Statics

Statics: CHAPTER 21
Composition and Resolution of

Learning Objectives concurrent forces

find resultant forces by parallelogram of forces.

 solve the problems related to composition and resolution of forces.
obtain resultant of coplanar forces/vector acting on a point.
solve the forces/vectors related problems using Lamis theorem.

Statics is the science which deals with the force acting on bodies when the body remains at
rest. A body is said to be rigid if distance between any two points in it remains invariable
during investigation. A body is said to be in rest which does not change its position relative
to the surrounding bodies.

Force:

A force is a vector quantity which changes or tends to change, the state of rest or uniform
motion of a body. Force acting or a body is related to the mass of the body and variation of
its velocity with time.
Tension and Thrust
If we tie one end of a string to a point of body and pull the other end of the string. We exert
a force on the body through the string and this force is called the tension (pull). If a body is
pushed by means of a rod then force exerted by on the body through the rod is called thrust
(push)
Resultant of Forces
If a number of forces act upon a body and if a single force can be found which has the
same effect upon the body as these forces, then this single force is called resultant of
these forces and these forces are called the component of resultants.

Equilibrium: When two or more forces acting on a body do not bring a change in its state
of rest or of its uniform motion, the body is said to be in equilibrium under the forces.

Resultant of two Forces acting in same or Opposite Direction
1.If two forces P and Q act on a body in the

same direction, then their resultant R = P + Q and act in the direction of either forces.
2.If two forces P and Q (P > Q) act on a body alongthe same straightline on opposite

direction, then their resultant R = P – Q and acts in the direction of the greater of the two forces.

Resultant of two forces acting at a point
Parallelogram law of forces: If two forces acting at a point are represented in
magnitude and directly the two sides of a parallelogram drawn from the point,
their resultant is represented in magnitude and direction by the diagonal of the
parallelogram drawn from the same point.
Let →P and →Q be two forces acting on a particle, whose directions include an angle of .
They may have the same line of action. If →RI s the resultant of the forces then

 →R = →P + →Q . . . ( )

R2 = →R  →R =(→P + →Q )  (→P + →Q )

= P2 + Q2 + 2→P  →Q

= P2 + Q2 + 2PQ cos

 R = + + cos ,

 From (i), we also have

→P →R= →P →P+ →P →Q

 PR cos = P2 + PQ cos

 R cos = P + Q cos . . . (ii)

 Again, using (i), we obtain

|→P × →R | = |→P × →P + →P × →Q |

 PR sin = PQ sin Thus, from (ii) and (iii), we obtain tan = Q sin

 R sin = Q sin . . . (iii) P + Q cos
Which gives the direction of the resultant R

Some Important results

Cor. 1. If = , ℎ = 2 cos =
2
2

Cor.2. If = , ℎ = 2 + 2 tan =
2

Cor.3. If the forces P and Q act in the same direction , then

= + ∵ = 0

Cor.4. If the forces P and Q act in the opposite direction ,then

= − ∵ = 180

Cor.5. If = = , ℎ ( . 1 ) = 120

Some Important results = tan–1 P sin
(i) When P = Q then P + P cos

R2 = P2 + Q2 + 2PQ cos = tan–1 sin 
 R2 = P2 + P2 + 2PP cos 1 + cos
 R2 = 2P2 + 2P2 cos
 R2 = 2P2 (1 + cos) = tan–12sin2  cos2

 R2 = 2P2 × 2cos2 /2 2cos2 /2

 R2 = 4P2 cos2 /2 = tan–1tan2

 R= 2Pcos /2 Q sin = 
Also, + Q cos 2
= tan–1
 P  The resultant R bisects the angle  between

the forces if they are equal.

