# Forces and moments Resolving forces.

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Forces and moments Resolving forces

Forces and moments Example 1 Drawing to scale

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

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

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

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

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

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

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

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

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

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

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

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

Restoring force of two forces
F3 = 74.80N 41.47N

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

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

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