Resultant and Equivalent
of Force Systems
A force on a body produces translation while a couple produces rotation. The forces
acting on a body may be in the same plane (coplanar) or in different planes (non-coplanar) or
in space ; see plate 1 and plate 2.
The coplanar forces may be parallel (like or unlike) or concurrent. If a number of forces
are acting on a body, Fig. 1.1, to find their resultant R (in magnitude, direction ‘θ’ with
horizontal and line of action), find first their algebraic sum of horizontal and vertical
components i.e., ΣX and ΣY, then
y2
F
2
F3
x2
R
x3
ΣY
Rigid
body
R
θ
ΣX
O
y
a1
a3
F1
y3
a
a2
a4
x
A
y1
x1
y4
F4
x4
Fig. 1.1
R=
( ΣX ) 2 + ( ΣY ) 2 ,
tan θ =
ΣY
ΣX
...(1.1)
To find the line of action of R (at ‘a’ w.r.t. a point A or a line), find the algebraic sum of
the moments of all the forces about A, then
ΣMA = ΣF1 . a1 = R . a ;
a=
ΣM A
R
...(1.2)
The force systems are illustrated in A, B & C (graphical methods) and D in space, as
follows:
3
4
Engineering Mechanics
20 N
A B
N = 15 N
40 N
C
D E
c
a
R
0
a
30 N E
0
c
0
0
b
15 N
0
e
d
e
b
d
Force or vector
diagram
Space diagram
(ii) Unlike II forces
kN
20
=
Q L – irons
Gusset
plate
(angle-irons)
Q
R
= 46 kN
20° L ir
s ons
P=3
0 kN
=
20
kN
R=
20° 46 kN
P=3
0 kN
(i) Parallelogram
law of forces
R = 46 kN
kN
20°
20
P=3
=
0 kN
Q
Resultant of
two forces
(ii) Triangle of forces
0
B
300
b N
a
C
Block A
c
0
a
N
110
R= aa¢
=
b
R
e
O
250 N
200 N
Aa 0
0
a
100
N
0 60°
e
E
30°
d
0 D
d
c
Force polygon
(will not close)
R = aa
200 N
No. of forces on Block A
=
67
kN
a
R
=
oe
e
4
a
25 kN
25 kN
25 kN
3
D
C
A
60° B
16 m
22 m
0
47°
0
45°
E
d
x
0
= 43 m 25 kN
c
0
0
a
b
Space diagram
O
pole
b
c
O
R = 67 kN
e
d
Vector or force diagram
See Example 1.3, Fig. 1.3 & 1.4, page 8 (single equivalent tug boat, R)
A . Resultant of Coplanar Non Concurrent Forces
5
180 – b
b
R
A
B
g
g
A
b
Block A
B
90°
= 2 60° W = 400 N
00
N
30° 60°
of forces
Spherical ball of
mass 40 kg
rests in a smooth Dr groove
B
C
R = 500 N
RB
RA
180 – g
180 – a
a
Lami’s Theorem:
R
W
C
a
W = RA = RB
Sin 90° Sin 60° Sin 30°
B
C
A
=
=
Sin a Sin b Sin g
Block A in equilibrium under three forces
A
2m
C
jib
10 kN
kN
B
D
E
180 N
b
a
C
of forces at A
160 N
A
150 N
27
B
30°
30°
B
18
10 kN
45°
(C)
kN
(t)
A
tie
30°
=3
Q = 300 N
A
P = 400 N
40
N
Resultant and Equivalent of Force Systems
E
R
200 N
R = 150 N
= aa
c
a
Jib crane
d
No. of concurrent forces
R = Resultant = aa
E = Equilibrant = aa
PLATE 1
B. Equilibrium Under Coplanar Concurrent Forces
6
Engineering Mechanics
f
I
1.1
NA
.7 k
1.