(ii) If  =  then,
2

R2 = P2 + Q2

 R = P2 + Q2

   = tan–1QP
 Q sin2 
tan–1 cos2
and  = P

+ Q

(iii) If  = 0,

then, R2 = P2 + Q2 + 2PQ cos

= P2 + Q2 + 2PQ

= (P + Q)2

 R = P + Q,
which is maximum resultant of the forces P and Q. So, the forces obtains

maximum resultant. If they act along same direction (collinear or parallel).

(iv) If  = 180 acts opposite direction (unlike

then, R2 = P2 + Q2 + 2PQ cos180
= P2 + Q2-2PQ
= (P – Q)2

 R = |P – Q|
Which is minimum resultant force P and Q.
So, forces obtains minimum resultant if they
parallel).

1. i. Find the resultant of following pairs of forces in magnitude and direction;
P = 24 N, Q = 7 N,  = 90°
Soln: Since R N is the resultant of the forces P N and Q N.
Hence by parallelogram law of forces R2 = P2 + Q2 + 2PQ cos 
or,R2 = (24)2 + (7)2 + 2 . 24 . 7 cos 90°
or,R2 = 576 + 49 = 0 = 625
 R = 25 N Ans.

1.ii. Find  if P = 3 N, Q = 5 N, R = 7 N

Soln: Since, R2 = P2 + Q2 + 2PQ cos 
or,72 = 32 + 52 + 2 × 3 × 5 cos 
or,30 cos  = 15
 cos  = 1  cos  = cos 60
2
  = 60° Ans.

4. Two forces whose magnitudes are P N and N act on a particle in directions
inclined at an angle of 135° to each other; find the magnitude and direction of

the resultant.

Soln:
Let R N is the resultant of the two forces P N and 2 N incline on angle of 135°. Then

by parallelogram law of forces, we have
2
Then, R2 = P2 + ( + 2P. 2. cos135°
2)
, 2 = 2 + 2 2 + 2 2 2. 1
2

or,R2 = 3P2 – 2P2 = P2

R=P

If  is the angle made by R with P, then by applying formula :

tan  = P Q sin  
+ Q cos

 1
 
P 2 . sin 135° P 2 . 
+ P 2 cos 135° + P 2
or,tan  = =
P – 1 
P 2
2

= P P P= 
–

or,tan = tan90°  = 90° Ans.

5. Two forces acting at an angle of 45° have a resultant equal to
10N; if one of the force be N, find the other force.

Solution: or,Q2 + 2Q – 8 = 0
Let the other force be Q N, then or, Q2 + 4Q – 2Q – 8 = 0

P = 2 N, Q = Q N, R = 10 N, or, Q(Q + 4) – 2(Q + 4) = 0
 = 45° or, (Q + 4) (Q – 2) = 0
Q = 2 N Ans.
Now, applying the parallelogram
law of forces we have (∵ Q can not be negative)
R2 = P2 + Q2 + 2PQ cos 

Or, 10 = 2 + 2 + 2 2 cos 45

7. Find the magnitude of two forces such that if they act at right
angles, their resultant is N, if they act an angle of 60°,
their resultant is N.

Solution: , 13 = P2 + Q2 + 2PQ cos60
Let P and Q be the two forces, then
, 13 = P2 + Q2 + 2PQ 1
Case I : If they act at right angle , then 2
or, (P2 + Q2) + PQ = 13
R = 10 N
we know that, R2 = P2 + Q2 + 2PQ cos  or,10 + PQ = 13
, 10 = P2 + Q2 + 2PQ cos 90 PQ = 3 ............. (ii)

, 10 = P2 + Q2 … … … . ( ) We know that ,
Case II :If they act an angle 60° , then (P + Q)2 = (P2 + Q2) + 2.PQ = 10 + 2.3 = 16

R = 13 N  P + Q = 4 ..................... (iii)
we know that, R2 = P2 + Q2 + 2PQ cos 
and
(P – Q)2 = P2 + Q2 – 2.PQ = 10 – 2.3 = 4
 P – Q = 2 ...................... (iv)

Adding and subtracting (iii) and (iv), we get

2P = 6 and 2Q = 2

 P = 3 N, Q = 1 N Ans

9. The sum of the forces is 18 N and their resultant which is perpendicular
to the smaller of the two forces is 12N. Find the magnitude of the forces.

