Law of Sine
A
B
C
=
=
… … … … … … … … … … . . 2.3
sin α sin β sin γ
γ
A
β
Law of Cosine
B
α
C = A + B − 2AB cos γ … … … … … … … . . 2.4
C
2/4 Resultant of Several Concurrent Force
Consider a particle A acted upon by several coplanar forces Fig. (2-6a). Since the
forces consider here all pass through A, they are also said to be concurrent. The
vectors P, Q, and S representing the forces acting on A may be added by the polygon
rule Fig. (2-6b).
Q
P
P
S
S
R
A
Q
A
a
b
Fig. (2-6)
2/5 Resolution of a force into components
We have seen that two or more forces acting on a particle may be replaced by a single
force which has the same effect on the particle. Conversely, A single force F acting
on a particle may be replaced by two or more forces which, together, have the same
effect on the particle. These forces are called the components of the original F, and
the process of the substituting them for F is called resolving the force F into
components Fig. (2-7).
11
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
P
F
P
P
F
F
A
A
Q
(a)
A
Q
Q
(b)
(c)
Fig. (2-7)
EXAMPLE 1. The two forces P and Q act on a bolt A. Determine their resultant.
SOLUTION:
1.Graphical solution
i.
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,
= 98 = 35°
= 98∡35°
ii.
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,
12
= 98 = 35°
Chapter two
A. Lecturer Saddam K. Kwais
= 98∡35°
2. Trigonometric solution
Apply the triangle rule:
From the Law of Cosines
R = P + Q − 2PQ cos B
= 40N + 60N − 24060 cos B
R = 97.73NAns.
From the Law of Sines
*+, .
=
*+, /
0
Q
sin A = sin B 1 2R3 = sin 155° 160N297.73N3
A = 15.04° ⟹ α = 20° + A ⟹ α = 35.04° Ans.
3. Alternative Trigonometric Solution.
CD = 60N sin 25° = 25.36N
We construct the right triangle BCD and compute:
BD = 60N cos 25° = 54.38N
25.36N
A = 15.04°
94.38N
25.36
R=
R = 97.73N
sin A
Then, using triangle ACD, we obtain
tan A =
α = 20° + A ⟹ α = 35.04°
R = 97.7N∡35.0°
13
Statics of Particles
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
EXAMPLE 2. A barge is pulled by two
tugboats. If the resultant of the forces
exerted by the tugboats is 5000 lbf
directed along the axis of the barge,
determine
a) the tension in each of the ropes for α = 45o,
b) the value of a for which the tension in rope 2 is a minimum.
SOLUTION
(a) tension for α=45◦
1. Graphical solution
Parallelogram Rule with known resultant direction and magnitude, known directions
for sides
T1 = 3700 lbf
T2 = 2600 lbf
2. Trigonometric solution
Triangle Rule with Law of Sines
T1
T2
5000 lbf
=
=
sin 45° sin 30° sin 105°
T1 = 3660 lbf
T2 = 2590 lbf
(b) The angle for minimum tension in rope 2 is determined by applying the Triangle
Rule and observing the effect of variations in α.
Note : The minimum tension in rope 2 occurs when T1 and T2 are perpendicular.
14
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
T2 = (5000 lbf ) sin 30° T2 = 2500 lbf
T1 = (5000 lbf ) cos 30° T1 = 4330 lbf
α = 90° − 30° α = 60°
2/6 Rectangular Components of a Force
In many problems it will be found desirable to resolve a force into two components
which are perpendicular to each other. In Fig. (2-8), the force F has been resolve into
component Fx along x-axis and a component Fy along y-axis. The parallelogram
drawn to obtain the two components is a rectangle, and Fx and Fy are called
y
rectangular components.
F
Fy
x
θ
Fx
0
Fig. (2-9)
Fig. (2-8)
15
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
Note : The axes may be chosen any two perpendicular axes as shown in Fig. (2-9).
