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

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

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Using the sine rule (if you know the angles) 20 o 55 o 105 o a b 2000 N (c) a/sin A = c/sin C therefore a/sin 20 o = 2000/sin105 o a = 2000 x sin 20 o /sin105 o N b/sin B = c/sin C therefore b/sin 55 o = 2000/sin105 o a = 2000 x sin 55 o /sin105 o N

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

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Using the cosine rule ( if you know one angle and two sides) 70 o F2 = 60N (C) F1 = 30N (B) F3 (A) A =110 o A 2 = B 2 + C 2 -2BCcosA (F3) 2 = – 2x60x30x cos110 o = 75.7N

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Vertical and horizontal components of forces F FvFv FHFH θ Sketch the diagram F v can be drawn at the other end of the sketch

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Vertical and horizontal components of forces F FvFv FHFH θ Sketch the diagram FvFv sin θ = F v /F F.sin θ = F v cos θ = F H /F F.cos θ = F H

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

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

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Restoring force of two forces 25 o 70 o F1(55N) F2 (25N) F3 F3 is the restoring force of F1 and F2 F1v = F1.sin70 o 55sin70 o = 51.68N F1h = F1.cos70 o 55cos70 o = 18.81N

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Restoring force of two forces 25 o 70 o F1(55N) F2 (25N) F3 F2v = F2.sin25 o 25sin25 o = 10.57N F2h = F2.cos25 o 25cos25 o = 22.66N

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

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Restoring force of two forces F N 41.47N (F3) 2 = (F3) 2 = F3 = 74.80N

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

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

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

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