# Equilibrium of a Particle

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## 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

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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.

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

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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.

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

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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.

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

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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.

### Tension (T)

The force acts along a string, wire or rope.

n

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.

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Example

Resolve:  12

### 

Ff = 0  Ff = 12 N

Resolve:  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 Ff Rn 10 sin 60 6g N Ff Rn 6g 10 sin 60 10 cos 60

The system is in equilibrium, find Rn

and Ff. 2 kg Rn 12 N Ff 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.

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### 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 Ff Rn 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 = 0

R + 8.66 2.5

### 

58.8 = 0  R = 52.6 N.

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3 2 2 2 45 30 0 0 Re s :S cosT cos   ST  [1] 2 1 2 2 45 30 8 0 78 4 Re s : S sin T sing   ST. [2] [2]

### 

[1]: 78.4 2 3 1   T  T = 57.4 N and S = 70.3 N

### 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 Rn N P N 30 Rn 5g cos30 5g sin30 P

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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 163 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

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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

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 163 sin30 P N 10g sin 30 10g N Rn 163 N R n 163 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

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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

References

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