The document summarizes composition of forces. It discusses concurrent forces and how to find their resultant using geometric and analytical methods. Geometrically, the resultant of concurrent forces can be found using parallelograms or parallelepipeds. Analytically, the resultant is equal to the sum of the components of the individual forces in any given direction. Examples are provided to demonstrate how to use these methods to calculate the resultant of different systems of concurrent forces.

2. Presented By:
Amenah Gondal
Presented To:
Ma’am Mehwish
Class:
B.S.Ed (I)
Topic:
Composition of Forces

4.  Composition of forces
 Components of Forces
 Composition of Concurrent Forces
 Geometrical Methods
 Analytical Method
 (𝜆, 𝜇) Theorem
 Examples

5. The basic problem of statics consists in the reduction of a
given system of forces to the simplest system which will be
equivalent to the original system. Such a reduction is called
composition of forces.

6.  When two forces 𝑃 and 𝑄 act on a particle then their
resultant 𝑅 can be find out by using Law of Parallelogram.
The forces 𝑃 and 𝑄 are called components of the resultant.
O
𝑃
𝑄
R

7.  Forces whose lines of actions intersect at one point called
concurrent forces.
 Such forces always have a resultant through the point of
concurrence.
 This resultant can be found either geometrically or
analytically.
 The two methods are:
 Geometrical method
 Analytical method

8. It consists of three types of forces:
 Two Forces
 Three non-coplanar Forces
 System of Forces

9.  Two Forces 𝑃1 𝑎𝑛𝑑 𝑃2
 Find Resultant 𝑅 using
parallelogram law of forces
 From figure
𝑂𝐶 = 𝑂𝐴 + 𝐴𝐶
∴ 𝑅 = 𝑃1 + 𝑃2
 i.e., the resultant of two concurrent forces is given by their
vector sum
𝑅
𝑃1
𝑃2
O A
C

10.  Alternatively, Resultant 𝑅can be calculated using law of
trigonometry:
Using law of cosine in ∆𝑂𝐴𝐵
c2= 𝑎2+𝑏2 − 2𝑎𝑏cos α
R2= 𝑃1
2
+𝑃2
2
-2P1 P2cos (180°-α)
R= 𝑃1
2
+ 𝑃1
2
− 2𝑃1 𝑃2cos (180° − 𝛼)
R= 𝑃1
2
+ 𝑃1
2
+ 2𝑃1 𝑃2cos𝛼
𝑅
𝑃1
𝑃1
𝑃2
𝑃2
O A
BC
𝜶 𝜷
𝜸

11. The angles 𝛾 𝑎𝑛𝑑 𝛽 which the resultant makes with
component forces are found by the law of sine. In ∆𝑂𝐴𝐵
𝒂
𝒔𝒊𝒏𝜶
=
𝒃
𝒔𝒊𝒏𝜷
=
𝒄
𝒔𝒊𝒏𝜸
𝑷 𝟏
𝒔𝒊𝒏𝜸
=
𝑷 𝟐
𝒔𝒊𝒏𝜷
=
𝑹
𝒔𝒊𝒏(𝟏𝟖𝟎° − 𝜶)
⇒
𝑷 𝟏
𝒔𝒊𝒏𝜸
=
𝑷 𝟐
𝒔𝒊𝒏𝜷
=
𝑹
𝒔𝒊𝒏𝜶

12.  Non-coplanar Forces are 𝑃1, 𝑃2 𝑎𝑛𝑑 𝑃3.
 To find Resultant construct parallelepiped whose edges are
𝑂𝐴,𝑂𝐵 𝑎𝑛𝑑 𝑂𝐶.
 By the law of parallelogram
of Forces
𝑂𝐴 + 𝑂𝐵=𝑂𝐸
And applying the same law again
𝑂𝐸 + 𝑂𝐶= 𝑂𝐷
𝑃1
𝑃2
𝑃3
A
B
C
D
E
O
𝑅

