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Force Systems and Resultants
2.1. IntroductIon
The study of bodies at rest or in equilibrium under the action of forces requires the concept of force, moment and couple.
2.2. Force
Force is a vector quantity hence it is described by magnitude and direction. In addition, the force has point of application.
The characteristics of a force can be stated as:
(i) Magnitude – F (ii) Direction (iii) Point of application
which have been shown in Fig. 2.1.
When we say the direction of a force, it implies the line of action along which the force acts and the sense indicated by the arrow. Let us consider a bar acted upon by forces F 1 and F 2 as shown in Fig. 2.2 and the bar is in equilibrium.
When we consider the equilibrium of a rigid body, the point of application force is not important as evident from Figs. 2.2(a) and (b).
Fig. 2.1 Force designation.
Point of application
Sense
Body
Line of action
F
Force Systems and Resultants 15
(ii) By constructing a triangle using the directions along which components are required, may be obtained as shown in Fig. 2.6.
F 2 F
F 1
F F 1
F
O
(a) (b)
O
α α
β
β
Fig. 2.6 Triangle of forces.
In this method, the line of action of F 2 is not preserved as shown in Fig. 2.6. (a) as well as F 1 which can be seen in Fig. 2.6 (b).
F 1 F 2 F sin β sin α (^) sin 180 α β
− ( + )
F = F 1 2 + F 2 2 − 2 F F (^) 1 2cos 180 − (^) ( α +β (^) ) (2.7)
Resolution of forces is the inverse operation of composition of forces. Resolution and composition of forces are two essential concepts which find frequent applications for the solution of mechanics problems.
2.4. MoMent oF a Force
Let us calculate the moment of a force F about an axis passing through the point ‘ A ’. The tendency of a moment is to cause rotation about the point ‘ A ’. The moment is sometimes referred to as torque.
x^1
y^1
F
O
β α
Fig. 2.4 Force components in specified directions.
x^1
y^1
F
O
F 2
F 1
β α
Fig. 2.5 Parallelogram of forces.
18 Engineering Mechanics
2.7. representatIon oF Force In three-dIMensIonal space
Force vector F in three-dimensional space is shown in Fig. 2.11 using right hand coordinate system.
The force F is written as
F = F x i + F y j + F z k (2.21)
where F x , F y , F z^ are rectangular components of the force F and i , j , k are unit vectors along x , y , z axes respectively.
α is the angle between force vector F and x -axis. β is the angle between force vector F and y -axis.
γ is the angle between force vector F and z -axis.
F F L
F F L
F F L
x
y
z
x l
y (^) m
z n
cos , cos
cos , cos
cos , cos
α α
β β
γ γ (2.22)
F = (^) ( F x^ ) + ( F y^ ) + ( F z )
2 2 2 (2.23)
If F is represented by length L to some scale and similarly F x^ by x , F y^ by y and F z^ by z and l , m , n are called direction cosines.
l^2 + m^2 + n^2 = 1 (2.24)
F F F F L
x y z
x y z
F
L
is known as force multiplier.
Force is also written as
F = F( l i + m j + n k ) (2.26)
Let n be the unit vector in the direction force
n (^) F = l i + m j + n k (2.27)
F = F n (^) F (2.28)
O x
y
z
F z
F^ F
y β γ α
F x
Fig. 2.11 Three-dimensional force.
Force Systems and Resultants 19
Resultant force F given the rectangular components F x , F y , F z :
(i) Let us take F y^ and F z^ and find the resultant of these components using parallelogram law in YOZ plane. (Fig. 2.12) (ii) Adding F x^ component to R vectorially the resultant force F is obtained (Fig. 2.13).
F y
Y
O
Z
R
F z
Fig. 2.12 Resultant in YOZ plane.
O X
Y
Z
F z
F y
F x
R F
Fig. 2.13 Resultant in three-dimensional space.
2.8. resultant oF coplanar concurrent Forces
R = F 1 + F 2 + F 3 + F 4 (2.29)
F 2
F 3
F 4
F 1
Fig. 2.14 Coplanar concurrent forces.
F 3
F 4 F 2
F
R
O
Fig. 2.15 Polygon of forces.
Resultant of forces with be obtained by adding the vectors graphically and the closing side of the polygon will be the resultant R.
2.9. resultant oF non-concurrent coplanar Forces
In general a system of forces acting in a plane on a body may be non-concurrent. The system of forces acting on the body must be reduced to a simple force system which does not alter the external effect of the original force system on the body. The equivalent of the forces acting on the body is the resultant force F acting at an arbitrary point and a couple. The choice of this arbitrary point will depend upon the convenience for the particular problem under consideration.
