Equilibrium of a particle
If all the forces acting on a particle cancel each other out, so that
nothing happens at all, the forces are said to be in equilibrium.
 The algebraic sum of the horizontal components is zero.
 The algebraic sum of the vertical components is zero.
Example 1
Find the value of P and Q if the following system of forces is in equilibrium.
8N
30
40
PN
2N
QN
Solution
Components diagram
8 sin 30
8 cos 30
Q cos 40
P
2 N Q sin 40
Horizontal component: : 8 cos 30 + Q cos 40  P = 0
[1]
Vertical component:  :
8 sin 30  2  Q sin 40 = 0
 Q sin 40 = 8 sin 30  2
 Q = 3.1114…
[2]
Sub Q into [1]:
8 cos 30 + 3.114 cos 40  P = 0
 P = 8 cos 30 + 3.114 cos 40 = 9.31
Force P is 9.31 N and force Q is 3.11 N
Mathematics made simple
M1 Equilibrium
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Equilibrium of a particle
*Example 2
Find the value of P and Q if the following system of forces is in equilibrium.
8N
PN
30
30
40
QN
Solution
Components diagram
P sin 30
8 sin 30
8 cos 30
Q cos 40
P cos 30
Q sin 40
Horizontal component: : 8 cos 30 + Q cos 40  P cos 30 = 0
 0.866025 P = 6.928203 + 0.766044 Q
 P = 8 + 0.884552 Q
[1]
Vertical component:  :
8 sin 30 + P sin 30  Q sin 40 = 0
 Q sin 40 = 8 sin 30 + P sin 30
 0.642788 Q = 4 + 0.5 P
[2]
Sub P into [2]:
0.642788 Q = 4 + 0.5 (8 + 0.884552 Q)
 0.642788 Q = 4 + 2 + 0.442276 Q
 0.200512 Q = 6
 Q = 29.9234
Sub Q into [1]:
P = 8 + 0.884552  29.9234 = 34.5
Force P is 34.5 N and force Q is 29.9 N
Mathematics made simple
M1 Equilibrium
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Equilibrium of a particle
Example
The following diagram shows a system of forces acting on a particle in a plane. A
third force is added so that the particle rests in equilibrium. Find the magnitude of this
force and the angle that make with the horizontal.
5N
30
6N
Solution
Components diagram
5 sin 30
5 cos 30
F cos
6
F sin
Horizontal component: : 5 cos 30 + F cos  6 = 0
 F cos = 6  5 cos 30
[1]
Vertical component:  :
5 sin 30  F sin = 0
 F sin = 5 sin 30
[2]
F sin 
5 sin 30

F cos  6  5 cos 30
Divide [2] by [1]:
 tan  
2.5
 1.49712
1.66987
  = 56.2591
Sub  into [2]:
F sin 56.2591 = 2.5
 F = 3.00
Added force is 3.00 N, acting at an angle of 56.3 below the horizontal.
Mathematics made simple
M1 Equilibrium
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Equilibrium of a particle
Exercise
1. Given that the following system of forces is in equilibrium, find the unknown force
P and Q.
PN
4N
20
50
38
Q
6N
2. A particle is in equilibrium under the action of three coplanar forces shown in the
diagram.
(i) Show that  = 60
(ii) Find the value of X.
2X N

XN
15 N
3.
20 N

12 N
PN
Three forces, of magnitude 20 N, 12 N and P N, act at a point in the direction shown
in the diagram. The forces are in equilibrium. Find
(i)
The value of  ,
(ii)
The value of P.
The force of magnitude 12 N is now removed.
(iii) Find the magnitude and direction of the resultant of the two remaining
forces.
1. P = 9.18 N and Q = 9.66 N 2.  = 60 and X = 8.66 N
Mathematics made simple
3. (i) 53.1 (ii) 16 N (iii) 12 N to right.
M1 Equilibrium
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Equilibrium of a particle
Types of force
Weight (W)
The weight of a body is the force with which the earth
attracts it.
W = mg: Weight always acts downwards.
The value of g is 9.8
W = mg
Tension (T)
T
The force acts along a string, wire or rope.