RA
1 kN
D 1.5 kN
a
0
c
0e
E
4m
RE = 3.8 kN
oc
los
ing
O
pole
c
Funicular
polygon
lin
e
b
4m 0
d
4
II t
RE
f, g, h,
i
a
d
e
b
Vector diagram or Force
polygon to find reactions
c
j
d
e
Superposition of force Maxwell or
polygons at each joint Force diagram
(i) Truss – Space diagram
(with Bow's notation)
[See Ex. 3.1 & Fig. 3.1, page 30]
RB = 8 kN
(c)
e
B
h
8
(C) = E
3.5
5.8
RA = 7 kN
A
G
(D)
3.5
H
I
1.7
6.3
3.5
J
1.1
RA
=1
1.7 0 H
0G
0
0
F 0
a
b
4m
4m
Cl
os
ing
lin
e
J
2.2
B
2.8
1 kN
1 kN
C
2.5
F
f
B
RA
E
j
a
3 kN
2 kN
R
i
g
(ii) Cantilever Truss – Space diagram
(with bow's notation)
Maxwell diagram
gives also reactions
in cantilever trusses
[See Ex. 3.3 & Fig. 3.14, page 34]
C. Equilibrium Under Coplanar Non Concurrent Forces
D
z
z
(x, y) of R = ?
40 kN
TA = ? T B = ?
TC = ?
O
R.C.C. mat
4m
20 kN
C
30 kN
15 kN
G
A
2mx
2m
y
2m
2m
B
W = 50 kN
(i) Concurrent forces
y
2m
3m
2.5
m
(ii) Parallel forces
m
x
2.5 Column
loads
Resultant and Equivalent of Force Systems
7
z
R – C at O = ?
z
B
50 kN (in x-y plane)
3m
E
m
3m
2m
f
1.5
A
D
x
1m
1m
20
1m 1m
20 kN
kN-m
60 kN
(in x-z plane)
(II to y-axis)
3m
m
1.5
30°
TE, TD = ?
RA = ?
Cables
C
20 kN crate
y
x
Forces for f = 0 :
Coplanar concurrent at C
Non Coplanar concurrent at B
y
(iv) Non concurrent,
Non parallel forces
(iii) Space frame-Derrick
D. Spatial Force System [Forces in three-dimensions]
PLATE 2
ΣMA = (ΣX)y – (ΣY)x
Also,
...(1.3)
This Eqn. can be solved for the ratio of y to x corresponding to any point ‘O’ along the
line of action of R as indicated by the dashed line, that is the force can be applied at any point
on its line of action. This is called the principle of tansmissibility of force.
From Eqs. (1.2) and (1.3), R . a = (ΣX )y – (ΣY ) x
...(1.4)
i.e., the moment of the resultant is equal to the algebraic sum of the moments of its components.
This is called ‘Varignon’s Theorem’.
Example 1.1. Determine the resultant of the forces in Fig. 1.1.
Fig. 1.1
→
Solution: + ΣX = 150 – 100 sin 45° = 79.29 N →
+ ↑ ΣY = 50 – 8 – 100 cos 45° = – 28.71 or 28.71 N ↓
R=
tan θ =
79.29 2 + 28.712 = 84.32 N
28.71
= 0.3621,
79.29
θ = 19.9°
8
Engineering Mechanics
Moment of the resultant (at a distance ‘a’ from C) = Moment of the
components (about C)
R . a = ΣMC
+
ΣMC = 150 × 1 + 100 × 1 – 50 × 1 – 8 × 1
= 192 N.m
a=
Fig. 1.1a
ΣMC
192
=
= 2.27 mm
R
84.32
Example 1.2. The resultant (100 kN) of four forces and three of these are shown in
Fig. 1.2. Determine the fourth force.
Fig. 1.2
Solution:
→
+ ΣX = – 70 cos 60° – 120 cos 60° + 50 cos 50°
= – 62.88
or 62.88 kN←
+ ↑ ΣY = 70 sin 60 – 120 sin 60° – 50 sin 50°
= – 81.6
or 81.6 kN ↓
RX = 100 cos 45° = 70.71 kN
RY = 100 sin 45° = 70.71 kN
F4
Y4
q
X4
X4 = 70.71 – (– 62.88) = 133.59 kN →
Y4 = 70.71 – (– 81.6) = 152.31 kN ↑
F4 =
tan θ =
Fig. 1.2a
133.59 2 + 152.312 = 202.6 kN, Fig. 1.2a
152.31
= 1.14, θ = 48.746°.