Force Systems and Resultants 21
M x^ = y F z^ − z F y M y^ = z F x^ − x F z M z^ = x F y^ − y F x M = M x i + M y j + M z k (2.32)
Magnitude M = h F 2 = (^) ( M x^ ) + (M y^ ) + ( M z )
2 2 2 where h is the perpendicular distance between the
forces and produces a counter-clockwise moment.
The following observations emphasize important characteristics of couple.
(i) The magnitude of moment is independent of the choice of the centre about which moments are taken. (ii) The moment vector M is independent of any particular origin and is thus considered a free vector. (iii) The forces of a couple may be rotated within their plane provided their magnitudes and distance between their lines of action are kept constant.
It can be shown that the relative positive vector R can be chosen connecting any two points on the line of action.
Resolving R into two components, R 1 perpendicular to the line of action of the forces and R 2 paral- lel to the line of action of F.
F 1 = − F 2 (2.33)
M = R × F = ( R 1 + R 2 ) × F 2
= R 1 × F 2 + R 2 × F 2 (2.34)
R 2 × F 2 = 0 since R 2 and F 2 are collinear vectors.
M = R 1 × F 2 (2.35)
2.11. equIvalent Force systeMs
In the study of mechanics, we are concerned many a time with equivalent force systems acting on a rigid body. The following considerations will form the basis for force equivalences for rigid bodies.
(i) The sum of a system of concurrent forces is a single force that is equivalent to the original system. (ii) A force may be moved along its line of action, i.e., forces are transmissible vectors.
F 1
F
A R B R (^1) R 2
Fig. 2.18 Position vector and forces forming couple.
22 Engineering Mechanics
(iii) The effect of a couple on a rigid body is to create a moment which is described by a free vector.
Translation of a force to a parallel position:
Let us consider a force acting in a plane and try to move it to a parallel position at point A as shown in Fig. 2.19 without changing the effect of force F on the rigid body.
This can be achieved by applying parallel forces F and −F passing through A. The net effect of the above operation leads to a force F and a moment M caused by couple whose magnitude is F d and sense is counter-clockwise. The moment vector M is perpendicular to the plane of the couple.
F d A
F
O (^) F F d
O A
Fig. 2.19 Translation of force.
M
M
F
F
F F
O A^ F
O A
Fig. 2.20 Force and couple.
It can also be inferred that a force F and couple moment M (Fig. 2.20) can be replaced by a single force F (Fig. 2.19).
2.12. the Wrench
If the force and the couple acting on the body are not coplanar, these can be replaced by an equiva- lent system which consist of a force and a couple vector lying along same line of action. This equiv- alent system consisting of the force F and a collinear couple C (^) F is known as wrench.
C
F A (^) B
F C^ F A
Fig. 2.21 Wrench.
When the moment representation of the couple and the force are in the same direction, the wrench is positive and otherwise negative if they are opposite in direction.
The following calculations are being suggested to calculate the equivalent system called wrench.
(i) Calculate the unit vector nF in the direction of the force. (ii) Calculate the dot product of couple vector C and unit vector n (^) F , i.e., ( C ⋅ n (^) F )
24 Engineering Mechanics
The location of the equivalent concentrated force F is
x
x
p L F
p L p L
L
L
A A A
2 2
3 3
Determination of the resultant force and its location involves the use of integral. Similar calculations are carried out in the case of distributed forces over areas and volumes.
summary In this chapter, basic knowledge of mechanics which includes understanding of force, moment and couple has been discussed. Determination of resultants is an essential element in the solution of mechanics problems. Resultant of a force system is a force acting at an arbitrary point and a couple. In addition, mathematical treatment of concentrated and distributed forces acting on a body has been presented.
ExamplEs
EXAMPLE 2.
Resolve the force 100 N into rectangular components
6
8 θ θ
8
6
100 N 100 N
10
Fig. 2.
Force is specified by the slope with vertical distance 8 and horizontal 6. The rectangular components are
F F N
F Fsin N
x
y
= = × =
= = × =
cos θ
θ
x
F
O A
Fig. 2.23 Resultant force.
F y
y
F
θ O F x x
Force Systems and Resultants 25
EXAMPLE 2.
Determine the rectangular components of the given force F.
Force is given in polar notation R = r θ =50 30°
R x^ = r cosθ = 50 cos(30º) = 43.3 N R y^ = r sinθ = 50 sin(30º) = 25
EXAMPLE 2.
Resolve the given force into components in the directions specified 1 and 2.