W = mg
The normal reaction (Rn)
R
When an object is in contact with a
surface, there is a force on the object at
right angles to the surfaces in contact. This
is called the normal reaction (Rn).
W
Friction (Ff)
This force acts due to roughness between an object
and a surface. It always acts against motion (or likely
motion). On a smooth surface the friction is zero.
R
P
Fr
W
Forces on an inclined plane
Forces are resolved parallel and perpendicular to plane.
Components of the weight
mg sin  N


mg cos N
w = mg N
Mathematics made simple
M1 Equilibrium
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Equilibrium of a particle
Example
The system is in equilibrium, find Rn
and Ff.
Rn
Ff
12 N
2 kg
Resolve:  12  Ff = 0  Ff = 12 N
Resolve:  Rn  2g = 0  Rn = 2  9.8 = 19.6 N
2g N
The friction force is 12 N and the normal reaction is 19.6 N.
Example
The forces acing are in equilibrium, find
the reaction Rn and the friction force Ff.
Rn
Ff
10 N
60
6 kg
Solution
Components diagram
Resolve:  10 cos60  Fn = 0
 Ff = 10 cos60 = 5 N.
Resolve: 
Rn + 10 sin60  6g = 0
Rn + 8.66  58.8 = 0
 Rn = 50.1 N.
Rn 10 sin 60
Ff
10 cos 60
6g N
Example
The forces acing are in equilibrium,
find the reaction R and the friction
force F.
R
F
6 kg
10 N
60
Solution
Components diagram
Rn
Resolve:  Ff  10 cos60 = 0
 Ff = 10 cos60 = 5 N.
Resolve: 
10 cos 60
Rn  10 sin60  6g = 0
Rn  8.66  58.8 = 0
 Rn = 67.5 N.
Mathematics made simple
M1 Equilibrium
Ff
6g 10 sin 60
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Equilibrium of a particle
Example
The forces acing are in equilibrium, find
the reaction R and the friction force F.
R
5N
30
F
10 N
60
6 kg
Solution
Components diagram
Rn 10sin60
Resolve:  10 cos60 + 5 cos30  F = 0
 F = 10 cos60 + 5 cos30 = 9.33 N.
10cos60
5 cos30
Ff
R + 10 sin60  5 sin30  6g = 0
R + 8.66 2.5  58.8 = 0
 R = 52.6 N.
Resolve: 
Example
Mass of 8 kg is suspended, in
equilibrium, by two light
inextensible strings which make
angles of 30 and 45 with the
horizontal. Calculate the tension in
the strings.
5 sin30 6g
S
T
30
45
8g N
Solution
Component diagram
T sin30 S sin45
S cos 45
T cos30
8g
Re s :  S cos 45  T cos 30  0 
2
2
Re s :  S sin 45  T sin 30  8g  0 
[2]  [1]:
S
2
2
T 0
[1]
S  12 T  78.4
[2]
3
2
1 3
T  78.4  T = 57.4 N and S = 70.3 N
2
Equilibrium of a particle
Mathematics made simple
M1 Equilibrium
-7-
Inclined plane
Force acting parallel to the plane
Example
A particle of mass 5 kg rests on a smooth plane inclined at 30 to the horizontal. A
Force P N acts on the particle up the plane along the line of greatest slope. Find the
magnitude of the normal reaction force and the magnitude of the force P.
Solution
Force diagram
Rn N
Parallel and Perpendicular components
Rn
PN
P
5g sin30
30
5g N
30
5g cos30
Parallel the plane to the plane:
P  4g sin30 = 0  P = 24.5 N
Perpendicular to the plane:
Rn  5g cos30 = 0  Rn = 42.4 N
The reaction force has magnitude 42.4 N and the force P has magnitude 24.5 N.
Horizontal force
Example
A particle of mass 10 kg rests on the surface of a smooth plane which is inclined at an
angle of 30 to the horizontal by means of a horizontal force P.
Find P and Rn.
Solution
Force diagram
Components diagram
Rn
P cos 30
Rn
PN
10g sin 30
30
10g N
10g cos30 P sin 30
Resolving parallel to the plane:
P cos30 – 10g sin30 = 0
 Pcos30 = 10gsin30  p = 10  9.8  tan30 = 56.58
Resolving perpendicular to the plane: Rn  10g cos30  P sin30 = 0
Rn = 10gcos30 + Psin30 = 113
The reaction force has magnitude 113 N and the force P has magnitude 56.6 N.
