Question 11.1.1: The cable shown in Figure 1 supports a 500-lb load at C and ...

Goal Find the vertical coordinate of point D (y_{D}) and the maximum tension in the cable.
Given Information about the cable geometry and applied loads, and a coordinate system with origin at A.
Assume The weight of the cable is negligible compared to the loads and the cable is uniform.
Draw Figure 2 is a free-body diagram of the entire system. The cable pulls on the supports at A and B and the supports pull back on the cable. This information allows us to draw the tension at A (T_{AC}) in a known direction (we know the slope because we know, based on Figure 1, that x_{C} = 8 ft and y_{C}= −3 ft). We also draw the tension at B (T_{BD}), but since we do not know y_{D}, we show it at an unknown angle β.
Formulate Equations and Solve based on the free- body diagram of the entire system (Figure 2), we write the equilibrium equation:

\sum{M_{z @ A} }\left(\curvearrowleft + \right) =0
– 500  lb(8  ft) – 250  lb(16  ft) + T_{BD} \sin\beta (24  ft) – T_{BD} \cos\beta (6  ft) =0
– 4000  lb.ft + T_{BD}\sin\beta(12  ft) – T_{BD} \cos\beta(3  ft) = 0                  (1)

Also, we can write:

\sum{M_{z @ C} }\left(\curvearrowleft + \right) =- 250   lb(8  ft)+ T_{BD}\sin\beta(16  ft) – T_{BD} \cos\beta(9  ft) = 0
– 2000  lb.ft+ T_{BD}\sin\beta(16  ft) – T_{BD} \cos\beta(9  ft) = 0                    (2)

Solving (1) and (2) simultaneously gives

β=36.9°                  and        T_{BD}= 834  lb

We now use this value of β to find the vertical coordinate of D (y_{D}). based on the geometry in Figure 3, we write

tan β= tan 36.9° =\frac{d_{D}}{8  ft}

Solving for d_{D}, we find

d_{D}=6.00  ft

If point D is 6 feet below anchor B, it is at the same level as anchor A. This means that the y coordinate of point D relative to the xy coordinate system defined in Figure 1 is

y_{D}=0.0  ft

To find the maximum tension in the cable, we could calculate the tension in the other cable segments. based on equilibrium  conditions applied to the free-body diagram in Figure 4 we find that T_{CD}=T_{AC}= 712  lb .Therefore the maximum tension in the cable is:

T_{max} =T_{BD} =834  lb

Alternatively, we could use the knowledge that when all of the loads are vertical the maximum tension occurs in the segment with the steepest slope relative to horizontal. The steepest segment is BD and therefore T_{BD}=T_{max} .
Check We can check our results by using the free-body diagram in Figure 2 and the calculated values of T_{BD} and T_{AC} to write the moment equilibrium equation about a moment center at D.
Note: because no horizontal forces are applied to the cable, the horizontal force component is constant along the cable. In other words, we can show that T_{CAx} = T_{CDx} = T_{BDx} = 667  lb . This is also a good check of our results.