1. Forces and moments
Resolving forces

2. Forces and moments
Example 1
Drawing to scale

3. Weight suspended by two ropes
Draw the perpendicular
Identify the angles between the forces A and B and the perpendicular
Draw the triangle using the angles A B
20o A
55o
55o
35o
70o
2000 Newtons
105o
The length of the sides of the triangle represent the magnitude of the forces NOT the length of rope
2000 N B
20o

4. Using the sine rule (if you know the angles)
a/sin A = b/Sin B = c/sin C
angle A = 20o angle B = 55o (opposites to sides a & b)
20o
55o
105o a
Angle C = 105o and side c represents 2000N
2000 N
(c) b

5. Using the sine rule (if you know the angles)
a/sin A = c/sin C therefore a/sin 20o = 2000/sin105o
a = 2000 x sin 20o/sin105o
708.17N
20o
55o
105o a b
2000 N
(c)
b/sin B = c/sin C therefore b/sin 55o = 2000/sin105o
a = 2000 x sin 55o/sin105o
1696.1N

6. Using the cosine rule ( if you know one angle and two sides)
F2 = 60N
70o
F1 = 30N F3

7. Using the cosine rule ( if you know one angle and two sides)
A2 = B2 + C2 -2BCcosA
70o
F2 = 60N (C)
F1 = 30N (B)
F3 (A)
A =110o
(F3)2 = – 2x60x30x cos110o
= 75.7N

8. Vertical and horizontal components of forces
Sketch the diagram
Fv can be drawn at the other
end of the sketch F Fv FH θ

9. Vertical and horizontal components of forces
Sketch the diagram
sin θ = Fv/F
F.sin θ = Fv F Fv FH θ Fv
cos θ = FH/F
F.cos θ = FH

10. Restoring force of two forces
F3 is the restoring force of F1 and F2
25o
70o
F1(55N)
F2 (25N) F3
25o
70o
Can be drawn to scale
74.8N

11. Restoring force of two forces
F3 is the restoring force of F1 and F2
25o
70o
F1(55N)
F2 (25N) F3
Can be solved by resolving the horizontal components of F1 and F2

12. Restoring force of two forces
F3 is the restoring force of F1 and F2
25o
70o
F1(55N)
F2 (25N) F3
F1v = F1.sin70o
55sin70o
= 51.68N
F1h = F1.cos70o
55cos70o
= 18.81N

13. Restoring force of two forces
F1(55N)
F2 (25N) F3
F2v = F2.sin25o
25sin25o
= 10.57N
F2h = F2.cos25o
25cos25o
= 22.66N

14. Restoring force of two forces
F1(55N)
F2 (25N) F3
F3v = F1v + F2v
= 62.25N
F3h = F1h +F2h
= 41.47N

15. Restoring force of two forces
F3 = 74.80N
41.47N

16. Resultant of two forces
Tan θ = opposite/adjacent
Tan θ = 62.25/41.47
Tan θ = 1.5
θ = 56.33o
62.25N θ
41.47N
Direction of F3 = = o

17. Moments of force 2m 4N 4m 2N
Total Anticlockwise moments = Total Clockwise moments
8Nm
8Nm

18. Moments of force 4m 3m 2m 2N 2N 4N
Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = Nm