Solution:
Let P and Q be the required
forces and R be their resultant
acting at point O.
Then by question.
P + Q = 18 and R = 12 ....... (i)

Since R is perpendicular to
smaller of the two forces say P.

tan 90° = P Qsin
+ Qcos

or,P + Qcos = 0 or,cos = – P
Q

Also, R2 = P2 + Q2 + 2PQ cos

or,144 = P2 + Q2 + 2PQ (– P )
Q

or,P2 – Q2 = – 144

or, (P + Q) (P – Q) = – 144

Then from (i) this gives,
18(P – Q) = – 144

or,P – Q = – 8.............. (ii)

Solving (i) and (ii), Adding and subtracting (i)
and (ii)

2P = 10 and 2Q = 26

 P = 5N, Q = 13N Ans.

10. Two forces of magnitude 3P, 2P respectively have a resultant
R. If the first force be doubled, the magnitude of the resultant is
doubled. Find the angle between the forces.

Solution: (2R)2 = (6P)2 + (2P)2 + 2.6P.2P cos
Suppose  is the required angle or,4R2 = 40P2 + 24P2 cos

between the forces 3P and 2P and R or,R2 = P2(10 + 6cos) ..................... (ii)
is their resultant.
Divi1di=ng(((1i1)30b++y16(2icic)oo,ssw)e) get
 R2 = (3P)2 + (2P)2 + 2.3P.2P cos or, 10 + 6cos = 13 + 12 cos
or,R2 = 13P2 + 12P2 cos
or, 6 cos = −3
 R2 = P2(13 + 12cos) ............... (i)
When first force 3P is doubled, the or, cos = −1 = cos 120
magnitude of the resultant is 2
doubled. Hence first force becomes
6P and resultant becomes 2R or, = 120

11.The resultant of two force P and Q is equal to Q and
making an angle of 30° with the direction of P; show that P is
either equal to Q or is double of Q.

Solution:
Let OA and OB represent the two forces P and Q

acting at point O.
Let OC be the diagonal of the parallelogram OACB

such that AOC = 30°
Now, = , = , = 3 Q
Since = , =

Applying cosine law in  AOC, we have
cos = (OA)2 + (OC)2 – (AC)2

2 ×

cos 30 = 2+3 2− 2

2 × 3

or, 3 = 2+2 2

2 2 3

or,2P2 + 4Q2 – 6PQ = 0

or, P2 – 3PQ + 2Q2 = 0

or,(P – Q) (P – 2Q) = 0

Either, P = Q or P = 2Q

Hence, P equals to Q or double of Q.

12. The resultant of two forces P and Q acting at an angle  is equal to (2m

+ 1) P2 + Q2. When they act at an angle (90° - ) the resultant is (2m – 1) P2 + Q2.

Prove that = −

+

Soln: Since (2m + 1) P2 + Q2 is the resultant of = 2m(2m – 2) (P2 + Q2)
P and Q acting at an angle  ∴ sin = 2m(m – 1) (P2 + Q2) ............. (ii)

 [(2m + 1) P2 + Q2 ]2 = P2 + Q2 + 2PQ cos Dividing (ii) by (i), we get
tan = −1
or,(2m + 1)2 (P2 + Q2) = (P2 + Q2) + 2PQ cos
+1
 2PQcos= {(2m + 1)2 – 1} (P2 + Q2)

= (2m + 2) (2m) (P2 + Q2)

∴PQcos = 2m(m + 1) (P2 + Q2 ..... (i)
Again (2m – 1) P2 + Q2 is resultant of P and

Q acting at angle (90 – )
 [(2m – 1) P2 + Q2 ]2

= P2 + Q2 + 2PQ cos(90 – )
or,(2m – 1)2 (P2 + Q2) = (P2 + Q2) + 2PQ sin
 2PQsin= {(2m – 1)2 – 1} (P2 + Q2)

Resolution of Forces

The component of force F making angle of  and  along the direction of OX

and

OY is given by = , =
+
+

where P and Q are components of force F along the direction of OX and OY.

