AQA AS Physics P6 Forces In Equilibrum Kerboodle Answers

6.1 Vectors and scalars AQA AS Physics C6 Force In Equlibrium Kerboodle Answers: Page No. 99 1.Ans

a 3.7N at 33° to 3.1N b 17.1N at 21° to 16N c 1.4N at 45° to 3N and 1N 2 i14.0N in the same direction ii 6.0N in the direction of the 10N force iii 10.8 N at 22° to the 10 N force 3. 6.1 kN vertically up, 2.2 kN horizontal 4. a 268 N b 225N

6.2 Balanced forces AQA AS Physics C6 Force In Equlibrium Kerboodle Answers : Page No. 102

1 a 7.3 N b 7.3 Nat 31.5° to the vertical 2. a Sketch a diagram and show the three forces acting on the object. b G Calculate i i 2.7 N ii 4.7 N 3 a139 N b 95N 4 .73° b the tension in each part of the string. 6.8N

6.3 The principle of moments AQA AS Physics C6 Force In Equlibrium Kerboodle Answers: Page No. 104

1. 300N 2 .a Sketch a diagram to represent this situation. b 6.2N 3.0.27m 4.6.75N

6.4 More on moments AQA AS Physics C6 Force In Equlibrium Kerboodle Answers : Page No. 106

1.0.51N at 100mmmark, 0.69N at the 800mmmark 2 . a at 1.0m end and 108 N at the other end, both vertically upwards b 122N at 1.0m end and 108N at the other end, both vertically downwards 3. 620kN, 640kN 4.a 100N, 50N b 150N

6.5 Stability AQA AS Physics C6 Force In Equlibrium Kerboodle Answers : Page No. 109

1 The centre of mass is higher if the upper shelves arc filled instead of the lower shelves. If tilted, it will topple over at a smaller angle with the upper shelves full, than if they were empty. 2 89N 3 a 48° b Yes, they will raise the overall centre of mass so it will topple on a less steep slope. 6.6 Equilibrium rules: Page No. 113 1 a 50N b 250N 2. a 1800Nm b 1800N 3 . a.6.0kN b.10.8kN 4 . 1.5 Nin the cord at 40° to the vertical, 1.9N in the other cord

6.7 Statics calculations AQA AS Physics C6 Force In Equlibrium Kerboodle Answers: Page No. 114-115

1 .a in the same direction 15N b in opposite directions 3.0N c .10.8N 2.7N 3 .a 6.8N b 52° 4 .18.0N 5. a 16.9kN b16.9kN 6 a.6.2N b.12.2N 7.Move it further 50mm away from the pivot. 8.a 6.8N b9.8N 9. a 2200N b3100N 10.a b950N at X, 750N at Y 11 a 2820kN b 1660kN and l 540kN (to 3sf) 12a 8.0N and 16.0N b 38N and 76N 13.a Sketch a free body force diagram of the girder in this position. b 11 kN, 11 kN 14.a Copy the diagram and mark the forces acting on the picture on your diagram. b 28.4N Banner 7

Practice questions: Page No. 116-117

1. (a) (i) . 1.30, 2.75, 4.35, 5.75, 7.15, 8.70 (ii) Plot a graph of S on they-axis against d on the x-axis. (5 marks) (b) (i) Taking moments about X gives: S D = W d + 0.5 D Wo , where Wo is the weight of the beam. Dividing each term by D gives the required equation. (ii) (c) (i) (ii) Repeat the measurements for the same distances several more times to obtain a more reliable value for the mean support force at each distance. 2. (a) magnitude of resultant = 13.6 (± 0.3) N required angle = 13 (± 2) ° (b) 3. (a) (i) horizontal component = 850 × cos 42 ° = 630 N (ii)vertical component = 850 × sin 42 ° = 570 N (iii)weight of girder = sum of vertical components of T = 2 × 570 = 1100 N (1140) (b) The weight acts vertically downwards at the centre of the girder 4 (a) the moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force. (b) (i) application of definition gives 46 = F × 0.25 cos 40 ° ∴ F = 240 N (ii)the moment of F increases to a maximum (when the crank is horizontal) and then decreases because the perpendicular distance increases and then decreases 5 .Ans- (a) Forces on plank: two arrows vertically downwards to represent weight of student and load one arrow vertically downwards in centre of plank to represent weight of plank one arrow vertically upwards from the log to represent the upward force (reaction) at the pivot (b) Taking moments about the pivot: clockwise moment = (25 × 9.81 × 1.0) + (L × 2.5) where L = load in N equated with anticlockwise moment = (65 × 9.81 × 0.50) gives load L = 29 N (c) as the student walks towards the log, the anticlockwise moment decreases the clockwise moment is now greater than the anticlockwise, so the plank rotates clockwise until the load touches the ground.

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