**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.**

**Solution**

Components diagram

Horizontal component: : 8 cos 30 + Q cos 40 P = 0 [1] Vertical component: : 8 sin 30 2 Q sin 40 = 0 [2]

Q sin 40 = 8 sin 30 2 Q = 3.1114…

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**

8 N
Q N
2 N
P N 30 _{40 }
8 cos 30
Q cos 40
8 sin 30
2 N Q sin 40
P

**Equilibrium of a particle**

***Example 2**

**Find the value of P and Q if the following system of forces is in equilibrium.**

**Solution**

Components diagram

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**

8 N Q N P N 30 40 30 8 cos 30 Q cos 40 8 sin 30 Q sin 40 P cos 30 P sin 30

**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.

**Solution**

Components diagram

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]
Divide [2] by [1]: 5 30
6 5 30
*F sin* *sin*
*F cos* *cos*
2 5 1 49712
1 66987
*.*
*tan* *.*
*.*
= 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.**
30
5 N
6 N
5 cos 30
F cos
5 sin 30
F sin
6

**Equilibrium of a particle**

**Exercise**

1. Given that the following system of forces is in equilibrium, find the unknown force

**P and Q.**

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.**

3.

**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.
4 N
6 N
P N
20
50
38
Q
2X N
X N
15 N
20 N
12 N
**P N**

1. P = 9.18 N and Q = 9.66 N 2. = 60 and X = 8.66 N 3. (i) 53.1 (ii) 16 N (iii) 12 N to right.

**Equilibrium of a particle**

### Types of force

**Weight (W)**

**Tension (T)**

The force acts along a string, wire or rope.

**The normal reaction (R**

**n**

**) **

**Friction (F**

**f**

**)**

**Forces on an inclined plane**

Forces are resolved parallel and perpendicular to plane.
**Components of the weight**

w = mg N mg cos N mg sin N W = mg T

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).**

R

W

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

W

P F

r

**The weight of a body is the force with which the earth **
attracts it.

W = mg: Weight always acts downwards.

**Equilibrium of a particle**

**Example**

Resolve: 12

###

Ff = 0 Ff = 12 NResolve: Rn

###

2g = 0 Rn = 2 9.8 = 19.6 N**The friction force is 12 N and the normal reaction is 19.6 N.**
**Example**
**Solution**
**Components diagram**
**Example**
**Solution**
** Components diagram**
10 cos 60
F_{f}
R_{n }10 sin 60
6g N
F_{f}
R_{n }
6g 10 sin 60
10 cos 60

The system is in equilibrium, find Rn

and Ff.
2 kg
R_{n}
12 N
F_{f}
2g N

The forces acing are in equilibrium, find the reaction Rn and the friction force Ff.

6 kg
Rn _{10 N}
Ff 60
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.

The forces acing are in equilibrium, find the reaction R and the friction

force F. _{6 kg}
R _{10 N}
F 60
Resolve: Ff

###

10 cos60 = 0 Ff = 10 cos60 = 5 N. Resolve: Rn###

10 sin60 6g = 0 Rn###

8.66 58.8 = 0 Rn = 67.5 N.**Equilibrium of a particle**

**Example**

**Solution**Components diagram

**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.
**Solution**
** Component diagram**
10cos60
5 cos30
F_{f}
R_{n }10sin60
5 sin30 6g
45
30
S
T
8g N

The forces acing are in equilibrium, find the

reaction R and the friction force F. 5 N

6 kg

R _{10 N}

F 30 60

Resolve: 10 cos60 + 5 cos30

###

F = 0 F = 10 cos60 + 5 cos30 = 9.33 N. Resolve: R + 10 sin60 5 sin30###

6g = 0R + 8.66 2.5

###

58.8 = 0 R = 52.6 N.3
2
2 2
45 30 0 0
*Re s :* *S cos* *T cos* *S* *T* [1]
2 1
2 2
45 30 8 0 78 4
*Re s : S sin* *T sin* *g* *S* *T* *.* [2]
[2]

###

[1]: 78.4 2 3 1 _{T}

_{ T = 57.4 N and S = 70.3 N}

**Equilibrium of a particle**

**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**

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**
T cos30
T sin30 S sin45
8g
S cos 45
5g N
30
R_{n }N
P N
30
R_{n}
5g cos30
5g sin30
P

**Force diagram** **Components diagram**

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**

**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**

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 163 N are applied to the particle, it rests in equilibrium.

Find P and Rn.

**Force diagram** **Components diagram**

30 P cos 30 10g cos30 P sin 30 P N 10g sin 30 10g N R n 30 P cos 45 10g cos30 P N 10g sin 30 10g N R n 45 Rn P sin 45 Rn

Resolving parallel to the plane: P + 163 cos30 10g sin30 = 0 P = 10g sin30 163 cos30 = 25

Resolving perpendicular to the plane: Rn 163 sin30 10g cos 30 = 0

Rn = 163 sin30 + 10g cos 30 = 98.7

**The reaction force has magnitude 98.7 N and the force P has magnitude 25 N.**

**Exercise**

**1.**

2.

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.
30
P
10g cos30 163 sin30
P N
10g sin 30
10g N
R_{n}
163 N
R
n
163 cos30

The forces acing are in equilibrium, find the reaction Rn and the friction force F.

6 kg

Rn

5 N

F 60

The forces acing are in equilibrium, find the

reaction R and the friction force F. 6 N

6 kg

R _{5 N}

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 2. 7.7 N Rn = 57.5 N 3. S =35.9 N T = 43.9 N

4. = 41 T = 38.4 N 5. P = 39.2 N Rn = 67.9 N 6. P = 28.3 N Rn = 56.6 N