Components or resolved forces

We have seen that two forces can be combined into a single force which is called their resultant.

There is the reverse process which consists of expressing a single force in terms of its components. These components are sometimes referred to as the resolved parts of the force.

F

x

y

O X

Y

OX = F cos

OY = F sin

Learning objectivesSplitting forces into their

componentsFinding the resultant of two or more forces

Example

30 Y

X

20 N

30°

20 N

X = 20 x Cos 30 =

17.3 N

Y = 20 x Sin 30 =

10.0 N

More examples

X = -15 Cos 42 =

-11.1 N

Y = 15 Sin 42 =

10.0 N

X = 35 Cos 62 =

16.4 N

Y = -35 Sin 62 =

- 30.9 N

x

y

42°

15 N

x

y

62°

35 N

x

y

32°10 N

X = -10 Sin 32 =

- 5.30 N

Y = -10 Cos 32 =-8.48 N

Finding the resultant of two forces using the component method

30°

8 N

5 N

X = 8 + 5 x Cos 30 =

12.33 N

Y = 5 x Sin 30 =2.5 N

12.33

2.5 N

Resultant

Resultant =

22 5.233.12 = 12.6 N

8 N 5cos30

5sin3030°

tan -1(2.5/12.33) 11.5

Draw a force diagram and, where appropriate, redraw each force into any two perpendiculardirections.

Find the resultant in each direction.

Find the resultant of the following forces:

2 N

6 N

30°

2 N

6 N

30°

6 cos 30° N

6 sin 30° N

2 N

Horizontal componentX = 6 cos 30° + 2

= 7.196…

Vertical componentY = 6 sin 30°

= 3

Find the overall resultant.

22 3...19.7 F= 7.79… 7.1… N

3 N F N

The resultant force has magnitude 7.8 N (2 s.f.)

Find the resultant of the following forces:

10 N

4 N

120°

Draw a force diagram and, where appropriate, redraw each force resolved into any twoperpendicular directions. 10 N

4 N

120°60°

4 sin 60° N

10 N4 cos 60° N

Find the resultant in each direction.

Find the overall resultant.

Horizontal componentX = 10 – 4 cos 60°

= 8

Vertical componentY = 4 sin 60° = 3.46…

22 8...46.3 F

= 8.71…

The resultant force has magnitude 8.7 N (2 s.f.)

8 N

3.46… N F N

More than two forces

22 YX

Split each force into components in these directions.

Choose two directions at right angles to each other.

For each direction, find the sum of the components.

Find the resultant.

Find the required angle.

Pythagoras: R

Angle with the X direction =

X

Y1tan

Example

22 YX

x

y20 N

70°

12 N

8 N

30°

Force X Y

20 N 20 cos 70 = 6.8404 20 sin 70 = 18.7939

12 N 12.0000 0

8 N -8 cos 30 =- 6.9282 -8 sin 30 = - 4.0000

Total 11.9122 14.7939

Resulta

nt

R =

R = 19.0 N

=

X

Y1tan

= 51.2°

Example

x

y

6 N

30 N

50°

20 N

35°5 N

X = 20 cos 35 – 30 sin 50 – 5 =

- 11.5983

Y = 6 – 20 sin 35 – 30 cos 50 =

- 24.7552

22 YX R = = 27.3 N

=

X

Y1tan = 64.9°

More exampleTwo forces act at a point. The magnitude of the forces are 3.95 N and 2.5 N, and angle between their direction is 90 + , where 0 << 90. The resultant of the two forces has magnitude R N and its line of action makes an angle of 90 with the force of magnitude 2.5 N, as shown in the diagram. Find the value of R and .

R N

2.5 N

Resolving

3.95 N

3.95 sin = 2.5 : sin = 2.5/3.95 : Giving = 39.3°

Resolving R =

3.95 x cos 39.3 = 3.06 N

Three forces, of magnitudes 8 N, 10 N, and 8 N, act at a point P in the direction shown in the diagram. PQ is the bisector of the acute angle between the two forces of magnitude 8 N.

Find

(i) the components of the resultant of the three forces a parallel to PQ (a) parallel to PQ (b) perpendicular to PQ

(ii) the magnitude of the resultant of the three forces,

(iii) the angle that the resultant of the three forces makes with PQ.

8 N

8 N

10 N

P

Q40°

40°

130°

150°

Resolving forces horizontally and vertically:

10cos 10° N

10sin 10° N 8sin 40° N

8cos 40° N

8cos 40° N

8sin 40° N

8 N

8 N

10 N

P

Q40°

40°

130°

10°

i) the components of the resultant of the three forces(a) parallel to PQ

Horizontal component X = 8 cos 40° + 8 cos 40° – 10 cos 10°

= 2.4086…= 2.41 N (3 s.f.)

Vertical component Y = 10 sin 10° + 8 sin 40° – 8 sin 40°

= 1.7364…= 1.74 N (3 s.f.)

(b) perpendicular to PQ

ii) the magnitude of the resultant of the three forces,

2 2F X Y 22 ...73.1...40.2

= 2.9693…

The resultant force has magnitude 2.97 N (3 s.f.)

iii) the angle that the resultant of the three forces makes with PQ.

...40.2

...73.1tan

The angle that the resultant makes with PQ is 35.8° (1 d.p.)

...789.35

Two forces act in a vertical plane. The forces have magnitudes 20 N and 7 N and make angles and respectively with the upward vertical, as shown in the diagram. The angles and are such that cos ≈ 0.96 and cos ≈ 0.6. [You are given that sin ≈ 0.28 and sin ≈ 0.8.]

i) Show that the resultant of the two forces acts vertically and find its magnitude.

ii) The two forces act on a particle of mass 2.5 kg. State, giving a reason, whether the direction of the acceleration of the particle is vertically upwards or downwards.

20 N

7 N

Resolving forces horizontallyand vertically:

20 N

7 N

7cos N

20sin N

7sin N

20cos N

i) Show that the resultant of the two forces acts vertically and find its magnitude.

Horizontal component

= 7(0.8) – 20(0.28) = 0

Vertical component

= 20(0.96) + 7(0.6) = 23.4

7 20sin sin 7 20cos cos

Since the resultant force horizontally is 0 N, the overall resultant of the two forces acts vertically and has magnitude 23.4 N.

(ii) The two forces act on a particle of mass 2.5 kg. State, giving a reason, whether the direction of the acceleration of the particle is vertically upwards or downwards.

23.4 N

2.5g N = 24.5 N

Since the weight is greater than the upward force (the resultant force), the direction of the acceleration is vertically downwards.