1. Problem 4-c 1.2 m y 1.5 m z x 5 kN A B C E D  1 m 2 m A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the line of action of the 5-kN force forms an angle  =30 o with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A.

2. 1.2 m y 1.5 m z x 5 kN A B C E D  1 m 2 m 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it. Solving Problems on Your Own Problem 4-c A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the line of action of the 5-kN force forms an angle  =30 o with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A.

3. 2. Write equilibrium equations and solve for the unknowns. For three-dimensional body the six scalar equations  F x = 0  F y = 0  F z = 0  M x = 0  M y = 0  M z = 0 should be used and solved for six unknowns. These equations can also be written as  F = 0  M O =  (r x F ) = 0 where F are the forces and r are position vectors. 1.2 m y 1.5 m z x 5 kN A B C E D  1 m 2 m A 3-m pole is supported by a ball- and-socket joint at A and by the cables CD and CE. Knowing that the line of actionof the 5-kN force forms an angle  =30 o with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A. Solving Problems on Your Own Problem 4-c

4. Problem 4-c Solution Draw a free-body diagram of the body. 1.2 m y 1.5 m z x 5 kN A B C E D  1 m 2 m 1.2 m y 1.5 m z x A B C E D 1 m 2 m 30 o 5 kN T CE T CD A y j A x i A z k

5. Write equilibrium equations and solve for the unknowns. 1.2 m y 1.5 m z x A B C E D 1 m 2 m 30 o 5 kN T CE T CD A y j A x i A z k 5 unknowns and 6 equations of equilibrium, but  M x = 0 does not involve any of the unknowns. r B/A = 2 ir C/A = 3 i Load at B, F B = _ ( 5 cos 30 o ) j + ( 5 sin 30 o ) k = _ 4.33 j + 2.5 k CD = _ 3 i+ 1.5 j + 1.2 k CD = 3.562 m T CD = T CD = ( _ 3 i + 1.5 j + 1.2 k) T CE = T CE = ( _ 3 i + 1.5 j _ 1.2 k) CD T CD 3.562 T CD 3.562 CE Problem 4-c Solution

6. 1.2 m y 1.5 m z x A B C E D 1 m 2 m 30 o 5 kN T CE T CD A y j A x i A z k  M A = 0: r C/A x T CD + r C/A x T CE + r B/A x F B = 0 i j k 3 0 0 _ 3 1.5 1.2 i j k 3 0 0 _ 3 1.5 _ 1.2 i j k 2 0 0 0 _ 4.33 2.5 T CD 3.562 + T CE 3.562 + = 0 Problem 4-c Solution

7. 1.2 m y 1.5 m z x A B C E D 1 m 2 m 30 o 5 kN T CE T CD A y j A x i A z k Equate coefficients of unit vectors to zero. j: _ 3.6 + 3.6 _ 5 = 0 _ 3.6 T CD +3.6 T CE _ 17.81 = 0 (1) T CD 3.562 T CE 3.562 k: 4.5 + 4.5 _ 8.66 = 0 4.5 T CD +4.5 T CE = 30.85 (2) T CD 3.562 T CE 3.562 (2) + 1.25 (1): 9T CE _ 53.11 = 0 ; T CE = 5.90 kN Eq. (1): _ 3.6T CD + 3.6 (5.902) _ 17.81 = 0 T CD = 0.954 kN Problem 4-c Solution

8. 1.2 m y 1.5 m z x A B C E D 1 m 2 m 30 o 5 kN T CE T CD A y j A x i A z k  F = 0: A + T CD + T CE + F B = 0 i: A x + ( _ 3) + ( _ 3) = 0 A x = 5.77 kN 3.562 0.954 3.562 5.902 j: A y + (1.5) + (1.5) _ 4.33 = 0 A y = 1.443 kN 3.562 0.954 3.562 5.902 k: A z + (1.2) + ( _ 1.2) + 2.5 = 0 A z = _ 0.833 kN 3.562 0.954 3.562 5.902 A = ( 5.77 kN) i + ( 1.443 kN ) j - ( 0.833 kN ) k Problem 4-c Solution