Problem 2.133

1. Problem 2.133 y A 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. 56 ft D a O B 20o 50o z C x

2. y Problem 2.133 A 56 ft D a O B 20o 50o z C x 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.

3. y Problem 2.133 A 56 ft D a O B 20o 50o z C x Fx Fy Fz cos qx= cos qy= cos qz= F F F 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.

4. y Problem 2.133 Solution From triangle AOB: 56 ft cos qy = = 0.86154 65 ft qy = 30.51o A Determine the direction cosines of the line of action of a force. 65 ft qy F Fy 56 ft Fx B O (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)(0.86154) Fy = + 3360 lb Fz = + (3900 lb) sin 30.51o sin 20oFz = + 677 lb

5. y Problem 2.133 Solution A 65 ft qy Fx _1861 lb F Fy (b) cos qx = = F 3900 lb Fx B O 20o Fz x Fz 677 lb cos qz = = + = + 0.1736 qz = 80.0o F 3900 lb Determine the direction cosines of the line of action of a force. 56 ft cos qx = _ 0.4771 qx = 118.5o From above (a): qy = 30.5o