# Components or resolved forces

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Components or resolved forces
Learning objectives Splitting forces into their components Finding the resultant of two or more 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 

Example 20 N X = 20 x Cos 30 = 17.3 N Y = 20 x Sin 30 = 10.0 N 30°
30 Y X 20 N X = 20 x Cos 30 = 17.3 N Y = 20 x Sin 30 = 10.0 N

More examples x y X = -15  Cos 42 = -11.1 N Y = 15  Sin 42 = 10.0 N
42° 15 N X = -15  Cos 42 = -11.1 N Y = 15  Sin 42 = 10.0 N x y X = 35  Cos 62 = 16.4 N 62° 35 N Y = -35  Sin 62 = N x y X = -10  Sin 32 = N 32° 10 N Y = -10  Cos 32 = -8.48 N

Finding the resultant of two forces using the component method
30° 8 N 5 N 8 N 5cos30 5sin30 30° X = 8 + 5 x Cos 30 = 12.33 N Y = 5 x Sin 30 = 2.5 N 12.33 2.5 N Resultant Resultant = = 12.6 N   tan -1(2.5/12.33)  11.5

Find the resultant of the following forces:
30° Draw a force diagram and, where appropriate, redraw each force into any two perpendicular directions. 6 cos 30° N 6 sin 30° N 2 N 2 N 6 N 30° Find the resultant in each direction. Horizontal component Vertical component X = 6 cos 30° + 2 Y = 6 sin 30° = 7.196… = 3 Find the overall resultant. 7.1… N 3 N F N = 7.79… The resultant force has magnitude 7.8 N (2 s.f.)

Find the resultant of the following forces:
120° Draw a force diagram and, where appropriate, redraw each force resolved into any two perpendicular directions. 10 N 4 N 120° 60° 4 sin 60° N 10 N 4 cos 60° N Find the resultant in each direction. Horizontal component Vertical component X = 10 – 4 cos 60° Y = 4 sin 60° = 3.46… = 8 Find the overall resultant. 8 N 3.46… N F N = 8.71… The resultant force has magnitude 8.7 N (2 s.f.)

More than two forces  Choose two directions at right angles to each other.  Split each force into components in these directions.  For each direction, find the sum of the components.  Find the resultant.  Pythagoras: R   Find the required angle.  Angle with the X direction  =

Example y x Resultant Force X Y  R = R = 19.0 N  =  = 51.2° 20 N
70° 8 N 30° 12 N Resultant Force X Y 20 N 20 cos 70 = 20 sin 70 = R = 12 N R = 19.0 N 8 N -8 cos 30 = -8 sin 30 =  = Total  = 51.2°

Example x y X = 20 cos 35 – 30 sin 50 – 5 = - 11.5983 Y =
50° 20 N 35° X = 20 cos 35 – 30 sin 50 – 5 = Y = 6 – 20 sin 35 – 30 cos 50 = R = = 27.3 N  = = 64.9°

More example Two 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 . 3.95 N R N 2.5 N Resolving  3.95 sin  = 2.5 : sin  = 2.5/3.95 : Giving  = 39.3° Resolving  R = 3.95 x cos = 3.06 N

the components of the resultant of the three forces a parallel to PQ
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 the components of the resultant of the three forces a parallel to PQ (a) parallel to PQ (b) perpendicular to PQ the magnitude of the resultant of the three forces, (iii) the angle that the resultant of the three forces makes with PQ. 8 N 10 N P Q 40° 130° 150°

Resolving forces horizontally and vertically:
P Q 40° 130° 10° 10cos 10° N 10sin 10° N 8sin 40° N 8cos 40° N 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° = …= 2.41 N (3 s.f.) (b) perpendicular to PQ Vertical component Y = 10 sin 10° + 8 sin 40° – 8 sin 40° = …= 1.74 N (3 s.f.)

ii) the magnitude of the resultant of the three forces,
= … The resultant force has magnitude 2.97 N (3 s.f.) iii) the angle that the resultant of the three forces makes with PQ. The angle that the resultant makes with PQ is 35.8° (1 d.p.)

Two forces act in a vertical plane
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.] 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 horizontally and vertically:
Show that the resultant of the two forces acts vertically and find its magnitude. Resolving forces horizontally and vertically: 20 N 7 N 7cos N 20sin N 7sin N 20cos N Horizontal component Vertical component = 7(0.8) – 20(0.28) = 20(0.96) + 7(0.6) = 23.4 = 0 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
(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.