R 1 R 2 R sin 75 ° sin 75 ° sin 30 °
R 1 R 2 80 N
sin sin
Check:
R R R R R
R
R N
= ( ) − ( )
1
2 2
2
2 2
cos
.. cos
EXAMPLE 2.
Find the resultant of the forces R 1 and R 2.
(i) Graphical method:
R R
R R
N
R R R
R
R
1
2 2
2
1 2
1
sin sin sin
sin
β α
β
β 53 13. °
R^ = 50 N30º
Fig. 2.
2
R^ = 80 N
1
R 75° 75°
45°
30°
15°
30° R 1
R 2
Fig. 2.
60 N y 80 N R 2 O
30° R 1
30°
Fig. 2.
R y
y
50 N
O R x^ x
30 °
R 1 = 80
R
30°
α
β (^) R 2 = 60
Force Systems and Resultants 27
(i) Resolution forces along x and y axes:
F N F N
F N
F N
F
1
2
2
2
3
x y
x
y
x
= × = ←
= × = ↑
3
4
4
× = ←
= × = ↓
×
×
N
F N
F N
F
y
x
y
== 240 N↓
R x^ = 259.8 − 320 − 75 + 100 = −35.2 N (←)
R y^ = 150 + 240 − 130 − 240 = 20 N ↑
R = (^) ( −35 2.) 2 + ( 20 )^2
tan . .
θ
θ
R
R
y x
(ii) Graphical approach:
R = F 1 + F 2 + F 3 + F 4
Closing vector R of polygon results in the resultant R.
O
30°
y
F1 y^ = 150 N^ 300 N
F 1 x^ = 259.8 N
F 2 y
F 2 x F3 x
F4 x
F 4 y
F3 y
4 3
60°
5
F 2 = 400 N
F 3 =
F 4 = 260
13 12
5
R x
R
R y θ
R
O
F 1 θ
F 2
F 4
F 3
28 Engineering Mechanics
EXAMPLE 2.
Resolve the vertical force and horizontal force acting on a body supported by an inclined plane along x and y -axes shown in Fig. 2.29.
Resolution of horizontal force F:
F x^ = F cosθ ( )
F y^ = + F sinθ ( ) F = F x i − F y j (^) O^ x
y θ
θ
θ
θ F F
W
F y
F x
Fig. 2.
Resolution of vertical for W: W x^ = + W sinθ ( )
W y^ = + W cosθ ( ) W = −W x i − W y j
The resolution of vertical and horizontal forces on an inclined plane will provide force components useful in many applications.
θ
F y
(^) = Fsin
θ
F x
(^) = Fcos
θ^
Wcos
θ^ = W
y
Wsin
θ^ = W
x
EXAMPLE 2.
A block is acted upon by its weight W = 500 N and a horizontal force F = 750 N and force exerted by the inclined plane is P. Determine the resultant R of these forces which is parallel to the incline and the magnitude of P. Let us define the coordinate axes x and y parallel and perpendicular to the inclined plane respectively.
(i) Resolution of force F parallel and perpendicular to the plane F x^ = +F cos45° = + 750 × 0.707 = +530.25 N (→)
F y^ = +F sin45° = + 750 × 0.707 = +530.25 N (↓) F = F x i – F y j = 530.25 i – 530.25 j
θ θ
W
W x W y
θ
O
x
y (^) 45° 60°
F = 750 N
W = 500 N
P
Fig. 2.
F y^ F x
F 45° (^) 45°
32 Engineering Mechanics
EXAMPLE 2.
Compute the moment of 1500 N force shown in Fig. 2.34 about points A and B
4 m
A
60°
Pcos 30° Psin 30°
60°
30° 5 m
B 5 m
P =1500 N C
X′, i′ A (^) B
Y′, j′
4 m
Fig. 2.
(i) (a) Resolving the force P along BC and perpendicular BC
Moment of force P about B
M (^) BP^ = ( ) (^5) ( P sin 30 ° (^) ) = ( )( (^5 750) ) = 3750 N m⋅ (clockwise moment)
(b) Vector method:
Position Vector r (^) BC = 5 j ′ in x′cy′ coordinate system Force Vector P = P sin30 i ′ + P cos30° j ′
M (^) BP^ r BC P P P
= × =
i ′ j ′ k ′ 0 5 0 sin 30 ° cos 30 ° 0
M BP^ =^ −5(Psin30°) k ′^ =^ +3750(− k ′) N·m (− k ′) indicates clockwise moment.
M (^) BP^ =^ 3750 N·m.