9: = 9 ;<= > … … … . . 2.59? = 9 sin > … … … 2.6
9
tan > = @ ?A9 B … . 2.8
:
9 = 9: + 9? … … … 2.7
• May resolve a force vector into perpendicular
components so that the resulting parallelogram is
DEF andC
DEH are referred to as rectangular
a rectangle. C
vector components and
IE = IEJ + IEK … … … … … … … … … … 2.9
Fig. (2-10)
• Define perpendicular unit vector LEandME
which are parallel to the x and y axes (Fig. 2-10).
• Vector components may be expressed as products
of the unit vectors with the scalar magnitudes of
the vector components (Fig. 2-11).
DE = FOEı + FO ȷDE … … … … … … … … … . 2.10
F
Fx and Fy are referred to as the scalar components of 9E
EXAMPLE 1. A force of 800 N is exerted on
a bolt A as shown in Figure . Determine
the horizontal and vertical components of the force.
SOLUTION : It is seen from Fig. b that
9R = −9;<= = −800;<=35° = −655
16
Fig. (2-11)
9S = −9=TU = +800=TU35° = +459
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
CR = −655VCS = +459W
The vector components of F are thus
DCE = −655V + 459W
and we may write F in the form
EXAMPLE 2. A man pulls with a force of 300 N
on a rope attached to a building, as shown in Fig. a .
What are the horizontal and vertical components of
the force exerted by the rope at point A ?
9: = +300;<=9? = −300=TU
SOLUTION : It is seen from Fig. b that
8X
8X
4
6X
6X
3
=
= sin =
=
=
YZ 10X 5
YZ 10X 5
We thus obtain
Observing that AB = 10 m, we find from Fig. a
cos =
4
9: = +300 = +240
5
3
9? = −300 = −180
5
and write
I = 240N[ − 180N\
When a force F is defined by its rectangular components Fx and Fy (see Fig. 2.8 ), the
9?
9:
angle θ defining its direction can be obtained by writing
tan > =
The magnitude F of the force can be obtained by applying the
Pythagorean theorem and writing
9 = ]9: + 9?
17
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
EXAMPLE 3. A force F = (700 lb) i + (1500 lb) j is
applied to a bolt A. Determine the magnitude of the
force and the angle θ it forms with the horizontal.
SOLUTION: First we draw a diagram showing the two
rectangular components of the force and the angle θ
( see Fig. a ). we write
tan > =
9? 1500Ib
=
9:
700Ib
(a)
Using a calculator, we enter 1500 lb and divide by 700 lb; computing the arc tangent
of the quotient, we obtain θ = 65.0°.
9=
`a
*+, b
=
cdeefg
*+, hd.e°
= 1655Ib
2/7 Addition of Forces by Summing x and y Components
• Wish to find the resultant of 3 or more concurrent forces (Fig. 2.12a),
R = P + Q + S…………………………..(2.11)
• Resolving each force into its rectangular components,
we write (Fig. 2.12b).
r
r
r
r
r
r
r
r
R x i + R y j = Px i + Py j + Q x i + Q y j + S x i + S y j
r
r
= (Px + Q x + S x )i + (Py + Q y + S y ) j...............( 2.12)
(a)
• The scalar components of the resultant are equal
to the sum of the corresponding scalar components
of the given forces.
R x = Px + Q x + S x
= ∑ Fx ......(2.13)
R y = Py + Q y + S y
(b)
= ∑ Fy .........( 2.14)
To find the resultant magnitude and direction (Fig. 2-12c),
R = R + R .........( 2.15)
2
x
2
y
θ = tan
−1
Ry
Rx
(c)
...........( 2.16)
Fig .(2.12)
18
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
EXAMPLE 1. Four forces act on bolt A as shown.
Determine the resultant of the force on the bolt.
SOLUTION:
• Resolve each force into rectangular components.
force mag
r
F1 150
r
F2
80
r
F3 110
r
F4 100
x − comp
y − comp
+ 129.9
− 27.4
+ 75.0
+ 75.2
0
− 110.0
+ 96.6
− 25.9
• Determine the components of the resultant by
adding the corresponding force components.
