TOPIC 9

The Equilibrium of Particles and Rigid Bodies

Resultant of Horizontal ForcesA PARTICLE is a very small object, it has a mass but no size.A RIGID BODY has both a mass and a size.EQUILIBRIUM is when a particle or rigid body does not move when acted on by two or more forces.The RESULTANT of a number of forces is a single force which would produce exactly the same effect acting alone as the other forces would when acting together.

Eg1

Eg 2

4N7N

3N

is equivalent to

2N10N 8NIs equivalent to

Resultant of Forces at an AngleThe resultant of two forces at right angles to one another is

represented by the diagonal of the rectangle whose sides

represent the two forces.

eg

F = √72 + 22 Tan θ = 2

= √49 + 4 7

= √53 θ = 15.95°

= 7.28N

θ

2NF

7N

is equivalent to

Resultant of Forces at an AngleSince any two forces at right angles may be combined to

produce a resultant, then any force can be resolved into

two mutually perpendicular forces called the components of

the force (we have already done this in Topic 5)

A single force F at an angle θ to the horizontal can be split

into the two mutually perpendicular forces Fcosθ in the

horizontal direction and Fsinθ in the vertical direction.

7N

is equivalent to

7sin27° = 3.18N

7cos27° = 6.24N

27°

Resultant of Forces at an Angle – Example 1

Demonstrate that the system of forces on the left is

equivalent to the single force on the right.

16N

10N

8N12N

60°30°

40.4°

4.74N

Resultant of Forces at an Angle – Answer 1

Draw a table to find the resultant of the forces.

Force Component in x-direction Component in y-direction

10N 10N O

16N 0N -16N

8N 8 cos 60° = 4N 8 sin 60° = 6.928N

12N -12 cos 30° = -10.392N 12 sin 30° = 6N

Total 3.608N -3.072N

Resultant of Forces at an Angle – Answer 1

The resultant is:

F = √3.6082 + 3.0722 Tan θ = 3.072

= √13.02 + 9.44 3.608

= √22.46 = 0.85

= 4.74N θ = 40.4°

3.608N

-3.072N

θ

F

Resultant of Forces at an Angle – Example 2

A particle of mass 4kg is attached to the lower end of a

light inextensible string. The upper end is fixed to a wall. A

horizontal force of P newtons is applied to the free end of

the string so that the string makes an angle of θ with the

downward vertical and experiences a tension of 200N. If

the 4kg mass rests in equilibrium, find P and θ.

Resultant of Forces at an Angle – Answer 2

Resolving vertically200Cosθ = 4gCosθ = 4g Resolving

horizontally 200 200sinθ = P= 4 x 10 200 x sin 78.46° = P 200 200 x 0.978 = P= 0.2 P = 195.96 N

θ = cos-1(0.2)θ = 78.46°

200 N

4g N

θ

P

θ

Resultant of Forces at an Angle – Example 3

A particle of mass mkg is attached to the lower end of a

light inextensible string. The upper end of which is fixed to

a wall. A horizontal force of 40N is applied to the free end

of the string so that the string makes an angle of θ with the

downward vertical and experiences a tension of 90N. If

the particle rests in equilibrium, find θ and m.

Resultant of Forces at an Angle – Answer 3

Resolving horizontally Resolving horizontally

90sinθ = 40 90cosθ = mgsinθ = 40 90 x cos26.39° = m x

10 90 90 x 0.896 = 10m = 0.444 80.64 = 10m

θ = sin-1(0.444) m = 80.64θ = 26.39° 10

m= 8.06 kg

90 N

mg

θ

40N

θ

Resultant of Forces at an Angle – Example 4A body of mass 5kg is supported by two inextensible strings,

the other ends of which are attached to two fixed points P and

Q in a ceiling. The 5kg mass rests in equilibrium with one

string experiencing a tension T newtons and inclined at 30° to

the horizontal and the other experiencing a force of S newtons

and inclined at 45° to the horizontal. Find T and S.

Resultant of Forces at an Angle – Answer 4

Resolving horizontally Substitute 1 into 2Scos45° = Tcos30° 1.225T x cos45° +Tcos60° = 5g S = Tcos30° 1.225T x 0.707 + T x 0.5 = 5x10

cos45° 0.866T + 0.5T = 50 S = 1.225T 1.366T = 50

T = 50 = 36.6NResolving vertically 1.366Scos45° + Tcos60° = 5g

Use 1 to find SS = 1.225 x 36.6 = 44.8N

S

5g

45° 60°

30°45°

T