**Problem 5.58**The homogeneous 860-kg bar \( AB \) is supported by a ball-and-socket joint at \( A \) and two cables attached to \( B \). Find the tension in cable \( CB \).**Diagram Description (Figure P5.58):**The diagram illustrates a three-dimensional setup where a bar \( AB \) is supported by a joint at point \( A \) and two cables extending from point \( B \).- The bar \( AB \) is leaning upwards and away from the coordinate plane, attached at the bottom by a ball-and-socket joint located at point \( A \).- Point \( B \) is situated above and slightly forward of the joint, at a height of 4 meters along the \( z \)-axis.- The base coordinates measure distances from \( A \) as follows: - \( B \) is located 4.5 meters along the \( y \)-axis.- Two additional points, \( C \) and \( D \), serve as anchoring positions for the cables: - Point \( C \) is 3 meters along the \( x \)-axis and 2.6 meters along the \( y \)-axis. - Point \( D \) is 3 meters along the \( x \)-axis and 2.2 meters along the \( y \)-axis. This configuration shows the spatial relationship and measurement points necessary to calculate the tension in cable \( CB \).

Co-ordinates of point A = (0,0,0)m Co-ordinates of point B = (-4.5,0,4)m Co-ordinates of point C = (3,-2.6,0)m Co-ordinates of point D = (2.2,3,0)m Total weight of the bar W is acting at center of bar thus Co-ordinates of point W = (0,1.83,2.4)mPosition vector of W - rAW→ = rW→⇒rAW→ = -2.25 - 0i^ + 0 - 0j^ + 2 - 0k^⇒rAW→ = -2.25 i^ + 2k^ Position vector of AD - rAB→ = -4.5 - 0i^ + 0 - 0j^ + 4 - 0k^⇒rAB→ = -4.5 i^ + 4k^ Unit vector along AD - λAD = AD→AD→⇒λAD =2.2 - 0i^ + 3 - 0j^ + 0 - 0k^2.22 + 32⇒λAD = .591 i^ + .806j^Unit vector along BC - λBC = BC→BC→⇒λBC =3 --4.5i^ + -2.6 - 0j^ + 0 - 4k^7.52 + 2.62+ 42⇒λBC = .844i^ - .293j^ - .45k^ Cartesian vector form of tension force DF - TBC→ = TBC.λBC⇒TBC→ = TBC .844i^ - .293j^ - .45k^Sum of moment about AD - ∑MAD = 0rW→ ×W→ + rAB→ ×TBC→.λAD = 0⇒-2.25 i^ + 2k^×-860×9.81k^ + -4.5 i^ + 4k^×TBC .844i^ - .293j^ - .45k^×.591 i^ + .806j^ = 0Solving this, we get1.78TBC - 15299.774 = 0⇒TBC = 8595.3787 KN Tension in cable CB.