• Calculate the magnitude and direction of the
resultant.
R = 199.12 + 14.32
tan α =
14.3 N
199.1N
R = 199.6N
α = 4.1°
2/8 Equilibrium Of Particles
• 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.
19
Chapter two
A. Lecturer Saddam K. Kwais
• Particle acted upon by two
forces:
- equal magnitude
- same line of action
- opposite sense
Free-Body Diagrams
Statics of Particles
• Particle acted upon by three or more
forces:
- graphical solution yields a closed
polygon
- algebraic solution
r
r
R = ∑F = 0
∑F
x
=0
∑F
y
=0
Fig. (2-13)
Space Diagram: A sketch showing the
Free-Body Diagram: A sketch
showing only the forces on the selected
particle.
physical conditions of the problem.
EXAMPLE 1. In a ship-unloading operation,
a 3500-lb automobile is supported by a cable.
A rope is tied to the cable and pulled to center
the automobile over its intended position.
20
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
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.
SOLUTION:
• Construct a free-body diagram for the particle at A.
• Apply the conditions for equilibrium.
• Solve for the unknown force magnitudes.
TAB
T
3500 lb
= AC =
sin 120° sin 2° sin 58°
TAB = 3570 lb
TAC = 144 lb
EXAMPLE 2. It is desired to determine the drag force
at a given speed on a prototype sailboat hull. A model
is placed in a test channel and three cables are used to
align its bow on the channel centerline. For a given
speed, the tension is 40 lb in cable AB and 60 lb in cable AE.
Determine the drag force exerted on the hull and the tension in cable AC.
SOLUTION:
• Choosing the hull as the free body, draw a free-body diagram.
• Express the condition for equilibrium for the hull by writing that the sum of all
forces must be zero.
• Resolve the vector equilibrium equation into two component equations. Solve for
the two unknown cable tensions.
21
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
SOLUTION:
• Choosing the hull as the free body,
draw a free-body diagram.
7 ft
= 1.75
4 ft
α = 60.25°
tan α =
1.5 ft
= 0.375
4 ft
β = 20.56°
tan β =
• Express the condition for equilibrium for the hull
by writing that the sum of all forces must be zero.
r r
r
r
r
R = TAB + TAC + TAE + FD = 0
• Resolve the vector equilibrium equation into
two component equations. Solve for the two
unknown cable tensions.
r
r
r
T AB = −(40 lb)sin 60.26° i + (40 lb )cos 60.26° j
r
r
= −(34.73 lb )i + (19.84 lb ) j
r
r
r
T AC = T AC sin 20.56° i + T AC cos 20.56° j
r
r
= 0.3512T AC i + 0.9363T AC j
r
r
T AE = −(60 lb)i
r
r
FD = FD i
r
R=0
r
= (− 34.73 + 0.3512T AC + FD ) i
r
+ (19.84 + 0.9363T AC − 60) j
This equation is satisfied only if each component of the resultant is equal to zero
(∑ Fx = 0)
(∑ Fy = 0)
0 = −34.73 + 0.3512 T AC + FD
0 = 19.84 + 0.9363T AC − 60
T AC = +42.9 lb
Ans.
FD = +19.66 lb
Ans.
22
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
Forces In Space (3D) : 2/9 Rectangular Components Of A force In Space
In this section, we shall discuss problems involving the three dimensions of space.
r
Resolve F into horizontal
Resolve Fh into
contained in the
and vertical components.
rectangular Components
plane OBAC.