Equilibrium of a particle
Mathematics made simple
M1 Equilibrium
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Force acting at an angle to the plane
Example
A particle of mass 10 kg rests on the surface of a smooth plane which is inclined at an
angle of 30 to the horizontal by means of a force P which is acting at angle of 45 to
the plane.
Find P and Rn.
Force diagram
Components diagram
Rn
PN
Rn P sin 45
P cos 45
45
10g sin 30
30
10g N
10g cos30
Resolving parallel to the plane:
P cos45– 10g sin30 = 0
 Pcos45 = 10gsin30  p = 10  9.8  sin30  cos45 = 69.296
Resolving perpendicular to the plane: Rn + P sin45  10g cos 30 = 0
Rn = 10gcos30  P sin45= 35.9
The reaction force has magnitude 35.9 N and the force P has magnitude 69.3 N.
Example
A particle of mass 10 kg rests on the surface of a smooth plane which is inclined at an
angle of 30 to the horizontal. When a force P acting up the plane and a horizontal
force of 163 N are applied to the particle, it rests in equilibrium.
Find P and Rn.
Force diagram
Components diagram
Rn
Rn
PN
P
163 cos30
163 N
10g sin 30
30
10g N
10g cos30 163 sin30
Resolving parallel to the plane:
P + 163 cos30  10g sin30 = 0
 P = 10g sin30  163 cos30 = 25
Resolving perpendicular to the plane: Rn  163 sin30  10g cos 30 = 0
Rn = 163 sin30 + 10g cos 30 = 98.7
The reaction force has magnitude 98.7 N and the force P has magnitude 25 N.
Exercise
Mathematics made simple
Rn
M1 Equilibrium
-9-
1.
The forces acing are in equilibrium, find
the reaction Rn and the friction force F.
5N
F
2.
The forces acing are in equilibrium, find
the reaction R and the friction force F.
R
6N
F
60
6 kg
30
6 kg
5N
60
3. Mass of 5 kg is suspended, in equilibrium, by two light inextensible strings which
make angles of 30 and 45 with the horizontal. Calculate the tension in the strings.
4. A particle of mass 4 kg is attached to the lower end of an inextensible string. The
upper end of the string is fixed. A horizontal force 25 N and upward vertical force of
10 N acts upon the particle, which is in equilibrium with the string making an angle 
with the vertical. Calculate the tension in the string and the angle .
5.A particle of mass 8 kg rests on a smooth plane inclined at 30 to the horizontal. A
Force P N acts on the particle up the plane along the line of greatest slope. Find the
magnitude of the normal reaction force and the magnitude of the force P.
6.A particle of mass 5 kg rests on the surface of a smooth plane which is inclined at
an angle of 30 to the horizontal by means of a horizontal force P.
Find P and Rn.
7.A particle of mass 10 kg rests on the surface of a smooth plane which is inclined at
an angle of 30 to the horizontal by means of a force P which is acting at angle of 45
to the plane.
Find P and Rn.
8. A particle of mass 6 kg rests on the surface of a smooth plane which is inclined at
an angle of 30 to the horizontal. When a force P acting up the plane and a horizontal
force of 12 N are applied to the particle, it rests in equilibrium.
Find P and Rn.
9. A particle of mass 6 kg rests on the surface of a smooth plane which is inclined at
an angle of 30 to the horizontal. When a force P acting at an angle 40 to the plane
and a horizontal force of 12 N are applied to the particle, it rests in equilibrium.
Find P and Rn.
1. F = 2.5 N
Rn = 63.1 N
4.  = 41
T = 38.4 N
7. P = 69.3 N
Rn = 134 N
Mathematics made simple
2. 7.7 N Rn = 57.5 N
5. P = 39.2 N
8. P = 19.0 N
Rn = 67.9 N
Rn = 56.9 N
M1 Equilibrium
3. S =35.9 N T = 43.9 N
6. P = 28.3 N
Rn = 56.6 N
9. P = 7.18 N
Rn = 52.4 N
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