Proof:

Let be a given force making an angle of
 and  resolved along OX and OY
respectively. If Ԧ and be two components
of Ԧ along OX and OY, then

Ԧ = Ԧ +
⇒ Ԧ × Ԧ = Ԧ × ( + ) = Ԧ × (∵ Ԧ × Ԧ = )
⇒ Ԧ × Ԧ = Ԧ ×

⇒ PFsin = sin +
⇒ = sin

sin +

Again,
Ԧ = Ԧ +

⇒ Ԧ × = ( + ) × = Ԧ × (∵ Ԧ × Ԧ = )
⇒ Ԧ × = Ԧ ×

⇒ FQsin = sin +

⇒ P= sin

sin +

NOTE:

If + = , ℎ
2
2 −
= sin = F cos
sin
2

and = sin

Coplanar Forces:

Two or more forces are said to be coplanar if their line of action lies on
the same plane. If a number of coplanar forces act at a point, then each
of forces can be resolved along two mutually perpendicular directions
having same effect.

Theorem:
The algebraic sum of resolved part of any number of forces acting at a point of
a rigid body and in a given direction is equal to the resolved part of their
resultant in the same direction.

1 . A force equal to 10 N is inclined at an angle of 30° to the

horizontal; find its resolved parts in a horizontal and

vertical directions.

Solution:

Let F = 10 N is inclined at an angle

30° and P and Q are resolved port of F in

horizontal and vertical direction.

Then, P= sin and = sin
sin + sin +

P = F sin60 °and Q = F sin 30°

or, = 10 × 3 and = 10 × 1
2 2

or, = 5 3 = 5

4. A force of 20 N acting vertically upwards is resolved into two forces
one being horizontal and equal to 10 N. What is the magnitude and
direction of the other force ?

Solution:

Suppose F1 = 10 N and F2 be the resolved
forces of the given force F = 20N. Let  be
the angle between the force F2 and F. Again
given that angle between F and F1 is 90°. We
shall find the force F2 and .

Now, from figure,

F2 = 20 sin90° and 1 = 20 sin
sin(90° + ) sin 90+

or,F2 =c2o0s×1 and 10 = 20 tan 
and
or,F2 = 20 sec  ..... (i) tan  =21 … … … . . ( )

From (i) and (ii), we get

F2 = 20 sec 
= 20 1 + tan2

12 20 5
= 20 1 + 2 = 2

= 10 5 N

From (ii), tan  = 1   = 1
2 tan -12.

6. Forces 1 N, 2 N and 3 N act at a point in direction parallel to the side of
an equilateral triangle taken in order. Find their resultant.

Solution:

Let the forces 1N, 2N

and 3N act at O parallel

to the sides BC, CA

and AB of an equilateral

triangle ABC respectively.

Resolving the forces along BX and perpendicular to BX, we have
X = 1.cos 0° + 2.cos 120° + 3.cos 240°

= 1 + 2 × −1 + 3 × −1
2 2
= −3
2

Y= 1.sin 0° + 2.sin 120° + 3.sin 240° • Again,
tan =
=2× 3+3×− 3

22
−3

= −3 = 2
2 −3

If R be the resultant making an = 21

angle  with BC 3

Then, o , tan = tan 180 + 30

= 2 + 2 ∵XandYare both negative
∴ = 210
2 −32
= −3 2 Hence, the resultant is inclined at an
2 + angle 180° + 30° = 210° with BC.