13. Put value of 𝑂𝐸
𝑂𝐴 + 𝑂𝐵+𝑂𝐶=𝑂𝐷
⇒ 𝑃1 + 𝑃2+𝑃3= diagonal
of the parallelepiped through O.
Hence the resultant 𝑅 of three concurrent non-coplanar
forces, acting at O, is represented by the diagonal, drawn
through O, of a parallelepiped with the given forces.
We can express the resultant of construction for the resultant
by the vector equation
𝑖=1
3
𝑃𝑖 = 𝑅

14.  For more than two forces say n .
 The resultant will be obtained by repeated application of
parallelogram law of forces.
𝑖=1
𝑛
𝑃𝑖 = 𝑅
where i= 1,2,3,…………….,n
O
𝑃3
𝑃1
𝑃2
𝑃4

15. The analytical method is based on the following
theorem:
“If R is the resultant of the forces 𝑃1, 𝑃2, 𝑃3……………..𝑃𝑛
then the component of R in any direction is equal to the
algebraic sum of the components of 𝑃1 +
𝑃2+𝑃3……………..𝑃𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒
𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛.”
O y
x
z
𝑃1
𝑃2
𝑃𝑛
𝑃3
𝑃4
R

16. To prove the theorem we take a unit vector 𝑢 along the
direction in which the components are taken. Since
R= 𝑖=1
𝑛
𝑃𝑖
We multiply this equation scalarly by 𝑢
R. 𝑢=(𝑃1 + 𝑃2 + 𝑃3……………..+𝑃𝑛). 𝑢
By using distributive property of vectors
R. 𝑢=𝑃1. 𝑢 + 𝑃2. 𝑢 + 𝑃3. 𝑢+……………..+𝑃𝑛. 𝑢
=(sum of components of component vectors along 𝑢)
R . 𝑢 = ( 𝑖=1
𝑛
𝑃𝑖). 𝑢 = 𝑖=1
𝑛
𝑃𝑖 𝑢
which completes the proof.

17.  x-component of 𝑅= sum of x-components of forces 𝑃1 +
𝑃2+𝑃3……………..𝑃𝑛
Rcos𝛼 =X = 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛼𝑖 (i)
 y-component of 𝑅= sum of y-components of forces 𝑃1 +
𝑃2+𝑃3……………..𝑃𝑛
Rcos𝛽 =Y = 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛽𝑖(ii)
 z-component of 𝑅= sum of z-components of forces 𝑃1 +
𝑃2+𝑃3……………..𝑃𝑛
Rcos𝛾 =Z = 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛾𝑖(iii)

18. Squaring and adding eq. i. ,ii and iii
 R2cos𝛼 + 𝑅2 𝑐𝑜𝑠𝛽+ 𝑅2cos𝛾
=( 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛼𝑖)2+( 𝑖=0
𝑛
𝑃 𝑖 𝑐𝑜𝑠𝛽𝑖)2+( 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛾𝑖)2
 R2(cos𝛼 + 𝑐𝑜𝑠𝛽+ cos𝛾)
= ( 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛼𝑖)2+( 𝑖=0
𝑛
𝑃 𝑖 𝑐𝑜𝑠𝛽𝑖)2+( 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛾𝑖)2
 R2=X2+Y2+Z2
 R= 𝑋2 + 𝑌2 + 𝑍2
which determine the magnitude of the resultant R.

19. Resultant Angle of 𝛼,𝛽 and 𝛾 can be calculated by the
following results:
As Rcos𝛼 =X ⇒ 𝑐𝑜𝑠𝛼 =
𝑋
𝑅
Similarly Rcos𝛽 =Y ⇒ 𝑐𝑜𝑠𝛽 =
𝑌
𝑅
Rcos𝛾 =Z ⇒ 𝑐𝑜𝑠𝛾 =
𝑍
𝑅
which determines the direction of the resultant R.

