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Forces and moments Resolving forces

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Forces and moments Example 1 Drawing to scale

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**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

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**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

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**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

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**Using the cosine rule ( if you know one angle and two sides)**

F2 = 60N 70o F1 = 30N F3

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**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

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**Vertical and horizontal components of forces**

Sketch the diagram Fv can be drawn at the other end of the sketch F Fv FH θ

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**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

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**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

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**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

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**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

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**Restoring force of two forces**

F1(55N) F2 (25N) F3 F2v = F2.sin25o 25sin25o = 10.57N F2h = F2.cos25o 25cos25o = 22.66N

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**Restoring force of two forces**

F1(55N) F2 (25N) F3 F3v = F1v + F2v = 62.25N F3h = F1h +F2h = 41.47N

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**Restoring force of two forces**

F3 = 74.80N 41.47N

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**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

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

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**Moments of force 4m 3m 2m 2N 2N 4N**

Total Anticlockwise moments = Total Clockwise moments 8Nm + 4Nm = 12Nm = Nm

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