…………… The angle made fboyrcthee=re2s1u0l°ta–nt1R20°
w= i9th0°t.he second
= 3

Triangle Law of Forces

If three forces acting at a point can be represented in magnitude and direction by the
sides of a triangle, taken in order, then the forces are in equilibrium.

Let the three forces P, Q, R acting
at the point O be represented in
magnitude and direction by the
three sides AB, BC, CA
respectively of the triangle ABC.
Then + + = 0

Lami's Theorem

If three coplanar forces acting at a point are in equilibrium then each is

proportional to the sine of the angle between the other two.

In other words; if a particle is in equilibrium under the action of three forces

P, Q, R then = = , ,

.

Let , , ℎ ℎ
, .

, + + = 0 … … ( )
⇒ × + × + × = 0
⇒ × = × … … ( ) ( ∵ × = 0 )

Again from (i)

× + × + × = 0

⇒ × = × … … ( )( ∵ × = 0 )

From (ii) and (iii)

× = × = ×

⇒ × = × = ×

⇒ PQsin = sin = sin

ℎ , , , ,

Dividing by PQR and taking reciprocal we get

= =
sin sin sin

1.Three forces acting on a particle are in equilibrium; the angle
between the first and second is 90° and that between the second

and third is 120°; find the ratio of the forces.

Solution: or, = =
1
Let the three forces P, Q, R acting at O 3 1
are in equilibrium. Then we have
POQ = 90°, QOR = 120° and 22
ROP = 150°. We have to find the
ratio of the forces i.e. P : Q : R. or, = =

By Lami's Theorem we have 312

or, : : = 3: 1: 2

= =
sin 120 sin 150 sin 90

2. The sides AB and AC of a triangle ABC are bisected in D and E;

show that the resultant of forces represented by BE and DC is

represented in magnitude and direction by

Solution:

Since AB and AC are bisected at D and E

so, = and =
2 2

Now, resultant force represented by BE and

DC

= +

= + + +

= + + +

22

= 4 + +

2

= 4 − = 3

22

5. A heavy chain has weights of 10 and 16 kg. attached to its ends and
hangs in equilibrium over a smooth pulley. If the greatest tension of
the chain is 20 kg. wt., find the weight of the chain.

Solution:

Let w1 be the weight of the part of the chain (from top to bottom)
containing the weight of

10 kg at its end. Then,
T = w1 + 10 or,20 = w1 + 10
or,w1 = 10 kg

Again, 2 be the weight of the part of the chain, containing the
weight of 16 kg at its another end. Then,

T = w2 + 16 or,20 = w2 + 16
or,w2 = 4 kg

 Total weight of the chain = w1 + w2 = 10 + 4 = 14 kg Ans.

6. Two men carry a weight 50N between two strings fixed to the weight; one string is inclined
at 30° to the vertical and the other at 60°, find the tension of each string.

Let T1 and T2 be the tension of the strings OA
and OB respectively, W = 50 N, be the

weight of the weight along OC. Such that
AOD = 30° and BOD = 60°.

Using Lamils theorem.

T1 = T2 = W

sin AOC sin BOC sinAOB

or, T1 1=50° T2 1=20° 50
sin sin sin90°
Ans: 25N and 25 2N

8. A body of weight 65 N is suspended by two strings of lengths 5
and 12 m attached to two points in the same horizontal line
whose distance apart is 13 m. find the tensions of the strings.

Solution: Suppose AC and BC be two strings
of lengths 5 m and 12m respectively and the
horizontal line = 13 .

Since (AC)2 + (BC)2 = (AB)2
i.e. 52 + 122 = 132
Hence, angle ACB = 90°
Suppose the weight of 65N be acting vertically
downwards along the line CD.
We produce DC to meet AB at E.
Then, CE  AB

Let CBA = 
 ACE = 90° - BCE =90-(90- ) =

If T1 and T2 are the tensions along the strings CA and CB, then three
tensions T1, T2 and 65N acting at C form in equilibrium.