133.59
Example 1.3. (a) State ‘Varignon’s Theorem’.
(b) Four tugboats exert 25 kN each (as shown in Fig. 1.3) to bring an ocean liner to the
pier. Determine the point on the hull where a single, more powerful tugboat should push to
produce the same effect as the original four boats.
Resultant and Equivalent of Force Systems
9
Boat - 1
Boat - 2
Boat - 3
4
3
60°
16 m
22 m
36 m
63 m
30 m
66 m
Hull of ship
30 m
1
Boat - 4
1
Fig. 1.3
Solution: (a) The Varignon’s (French Mathematician) Theorem states that “if a number
of coplanar forces acting on a body, the algebraic sum of the moments of all the forces about
any point is equal to the moment of their resultant about the same points, i.e. ΣF1. a1 = R.a” or
simply ‘‘the moment of a force about any point equals the algebraic sum of the moments of its
components about the same point, i.e. F . a = X . x + Y . y, where F =
x 2 + y 2 ”.
(b) To find the resultant ‘R’ due to the four forces acting on the hull of the ship, Fig. 1.4.
21.65
F1
60°
12.5
16 m
O
25 kN
F2
5
4
3
15
X = 45.18
= 47.3
25 kN F3
Y
= 48.97
R = 66 kN
22 m
66 m
63 m
43 m
a
X
Y
x
R
30 m
30 m
22 m
17.68
25 kN 1
1
F4
45°
17.68
25 kN
20
Y
Fig. 1.4
→
+ ΣX = 12.5 + 15 + 17.68 = 45.18 kN →
+ ↑ ΣY = – 21.65 – 20 – 25 + 17.68
= – 48.97 or 48.97 kN ↓
R=
45.18 2 + 48.97 2 = 66.63 kN, Fig. 1.4(a)
X
Fig. 1.4(a)
10
Engineering Mechanics
θ = tan–1
48.97
= 47.3°.
45.18
Let the line of action of R cut the X-axis at x. Taking moments about ‘O’ (and using
Varignon’s theorem),
ΣM0 = ΣX.y + ΣY.x,
y=0
ΣM0 = 48.97 × x
+
+ ΣM0 = 12.5 × 16 + 15 × 22 + 20 × 63 + 25 × 159
– 17.68 × 22 – 17.68 × 129 = 3095.32 kN.m
48.97 x = 3095.32,
x=
3095.32
= 63.2 m
48.97
From Fig. 1.4, the resultant R will intersect the hull at a distance of (x – a) from O :
tan θ =
22
= 1.084,
a
63.2 – 20.3 ≈ 43 m
a = 20.3 m,
i.e., where a single, more powerful tug boat should push, exerting a force of 66.63 kN. See the
graphical solution in Plate 1–A, page 5.
Example 1.4. Find the resultant of the four forces acting as shown in Fig. 1.5.
200 kN
100
50
70.7 kN
45°
30°
50
70.7 kN
173.2
1m
309
q
200 kN
45°
R
1.5 m
30°
80
Fig. 1.6(a)
a
1.5 m
86.6
A
A
1m
80 kN
30°
x
100 kN
SY
80 kN
100 kN
50
Fig. 1.5
q
30°
1m
Fig. 1.6
Solution: See Fig. 1.6
→
+ ΣX = 50 + 173.2 + 86.6 = 309.8 kN →
+ ↑ ΣY = – 50 + 100 – 50 – 80 = – 80 or 80 kN ↓
SX
R
Resultant and Equivalent of Force Systems
R=
Σx 2 + Σy 2 =
11
309.8 2 + 80 2 = 320 kN, Fig. 1.6(a)
80
= 14.48°
309.8
θ = tan–1
To find the line of action of the resultant R, take moments about A (and applying
Varignon’s theorem),
ΣMA = R . a = ΣX.y + ΣY.x,
y = 0,
ΣMA = 50 × 1.5 – 100 × 1 + 173.2 × 1.5 + 50 × 1 = 284.8 kN
+
∴
a=
ΣM A 284.8
=
= 0.89 m,
R
320
x=
ΣM A 284.8
=
= 3.56 m
80
ΣY
Note: a = x sin θ = 3.56 sin 14.48° = 0.89 m
Example 1.5. A rigid bar AB is subjected to a system of parallel forces as shown in
Fig. 1.7. Reduce the given system of forces to an equivalent (a) single force, (b) force and moment
at A, (c) force and moment at D, and (d) force and moment at B.