Fy = F cos θ y
Fx = Fh cos φ
r
The vector F is
Fh = F sin θ y
= F sin θy cos φ
F=
]FO
+
Fi
+
Fj … … … … … … … . . 2.17
r
• With the angles between F and the axes,
Fx = F cos θ x
Fy = F cos θ y
Fz = F cos θ z ....................(2.18)
• the unit vectors i , j , and k , directed respectively
along the x , y , and z axes ( Fig. 2.14 ), we can express
F in the form
23
Fy = Fh sin φ
= F sin θy sin φ
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
r
r
r
r
F = Fx i + Fy j + Fz k.....................................................(2.19)
Fig. (2-14)
Substituting into (2.19) the expressions obtained for Fx , Fy , Fz into (2.18), we get
(
)
r
r
r
F = F cos θ x i + cos θ y j + cos θ z k .......... .( 2.20 )
r
F = Fλ.......... .......... .......... .......... .......... ..( 2.21)
r
r
r
r
λ = cos θ x i + cos θ y j + cos θ z k.......... .....( 2 .22 )
Clearly, the vector λ is a vector whose magnitude
is equal to 1 and whose direction is the same as that
Fig. (2-15)
of F ( Fig. 2-15)
r
r
λ is a unit vector along the line of action of F and cos θx , cos θ y , and cos θz are the
r
direction cosines for F .
The components of the unit vector λ are respectively equal to the direction cosines
of the line of action of F :
λO = cos θO λi = cos θi λj = cos θj … … … … … … … 2.23
the sum of the squares of the components of a vector is equal to the square of its
magnitude, we write
λO + λi + λj = 1
or, substituting for λx , λy , λz from (2.23),
cos θO + cos θi + cos θj = 1 … … … … … … … … … … … … … … … … … … 2.24
EXAMPLE 1. A force of 500 N forms angles of 60°, 45°, and 120°, respectively,
with the x , y , and z axes. Find the components Fx , Fy , and Fz of the force.
SOLUTION : Substituting F = 500 N, θx = 60°, θy = 45°, θz = 120° into formulas
Fx = F cos θx
Fy = F cos θ y
FO = 500 cos 60° = 250N
Fz = F cos θz
24
FO = 500 cos 45° = 354N
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
FO = 500 cos 120° = −250N
2/10 Force Defined By Its Magnitude And Two Points On Its Line Of Action :
In many applications, the direction of a force F is defined by the coordinates of two
DDDDDDDn، joining M and N and the same sense as F.
Consider the vector m
points, M ( x1 , y1 , z1 ) and N ( x2 , y2 , z2 ), located on its line of action ( Fig. 2.16 ).
Fig. (2-16)
Denoting its scalar components by dx , dy , dz , respectively, we write
DDDDDDDn = o: [ + o? \ + op q … … … … … … … … … … … … … … … … … … … … … . 2.25
m
DDDDDDDn by its magnitude MN. Substituting for DDDDDDDn
m، from
obtained by dividing the vector m
The unit vector λ along the line of action of F (i.e., along the line MN ) may be
DDDDDDDn 1
m
r=
= so [ + o? \ + op qt … … … … … … … … … … … … … … … … … . . 2.26
m o :
Recalling that F is equal to the product of F and λ, we have
(2.25) and observing that MN is equal to the distance d from M to N , we write
9
so [ + o? \ + op qt … … … … … … … … … … … … … … … … … . . 2.27
o :
from which it follows that the scalar components of F are, respectively,
I = 9r =
9: =
9o?
9o:
9op
9? =
9p =
… … … … … … … … … … … … … . . 2.28
o
o
o
Subtracting the coordinates of M from those of N , we first determine the components
of the vector MN، and the distance d from M to N :
o: = R − Rc o? = S − Sc op = u − uc
25
Chapter two
o = ]o: + o? + op
A. Lecturer Saddam K. Kwais
Statics of Particles
Substituting for F and for dx , dy , dz , and d into the relations (2.28), we obtain the
components Fx , Fy , Fz of the force.
The angles θx , θy , θz that F forms with the coordinate axes can then be obtained from
o?
o:
op
cos >? = cos >p = … … … … … … … … … … . 2.29
o
o
o
Eqs. (2.18). Comparing Eqs. (2.22) and (2.26), we can also write
cos >: =
DDDDDDDn.
the vector m
and determine the angles θx , θy , θz directly from the components and magnitude of
2/11 Addition Of Concurrent Forces In Space :
The resultant R of two or more forces in space will be determined by summing their
rectangular components. Graphical or trigonometric methods are generally not
practical in the case of forces in space.