(ii) (a) Resolving the force P parallel to AB and perpendicular AB.
Moment of force P about A
M (^) AP^ = (6.5)P sin 30° − (4.33)P cos 30°
M (^) AP^ = (6.5)(750) − (4.33)
M (^) AP^ = −749.67 N·m
M (^) AP^ = 749.67 N·m (clockwise moment)
A B
60°
C (^) Pcos 30°
Psin 30°
x , i
y , j
4 m 2.5m
4.33 m
Force Systems and Resultants 35
EXAMPLE 2.
What are the force components of 1000 N shown in Fig. 2.37. What are the direction cosines associ- ated with the 1000 N force.
Length of force Vector L
L L m
= ( ) + ( ) + ( )
2 2 2
.
Direction cosines:
l
x
m
y
n z
L
L
L
Fx^ = F l = −(0.19) (1000) = −190 N
F y^ = F m = −(0.31) (1000) = −310 N F z^ = F n = −(0.93) (1000) = −930 N.
EXAMPLE 2.
Determine the resultant of the system of concurrent forces having the following magnitudes passing through the origin and the indicated points: P = 300 N (+12, +6, −4), T = 500 N (−3, −4, +12) F = 250 N (+6, −3, −6)
Direction cosines: l
x m
y n
z = = = L L L
Force (N)
Coordinates L
Direction cosines x y z l m n
P = 300 12 + 6 − 4 14 0.857^ 0.43^ −0.
T = 500 − 3 − 4 12 13 −0.23 −0.3 0.
F = 250 6 − 3 − 6 9 0.66^ −0.33 −0.
Force component in x -direction = Force × l y- direction = Force × m z- direction = Force × n
O y
z
15 m
x
3 m 5 m
1000 N
Fig. 2.
Force Systems and Resultants 37
Length of force vector L:
AC
AB
A
= (^) ( − − ) + (^) ( − ) + (^) ( − ) =
= ( − − ) + ( − ) + −( − ) =
2 2 2
2 2 2
DD = (^) ( 6 − (^0) ) 2 + (^) ( 0 − (^12) ) 2 + (^) ( 4 − (^0) ) 2 = 14
Force (N)
Components of distance L
F m (N/m)
Force components x y z F x F y F z P = 280 − 4 − 12 6 14 20 − 80 −240 120 Q = 260 − 4 − 12 − 3 13 20 −80 −240 − R = 210 6 − 12 − 4 14 15 90 −180 60 ToTAL (^) − 70 − 660 120
Resultant force = 10 ( )^7 2 + (^66 )^2 + (^12 )^2 =^ 674 45.^ N
EXAMPLE 2.
A vertical boom AE is supported by guy wires from A to B, C, D. If the tensile load in AD = 252 N, find the forces in AC and AB so that the resultant force on A will be vertical.
Coordinates of points.
Point x y z A 0 12 0 B 0 0 − 9 C − 4 0 3 D 6 0 4 E 0 0 0
C
z
E
12 m
3 m
B
D
A
y
x
4 m 9 m
4 m
6 m
Fig. 2.
Force Systems and Resultants 39
Components of force P : along x -direction P x = P m X = 10(−4) = −40 N
along y -direction P y = P m Y = 10(8) = 80 N
along z -direction P z = P m Z = 10(6) = 60 N P = − 40 i + 80 j + 60 k
Position vector r CB = 4 i + 5 j
(i) M (^) CP^ = r CB × P = i j k −
i j k 4 5 0 40 80 60
Moment of force P about C = M (^) CP= 300 i − 240 j − 520 k
(ii) Position vector r DB = 4 i − 5 j + 3 k
Moment of force P about D = M (^) DP^ = r DB × P = − −
i j k 4 5 3 40 80 60
M (^) DP^ = − 540 i − 360 j + 120 k
(iii) Position vector r CD = 10 j − 3 k
Unit vector along CD = CD = n CD = 10 j^ −^3 k 10 44.
Moment about a line directed from C to D of P = M CP^ ⋅ n CD = − 379 3. N m⋅
EXAMPLE 2.
Replace the force and couple shown in Fig. 2.41 by a wrench passing through E in xz plane.
The direction cosines of force vector are
l
x
m y
n
z
L
L
L
where L is length of AB , L= (^) ( 0 − (^4) ) 2 + (^) ( 4 − (^0) ) 2 + (^) ( 0 − (^2) ) 2 = 6
O x
y
z
2
C
C = 72 j + 36 k
B (0, 4, 0)
F = 30N 4
A (4, 0, 2)
4
Fig. 2.