20. Forces P,Q, R act at a point parallel to the sides of triangle
ABC taken in the same order. Show that the magnitude of the
resultant is
(𝑃2
+ 𝑄2
+ 𝑅2
− 2𝑄𝑅𝑐𝑜𝑠𝐴 − 2𝑅𝑃𝑐𝑜𝑠𝐵 − 2𝑃𝑄𝑐𝑜𝑠𝐶)1 2
Sol:
Take rectangular axes Ox, Oy with Ox parallel to the
side BC of the triangle ABC
Now
< 𝑃𝑂𝑄 = 180° − 𝐶
< 𝑃𝑂𝑅 = 180° + 𝐵
180° − 𝐶180° + 𝐵
180° − 𝐴
𝑃
A
𝑄𝑅
CB
x
o
y
𝑃
𝑅
𝑄

21. If X and Y are the components of the resultant along axes,
X=P + Q cos(180° − 𝐶) + 𝑅𝑐𝑜𝑠(180° + 𝐵)
= P−Q cos𝐶 − 𝑅𝑐𝑜𝑠𝐵---------- I
Y=Q sin(180° − 𝐶) + 𝑅𝑠𝑖𝑛(180° + 𝐵)
= Qsin𝐶 − 𝑅𝑠𝑖𝑛𝐵---------------II
Hence squaring and adding I and II
𝑋2 + 𝑌2 =
(𝑃 − 𝑄 𝑐𝑜𝑠𝐶 − 𝑅𝑐𝑜𝑠𝐵)2+(𝑄𝑠𝑖𝑛𝐶 − 𝑅𝑠𝑖𝑛𝐵)2= 𝑃2 +
𝑄2 𝑐𝑜𝑠2 𝐶 + 𝑅2 𝑐𝑜𝑠2 𝐵 − 2𝑃𝑄𝑐𝑜𝑠𝐶 + 2𝑄𝑅( 𝑐𝑜𝑠𝐵𝑐𝑜𝑠𝐶 −

22. = 𝑃2+𝑄2 + 𝑅2 + 2𝑄𝑅𝑐𝑜𝑠(180° − 𝐴) − 2𝑅𝑃𝑐𝑜𝑠𝐵 − 2𝑃𝑄𝑐𝑜𝑠𝐶
= 𝑃2+𝑄2 + 𝑅2 + 2𝑄𝑅𝑐𝑜𝑠𝐴 − 2𝑅𝑃𝑐𝑜𝑠𝐵 − 2𝑃𝑄𝑐𝑜𝑠𝐶
Taking square root on both sides
𝑋2 + 𝑌2
= 𝑃2+𝑄2 + 𝑅2 + 2𝑄𝑅𝑐𝑜𝑠𝐴 − 2𝑅𝑃𝑐𝑜𝑠𝐵 − 2𝑃𝑄𝑐𝑜𝑠𝐶
= (𝑃2
+ 𝑄2
+ 𝑅2
− 2𝑄𝑅𝑐𝑜𝑠𝐴 − 2𝑅𝑃𝑐𝑜𝑠𝐵 − 2𝑃𝑄𝑐𝑜𝑠𝐶)1 2

23. "If two concurrent forces are represented by 𝜆OA and
𝜇OB then their resultant is given by (𝜆 + 𝜇)𝑂𝐶, where C
divides AB so that AC: CB= 𝜇: 𝜆.”
Proof:
In order to prove this theorem we proceed as follows:
In the ΔOAC
𝑂𝐴=𝑂𝐶+𝐶𝐴 ------------ (I)
In the ΔOBC
𝑂𝐵=𝑂𝐶+𝐶𝐵 ------------ (II)
𝐵
𝐴
𝑂
𝐶𝝀𝑂𝐴
𝜇𝑂𝐵
𝜇
𝝀