The method followed here is similar to that used in Sec. 2.8. with coplanar forces.
Setting
R = ∑F
we resolve each force into its rectangular components and write
: [ + ? \ + p q = ∑s9: [ + 9? \ + 9p qt
= ∑9: [ + s∑9? t\ + ∑9p q
: = ∑9: ? = ∑9? p = ∑9p … … … … … … … … … … … … … . 2.30
from which it follows that
The magnitude of the resultant and the angles θx , θy , θz that the resultant forms with
the coordinate axes are obtained using the method discussed in Sec. 2.10. We write
= ]: + ? + p … … … … … … … … … … … … … … … … … … … … … … . 2.31
cos >: =
?
:
p
cos >? = cos >p = … … … … … … … … … … 2.32
26
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
EXAMPLE 1. A tower guy wire is anchored by means
of a bolt at A. 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 θx, θy, θz defining the direction of the force
SOLUTION:
1. Determine the unit vector pointing from
A towards B.
The component of the point B(0,80,0)
The component of the point A(40,0,-30)
Denoting by i , j , k the unit vectors along the
coordinate axes, we have
DDDDDn
AB = wx/ − x- [ + y/ − y- \ + z/ − z- q{
DDDDDn = w0 − 40[ + 80 − 0\ + 0 + 30q{
AB
r
r
r
AB = (− 40 m ) i + (80 m ) j + (30 m )k
The total distance from A to B is
AB = d =
AB =
d +d +d
2
2
2
x
y
z
(− 40 m )
2
+ (80 m ) + (30 m ) = 94.3m
2
2
DDDDDn/AB, we write
Introducing the unit vector r = AB
r
r
r
r  − 40  r  80  r  30 
λ=
i
+
j
+
k
=
−
0
.
424
i
+
0
.
848
j
+
0
.
318
k


 

 94.3 
 94.3   94.3 
2. Apply the unit vector to determine the components of the force acting on A.
r
r
F = Fλ
(
r
r
r
= (2500 N ) − 0.424 i + 0.848 j + 0.318k
r
r
r
= (− 1060 N )i + (2120 N ) j + (795 N )k
Fx = - 1060 N
Fy = + 2120 N
)
Fz = + 795 N
27
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
3. Noting that the components of the unit vector
are the direction cosines for the vector,
calculate the corresponding angles
r
r
r
r
λ = cos θx i + cos θy j + cos θz k
r
r
r
= −0.424 i + 0.848 j + 0.318k
θx = 115.1o
θ y = 32.0o
θz = 71.5o
EXAMPLE 2. A wall section of precast concrete
C
is temporarily held by the cables shown. Knowing
27 ft
8 ft
that the tension is 840 lb in cable AB and 1200 lb
D
in cable AC , determine the magnitude and
B
11 ft
direction of the resultant of the forces exerted
16 ft
by cables AB and AC on stake A.
SOLUTION:
The component of the point B(0,8,0)
The component of the point C(0,8,-27)
The component of the point A(16,0,-11)
DDDDDn
AB = wx/ − x- [ + y/ − y- \ + z/ − z- q{
DDDDDn
AB = w0 − 16[ + 8 − 0\ + 0 + 27q{
r
r
r
AB = (− 16 ft ) i + (8 ft ) j + (11 ft )k
The total distance from A to B is
AB = d =
AB =
d +d +d
2
2
2
x
y
z
(− 16 ft )
2
+ (8 ft ) + (11 ft ) = 21ft
2
2
DDDDDn
AC = wx} − x- [ + y} − y- \ + z} − z- q{
28
A
DDDDDn
AC = w0 − 16[ + 8 − 0\ + −27 + 11q{
r
r
r
AC = (− 16 ft ) i + (8 ft ) j − (16 ft )k
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
The total distance from A to C is
AC = d =
AC =
d
2
x
2
(− 16 ft )
2
Force
Cable AB
Cable AC
9: =
+ d y + dz
2
+ (8 ft ) + (− 16 ft ) = 24ft
2
2
Distance
Component
(ft)
dx
dy
dz
-16
8
11
-16
8
-16
d
(ft)
21
24
Force Component
(Ib)
Fx
-640
-800
Rx=-1440
Fy
320
400
Ry=720
Fz
440
-800
Rz=-360
F
(Ib)
840
1200
9o?
