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**Problem 2.133 y A Cable AB is 65 ft long, and 56 ft**

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 qx, qy, and qz defining the direction of that force. D a O B 20o 50o z C x

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**Solving Problems on Your Own**

x y z 56 ft D B C A 50o 20o a Problem 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 qx, qy, and qz 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 qy and f, projections of F through these angles or their components will yield the components of F.

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**Solving Problems on Your Own**

x y z 56 ft D B C A 50o 20o a Problem 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 qx, qy, and qz 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 qx= Fx F cos qy= Fy F cos qz= Fz F

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**Fy = + F cos qy = (3900 lb)(0.86154) Fy = + 3360 lb **

Problem Solution A Determine the direction cosines of the line of action of a force. 65 ft qy From triangle AOB: cos qy = 56 ft 65 ft = qy = 30.51o F Fy 56 ft Fx O B (a) Fx = _ F sin qy cos 20o = _ (3900 lb) sin 30.51o cos 20o Fx = _1861 lb 20o Fz z x Fy = + F cos qy = (3900 lb)( ) Fy = lb Fz = + (3900 lb) sin 30.51o sin 20o Fz = lb

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**Fx F = _1861 lb 3900 lb cos qx = _ 0.4771 qx = 118.5o cos qz = Fz F**

y Problem Solution O x B A Fy Fz F Fx 20o qy 65 ft Determine the direction cosines of the line of action of a force. (b) cos qx = Fx F = _1861 lb 3900 lb 56 ft cos qx = _ qx = 118.5o From above (a): qy = 30.5o cos qz = Fz F = + 677 lb 3900 lb = qz = 80.0o

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Equivalent Systems of Forces

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