25 kN
A
10 kN
15 kN
C
1m
D
2m
20 kN
B
2m
Fig. 1.7
Solution: (a) For an ≡ single force on AB:
+ ↑ Resultant R = 10 – 25 + 20 – 15 = – 10 or 10 kN ↓
To find its location : ΣMA = R . a
+
∴
ΣMA = 25 × 1 – 20 × 3 + 15 × 5 = 40 kN.m
a=
ΣM A 40
=
= 4 m, Fig. 1.8
R
10
Fig. 1.8 ≡ Single Force on AB
To find ≡ force and couple at any point other than E, introduce equal and opposite forces R at the
required point, which will not affect the equilibrium:
12
Engineering Mechanics
(b) A force R at E may be treated as another force at A and a couple as shown in
Fig. 1.9(b)
Fig. 1.9. ≡ Force and Moment at A
(c) A force at E ≡ another force at D and a couple as shown in Fig. 1.10(b)
R
R = 10 kN
4m
E
D
A
3m
1m
(a)
R
R = 10 kN
D
A
3m
B
10 × 1 = 10 kN.m
(b)
Fig. 1.10 ≡ Force and Moment at D
(d) A force at E ≡ another force at B and a couple as shown in Fig. 1.11(b)
R = 10 kN
R
4m
E
A
(a)
B
1m
R
10 kN
A
B
(b)
10 × 1 = 10 kN.m = M
Fig. 1.11 ≡ Force and Moment at B
From Fig. 1.7, +
ΣMB = 10 × 5 – 25 × 4 + 20 × 2 = – 10 i.e., 10 kN.m
Thus, the equivalent system at any point other than E is the same resultant force R and
a moment which is equal to the algebraic sum of the moments of all the forces acting on the bar
AB about that point.
Resultant and Equivalent of Force Systems
13
Note: A single force F (acting at A in Fig. 1.12) can be looked
upon as a force of the same magnitude and direction shifted parallel
to itself (by AB = a) accompanied by a couple of moment (M) = force ×
shift (= F . a)
i.e.,
F at A = F at B + F . a
Contrariwise, a force, F and a couple of moment M
(= F . a) acting at point (B) is equivalent to a single force (F) of
the same magnitude and direction shifted parallel to its original
line of action, the shift being =
M
(i.e., = ‘a’ in Fig. 1.12).
F
Fig. 1.12
Example 1.6. Replace the system of a force and couple shown in Fig. 1.13 by a single
force.
100 N
50 N
0.4 m
50 N
x
100 N
Fig. 1.13
100 N
Fig. 1.14
Fig. 1.15
Solution: Since a force ≡ another force (shifted suitably) and a couple, Fig. 1.14, shift ‘x’
to counteract the existing couple :
100 × x
= 50 × 0.4
x = 0.2 m or 1.8 m from the support
The equivalent system is as shown in Fig. 1.15.
Example 1.7. Replace the system of a force and couple shown in Fig. 1.16 by a single
force on the line AB.
120 N
A
96
0
N
A
r=
10
0m
m
F
60°
O
50
mm
B
Fig. 1.16
120 N
O
60°
C
B
Fig. 1.17
Fig. 1.18
14
Engineering Mechanics
Solution: Since a force ≡ another force (shifted suitably) and a couple Fig. 1.17,
shift ‘x’ to counteract the existing couple:
960 × x
= 120 × 0.2
x = 0.025 m or 25 mm ; OC =
25
= 50 mm
cos 60°
The equivalent system is as shown in Fig. 1.18.