9o:
9op
9? =
9p =
o
o
o
For Cable AB
9: =
840 × −16
840 × 8
840 × 11
= −640.9? =
= 320.9p =
= 440
21
21
21
For Cable AC
9: =
1200 × −16
1200 × 8
1200 × −16
= 800.9? =
= 400.9p =
= −800
24
24
24
R = ]RO + Ri + Rj = −1440 + 720 + −360 = 1650Ib
cos >: =
: −1440
−1440
=
⟹ >: = cos c €
 ⟹ >: = 150.8°
1650
650
cos >p =
p −360
−360
=
⟹ >p = cos c €
 ⟹ >p = 102.6°
1650
650
From the equilibrium
cos >? =
?
720
720
=
⟹ >? = cos c €
 ⟹ >? = 64.1°
1650
650
29
Chapter two
A. Lecturer Saddam K. Kwais
Statics of Particles
2/12 Equilibrium Of A particle In Space
When the resultant of all forces acting on a particle is zero, the particle is in
∑9: = 0∑9? = 0∑9p = 0 … … … … … … … … … … … … … … … . . 2.30
equilibrium.
EXAMPLE 1. 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 A of
P and the tension in each cable.
SOLUTION:
Point A is chosen as a free body
Draw a free-body diagram
TAC
TAB
A
P
W
W = 1200 kg * 9.81 m/s2 = 1962 N
B(0,12,8)
C(0,12,-10)
A(1.2,2,0)
DDDDDn = w0 − 1.2[ + 12 − 2\ + 8 − 0q{ = −1.2[ + 10\ + 8q
AB
DDDDDn
AC = w0 − 1.2[ + 12 − 2\ + −10 − 0q{ = −1.2[ + 10\ − 10q
30
Chapter two
Force
TAB
TAC
P
W
9: =
Distance
Component
(m)
dx dy dz
-1.2 10
8
-1.2 10 -10
A. Lecturer Saddam K. Kwais
d
(m)
12.86
14.19
Force Component
(N)
Fx
-0.0933TAB
-0.0846TAC
+P
0
9o?
9o:
9op
9? =
9p =
o
o
o
For Cable AB
9: =
9? =
9p =
Statics of Particles
Fy
0.778TAB
0.705TAC
0
-1962
F (N)
Fz
0.622TAB
0.705TAC
0
0
TAB
TAC
P
-1962
For Cable AC
‚ƒ„ × −1.2
= −0.0933‚ƒ„
12.86
‚ƒ„ × 10
= 0.778‚ƒ„
12.86
9: =
‚ƒ… × −1.2
= −0.0846‚ƒ
14.19
9p =
‚ƒ… × −10
= −0.705‚ƒ
14.19
9? =
‚ƒ„ × 8
= 0.622‚ƒ„
12.86
‚ƒ… × 10
= 0.705‚ƒ
14.19
∑9: = 0 ⟹ −0.0933‚ƒ„ − 0.0846‚ƒ… + † = 0 … … … … … … … . 1
∑9? = 0 ⟹ 0.778‚ƒ„ 0.705‚ƒ… + 0 − 1962 = 0 … … … … … … … 2
∑9p = 0 ⟹ 0.622‚ƒ„ − 0.705‚ƒ… + 0 + 0 = 0 … … … … … … … . 3
Solving these equations, we obtain
† = 235‚ƒ„ = 1402‚ƒ… = 1238YU=.
31