Then by Lami's theorem, we have
1 = 2 = 65

sin< sin< sin<

1 = 2 = 65

sin(90+ ) sin(180− ) sin 90

1 = 2 = 65

cos sin 1

 T1 = 65 cos, T2 = 65 sin ............... (i)

But from ACB, we get
cos = BC = 12
AB 13

 sin = AC = 5
AB 13

 From (i), we get

1 = 65 × 12 = 60 and 2 = 65 × 5
13 13

9.The ends of an inelastic and weightless string 0.17 m long are attached to two
points 0.13 m apart in the same horizontal line and a weight of 4 N is attached
to string 0.05 m form one end. Find the tension in each portion of the string.
Solution:
Let A and B be two points such that AB = 0.13 m. Let a weight of 4N is

attached to the string at point C, which is 0.05 m from the end A. So that
AC = 0.05 m and BC = (0.17 - 0.05)= 0.12

Since, (0.05)2 + (0.12)2 = (0.13)2
 (AC)2 + (BC)2 = (AB)2
So the angle ACB is a right angle.

Let the line of action of the weight 4N be produced to meet AB at D.

Let CBA = . So that ACD = .

Let T1 and T2 be the tensions along the strings CA and CB respectively. Since three
forces T1, T2 and 4N acting at C are in equilibrium. Then by Lami's theorem we have

1 = 2 = 4

sin< sin< sin<

1 = 2 = 4

sin(90+ ) sin(180− ) sin 90

1 = 2 = 4

cos sin

cos =ABBC =00..1132
 sin =AABC 0.05
= 0.13

Hence 1 = 4 × 0.12 = 3.69N and 2 = 4 × 0.05 = 1.54
0.13 0.13

10. A uniform sphere of weight 3 N rests in contact with a smooth vertical wall. It is
supported by a string whose length equals the radius of the sphere, joining a point on
the surface of the sphere to a point of the wall. Calculate the tension in the string
and the reaction of the wall.

Solution:

Suppose the point of contact of the wall and the sphere be A and let GH be
the string where G and H are the point on the sphere and the point on the
wall respectively. Let the weight of 3N of the sphere acting at O and the
tension T in the string HG and the reaction R

at point A keeping the sphere in equilibrium. Hence, these forces must
pass through a point i.e. HG produced must passes through 0, the
intersection of 3N and R. Since HG = OG =OA and HAO = 90°.
Hence G is the middle point of OH.

Now in right angle triangle AOH, we have

AG = OG = HG

 AG = OG = OA.
Hence, the triangle AOG is equilateral.

 AOG = 60°

Hence by applying Lami's theorem.

= = 3
sin 90 sin(90+60) sin(180−60)

……………..

∴ = 3.46 = 1.73

11.Forces P, Q, R acting along OA, OB, OC where O is the circumcenter of

the triangle ABC, are in equilibrium, show that
(i) = =
= ( + − )
(ii) ( ( + − )
+ − ) =

Soln: Let O be the circumcenter of the triangle ABC.

Since the angles at O are double of the corresponding

angles at the circumference.

 BOA = 2A, COA = 2B and AOB = 2C .. (i)

First part :

Now by Lami’s Theorem we have

= =

sin sin sin

or, = =

sin 2 sin 2 sin 2

= =
2sin cos 2sin cos 2sin cos

or, 2Rsin cos = 2Rsin cos = 2Rsin cos

where R denote the radius of circumscribed circle of the triangle ABC such that a =
2Rsin .

∴ = = … … .

cos cos cos

proved.

Second part

We know that cos = 2 + 2 − 2 , cos = … , cos = …..

2

substituting the value of cos , cos , cos

……………………………….

= =
2( 2+ 2− 2) 2( 2+ 2− 2) 2( 2+ 2− 2)

12. O is the orthocenter of the triangle ABC. Forces P, Q, R acting along

OA, OB, OC are in equilibrium. Prove that = =

. . = =