Components or resolved forces

1. y Y F  O x X Learning objectives Components or resolved forces 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. OX = F cos  OY = F sin 

2. 20 N 30° 20 N Y 30 X Example X = 20 x Cos 30 = 17.3 N Y = 20 x Sin 30 = 10.0 N

3. y 15 N 42° x y x 62° 35 N y x 32° 10 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 = -10 Sin 32 = - 5.30 N Y = -10 Cos 32 = -8.48 N

4. 5 N 5sin30 30° 30° 5cos30 8 N 8 N Resultant 2.5 N  12.33 Finding the resultant of two forces using the component method X = 8 + 5 x Cos 30 = 12.33 N Y = 5 x Sin 30 = 2.5 N Resultant = = 12.6 N  tan -1(2.5/12.33)  11.5

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

6. 3.46… N F N 4 N 8 N 120° 10 N 4 sin 60° N 4 N 120° 60° 10 N 10 N 4 cos 60° N Find the resultant of the following forces: Draw a force diagram and, where appropriate, redraw each force resolved into any two perpendicular directions. Horizontal component Vertical component Find the resultant in each direction. X = 10 – 4 cos 60° Y = 4 sin 60° = 3.46… = 8 Find the overall resultant. = 8.71… The resultant force has magnitude 8.7 N (2 s.f.)

7. 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  =

8. y 20 N 70° x 30° 12 N 8 N Resultant  Example Force X Y 20 N 20 cos 70 = 6.8404 20 sin 70 = 18.7939 R = 12 N 12.0000 0 R = 19.0 N 8 N -8 cos 30 =- 6.9282 -8 sin 30 = - 4.0000  = Total 11.9122 14.7939  = 51.2°

9. y 6 N x 5 N 35° 50° 20 N 30 N  Example X = 20 cos 35 – 30 sin 50 – 5 = - 11.5983 Y = 6 – 20 sin 35 – 30 cos 50 = - 24.7552 R = = 27.3 N = 64.9°  =

10. 3.95 N R N  2.5 N 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 . Resolving  3.95 sin  = 2.5 : sin  = 2.5/3.95 : Giving  = 39.3° Resolving  R = 3.95 x cos 39.3 = 3.06 N

11. 8 N 130° 10 N 40° Q 40° P 150° 8 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 • 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.

12. Resolving forces horizontally and vertically: 8 N 10 N 10sin 10° N 8sin 40° N 130° 40° Q 8cos 40° N P 10° 40° 8cos 40° N 10cos 10° N 8 N 8sin 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° = 2.4086…= 2.41 N (3 s.f.) (b) perpendicular to PQ Vertical component Y = 10 sin 10° + 8 sin 40° – 8 sin 40° = 1.7364…= 1.74 N (3 s.f.)

13. ii) the magnitude of the resultant of the three forces, = 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. The angle that the resultant makes with PQ is 35.8° (1 d.p.)

14. 20 N 7 N • 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.

15. 20 N 7 N 20cosN 7cos N 7sin N 20sin N • Show that the resultant of the two forces acts vertically and find its magnitude. Resolving forces horizontally and vertically: Horizontal component Vertical component = 23.4 = 20(0.96) + 7(0.6) = 7(0.8) – 20(0.28) = 0 Since the resultant force horizontally is 0 N, the overall resultant of the two forces acts vertically and has magnitude 23.4 N.

16. (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.