statics 2

statics 2 - Problem 2.1 The magnitudes |FA | = 60 N and |FB...

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Problem 2.1 The magnitudes | F A | =6 0 N and | F B | =80 N. The angle α =45 . Graphically de- termine the magnitude of the sum of the forces F = F A + F B and the angle between F B and F . Strategy: Constructtheparallelogramfordetermining the sum of the forces, drawing the lengths of F A and F B proportional to their magnitudes and accurately measur- ing the angle α , as we did in Example 2.1. Then you can measure the magnitude of their sum and the angle between their sum and F B . F A F B Solution: The graphical construction is shown: The angle β is graphically determined to be about 26 and the angle θ is about θ =19 . The magnitude of the sum | F | = | F A + F B | is about | F | = 130 N. (These values check with a determination using trigonometry, β =25 . 9 , θ . 1 , and | F | = 129 . 6 N .) 45˚ F A F B F Problem 2.2 The magnitudes | F A | =4 0 N and | F A + F B | N. The angle α =60 . Graphically determine the magnitude of F B . Solution: Measuring, F B = 52 N F A F B 60 ° 0 40 50 80 100 N F A F A + F B F B
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Problem 2.3 The magnitudes | F A | = 100 lb and | F B | = 140 lb. The angle α =40 . Use trigonom- etry to determine the magnitude of the sum of the forces F = F A + F B and the angle between F B and F . Strategy: Use the laws of sines and cosines to analyze the triangles formed by the parallelogram rule for the sum of the forces as we did in Example 2.1. The laws of sinesandcosinesaregiveninSectionA.2ofAppendixA. Solution: The construction is shown. Use the cosine law to de- termine the magnitude | F | . | F | 2 = | F A | 2 + | F B | 2 2 | F A || F B | cos 140 | F | 2 = (100) 2 + (140) 2 2 (100) (140) cos 140 or | F | = p 5 . 1049 ... (10 4 ) = 225 . 94 = 225 . 9 lb Use the law of sines to determine the angle θ ., i.e., | F | sin 140 = | F B | sin θ . From which we get sin θ = | F B | | F | sin 140 =0 . 398 , and θ = 23 . 47 . The angle between F B and F is Thus β 23 . 47=16 . 53 40 ° 140 F A F B F Problem 2.4 The magnitudes | F A | N and | F A + F B | =80 N. The angle α =60 . Use trigonometry to determine the magnitude of F B . Solution: Draw the force triangle. From the law of sines | F A + F B | sin 120 = | F A | sin θ sin θ = | F A | | F A + F B | sin 120 sin θ = ± 40 80 ² (0 . 866) θ =25 . 66 θ + γ + 120 = 180 γ =34 . 34 From the law of sines, | F B | sin γ = | F A + F B | sin 120 | F B | = ± sin γ sin 120 ² (80 N ) | F B | =52 . 1 N F A F B 60 ° 120 ° F A F B F A B
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Problem 2.5 The magnitudes | F A | = 100 lb and | F B | = 140 lb. If α can have any value, what are the minimum and maximum possible values of the magni- tude of the sum of the forces F = F A + F B , and what are the corresponding values of α ? Solution: A graphical construction shows that the magnitude is a minimum when the two force vectors are opposed, and a maximum when both act in the same direction. The corresponding values are | F | max = | F A + F B | = | 100 + 140 | = 240 lb, and α =0 . | F | min = | F A + F B | = | 100 140 | =40 lb, and α = 180 . Problem 2.6 The angle θ =30 . What is the magni- tude of the vector r AC ? 60 mm 150 mm A C B r AB r BC r AC Solution: A 30 ° 60 mm 150 mm B r AC C From the law of sines BC sin 30 = AB sin α = AC sin γ We know and AB . Thus 150 sin 30 = 60 sin α α =11 . 54 Also 30 + α + γ = 180 γ = 138 . 46 Now, from the law of sines 150 sin 30 = AC sin 138 . 46 AC = | r AC | = 199 mm
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Problem 2.7 The vectors F A and F B
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statics 2 - Problem 2.1 The magnitudes |FA | = 60 N and |FB...

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