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Published byElaina Rawle Modified over 2 years ago

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Problem Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles x, y, and z defining the direction of that force. O x y z 56 ft D B C A 50 o 20 o

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Solving Problems on Your Own O x y z 56 ft D B C A 50 o 20 o Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles x, y, and z defining the direction of that force. 1. Determine the rectangular components of a force defined by its magnitude and direction. If the direction of the force F is defined by the angles y and, projections of F through these angles or their components will yield the components of F. Problem 2.133

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Solving Problems on Your Own Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles x, y, and z defining the direction of that force. 2. Determine the direction cosines of the line of action of a force. The direction cosines of the line of action of a force F are determined by dividing the components of the force by F. cos x = FxFx F cos y = FyFy F cos z = FzFz F O x y z 56 ft D B C A 50 o 20 o Problem 2.133

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Problem Solution z y Determine the direction cosines of the line of action of a force. O x B A FyFy FzFz F FxFx 20 o y 65 ft From triangle AOB: cos y = 56 ft 65 ft = y = o (a) F x = _ F sin y cos 20 o = _ (3900 lb) sin o cos 20 o F x = _ 1861 lb F y = + F cos y = (3900 lb)( ) F y = lb F z = + (3900 lb) sin o sin 20 o F z = lb 56 ft

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Problem Solution O x B A FyFy FzFz F FxFx 20 o y 65 ft y Determine the direction cosines of the line of action of a force. (b) cos x = FxFx F = _ 1861 lb 3900 lb cos x = _ x = o From above (a): y = 30.5 o cos z = FzFz F = lb 3900 lb = z = 80.0 o 56 ft

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