Example 1.8. An angle is subjected to a force-couple system as shown in Fig. 1.19. Reduce
the system (a) to an equivalent system at A, (b) as a single resultant anywhere.
Solution:
(a) At A,
ΣX = 80 cos 45° = 56.56 N →
ΣY = 50 + 80 sin 45° – 150 = – 43.44 or 43.44 N ↓
R=
56.56 2 + 43.44 2 = 71.32 N
θ = tan–1
at
+
43.44
= 37.53°
56.56
ΣM = – 300 – 80 sin 45° × 3 + 150 × 4 = 130.32 Nm
Hence, the equivalent system at A is a ‘force-couple system’ of R and M, Fig. 1.20.
50 N
50 N
B
3m
300 Nm
300 Nm
80 N
80 N
2.3 m
q
A
45°
3m
A
1m
a
q
M
R
R
45°
3m
150 N
Fig. 1.19
1m
150 N
Fig. 1.20
(b) The force-couple system at A can be reduced to a single resultant R acting at B,
Fig. 1.20, in the same direction at a distance,
a=
AB =
130.32 Nm
M
=
= 1.83 m
R
71.32 N
a
183
.
=
= 2.3 m
cos θ cos 37.53°
Principle of Transmissibility. The point of application of a force may be transmitted
along its Line of Action to another point, without changing its effect on any rigid body. This
Resultant and Equivalent of Force Systems
15
is known as the principle of transmissibility of force and it
enables to regard a force acting on a rigid body as a Sliding
Vector. In Fig. 1.21. the force P on the body can be applied
at A or B, without changing the reactions at the bearing
support at C or roller support at D.
P
B
P
A
C
Pinned
D
RCH
Roller
RD
RCY
Fig. 1.21
PROBLEMS
1.1. Six forces are acting along the sides of a regular hexagon of side 100 mm as shown in Fig. P 1.1.
Find the resultant and its distance from A.
[185. 2N, 57.32°
2
2
4
Loads in kN
2
2
3m
1
1
2
, 276.5 mm]
A
3m
3m
Fig. P 1.1
3m
B
Fig. P 1.2
1.2. For the loaded truss shown in Fig. P 1.2, find the resultant load and where its line of action will
intersect AB.
[15.46 kN, 75°
; 4.4 m from A]
1.3. Determine the resultant of the forces acting on the dam shown in Fig. P 1.3 and locate its intersection with the base AB. For safe design, the intersection should occur within the middle-third
of the base. Is the design safe ?
[137.12 kN,
79.92° ; 2.15 m from B ; Yes]
2m
MWL
Dam
c.g.
P = 50 kN
TWL
Q = 30 kN
2m
A
W = 120 kN
5m
Base of dam
Fig. P 1.3
1m
60°
B
16
1.4.
Engineering Mechanics
(a) Determine completely the resultant of the five forces shown in Fig. P 1.4. The squares are
100 mm × 100 mm.
[0.8 N →, Y = 67 m ; 0.8 N →, 53.6 Nm
(b) Give an equivalent system at ‘O’.
at O]
Y
20
0
N
100
N
20
0
100
N
N
X
O
282 N
Fig. P 1.4
1.5. A force of 500 N is acting at A in Fig. P 1.5 produces a moment of 1200 Nm about O. Find the
y-intercept of the force. Give an equivalent system with the same force acting at ‘O’.
[3 m ; 500 N and 1200 Nm
Fig. P 1.5
at O]
Fig. P 1.6
1.6. A 50 N force is applied to a corner plate as shown in Fig. P 1.6. Determine an equivalent forcecouple system at A.
[50 N, 3.08 Nm]
1.7. Find the single resultant of the forces acting on the pulley in Fig. P 1.7. Give its intercept with
the axes.
[1020 N ], (0.1, 0.02) m]
1.8. An electric light fixture weighing 15 N is hung from two strings from the roof and wall as shown
in Fig. P 1.8. Find the tension in the strings.
[7.8, 11 N]
Resultant and Equivalent of Force Systems
17
Roof
60°
Wall
45°
Light
15 N
Fig. P 1.7
Fig. P 1.8