24. Multiplying (I) by 𝝀 and (II) by 𝝁
𝜆𝑂𝐴=𝜆𝑂𝐶+𝜆𝐶𝐴 ---------------- (III)
𝜇𝑂𝐵=𝜇𝑂𝐶+𝜇𝐶𝐵 ----------------- (IV)
Now adding (III) and (IV)
𝜆𝑂𝐴 + 𝜇𝑂𝐵= (𝜆𝑂𝐶+𝜆𝐶𝐴 ) + (𝜇𝑂𝐶 + 𝜇𝐶𝐵) ----------- (V)
𝜆𝑂𝐴 + 𝜇𝑂𝐵=(𝜆 + 𝜇)𝑂𝐶+ 𝜆𝐶𝐴 + 𝜇𝐶𝐵-------------- (VI)
As , it is given that AC:CB=𝜇: 𝜆
𝐴𝐶
𝐶𝐵
=
𝜇
𝜆
⟹ 𝜆𝐴𝐶=𝜇𝐶𝐵
𝜇𝐶𝐵 − 𝜆𝐴𝐶=0
𝜇𝐶𝐵 + 𝜆𝐶𝐴=0
∴ 𝑠𝑖𝑛𝑐𝑒 − 𝜆𝐴𝐶 = 𝜆𝐶𝐴

25. Then eq (VI) becomes
𝝀𝑶𝑨 + 𝝁𝑶𝑩=(𝝀 + 𝝁)𝑶𝑪
which completes a theorem
In case
𝜆 = 𝜇 = 1
Then (𝜆, 𝜇) theorem becomes
𝑂𝐴 + 𝑂𝐵= 2𝑂𝐶
Where C is the middle point of AB .This in fact the result
given by the parallelogram of forces since OC is half of the
diagonal of the parallelogram.

26. Forces P,Q, R act at a point O and their resultant is R. If
any transversal cuts the lines of action of the forces in
the points A, B and C respectively, prove that
𝑷
𝑶𝑨
+
𝑸
𝑶𝑩
=
𝑹
𝑶𝑪
Sol:
We can use the (𝜆, 𝜇) theorem to get the required results:
As 𝑃 = 𝑃 𝑂𝐴 = 𝑃
𝑂𝐴
𝑂𝐴
=
𝑃
𝑂𝐴
𝑂𝐴
𝑄 = 𝑄 𝑂𝐵 = 𝑄
𝑂𝐵
𝑂𝐵
= (
𝑄
𝑂𝐵
)𝑂𝐵
𝑅 = 𝑅 𝑂𝐶 = 𝑅
𝑂𝐶
𝑂𝐶
= (
𝑅
𝑂𝐶
)𝑂𝐶
B
C
A
O
𝑄
𝑃
𝑅
L

27. Using the (𝜆, 𝜇) theorem, we have
𝑃 + 𝑄 =
𝑃
𝑂𝐴
𝑂𝐴 +
𝑄
𝑂𝐵
𝑂𝐵
=(
𝑃
𝑂𝐴
+
𝑄
𝑂𝐵
) 𝑂𝐴 + 𝑂𝐵
= (
𝑃
𝑂𝐴
+
𝑄
𝑂𝐵
) 𝑂𝐶
∴ (𝑂𝐴 + 𝑂𝐵 = 𝑂𝐶)
Where C is a point lie on AB such that
AC: CB =
𝑄
𝑂𝐵
:
𝑃
𝑂𝐴
As we know that
𝑃 + 𝑄 = 𝑅
Then
𝑃
𝑂𝐴
+
𝑄
𝑂𝐵
𝑂𝐶 = (
𝑅
𝑂𝐶
)𝑂𝐶

28. which gives
𝑃
𝑂𝐴
+
𝑄
𝑂𝐵
= (
𝑅
𝑂𝐶
)
Hence proved
𝑃
𝑂𝐴
+
𝑄
𝑂𝐵
=
𝑅
𝑂𝐶