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Forces and moments

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. 20 o. A. 55 o. 55 o. 70 o.

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Forces and moments

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  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 70o 35o 105o The length of the sides of the triangle represent the magnitude of the forces NOT the length of rope 2000 N B 20o 2000 Newtons

  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) a 55o 105o Angle C = 105o and side c represents 2000N 2000 N (c) b 20o

  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 a 55o 105o 2000 N (c) b/sin B = c/sin C therefore b/sin 55o= 2000/sin105o a = 2000 x sin 55o/sin105o 1696.1N b 20o

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

  7. Using the cosine rule ( if you know one angle and two sides) A2 = B2 + C2 -2BCcosA (F3)2 = 302 + 602 – 2x60x30x cos110o F2 = 60N (C) F3 (A) = 75.7N A =110o 70o F1 = 30N (B)

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

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

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

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

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

  13. Restoring force of two forces F1(55N) F2v = F2.sin25o 25sin25o = 10.57N F3 70o F2 (25N) F2h = F2.cos25o 25cos25o = 22.66N 25o

  14. Restoring force of two forces F1(55N) F3v = F1v + F2v 51.68 +10.57 = 62.25N F3 70o F2 (25N) F3h = F1h +F2h 18.81 + 22.66 = 41.47N 25o

  15. Restoring force of two forces F3 (F3)2 = 62.252 + 41.472 62.25N (F3)2 = 5594.82 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 F3 62.25N θ 41.47N Direction of F3 = 180 + 56.33 = 236.33o

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

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

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