Question 11.2.4: A 150-ft foot bridge crossing a gorge is suspended from two ...

Goal For each cable find the maximum and minimum cable tensions, the angle between the cable and the horizontal at B, and the total length.
Given Information about the cable geometry and loading.
Assume The cable is inextensible and the cable weight is negligible relative to the loading.
Draw We define a coordinate system with its origin at the lowest point of the cable, the location of which is unknown (Figure 2). We define x_{B} as the x coordinate of support B; then the coordinate x_{A}= x_{B}− 150  ft .
Formulate Equations and Solve The load is shared equally by the two suspension cables; therefore ω = 30 lb/ft for each cable. We locate the origin by applying (11.3A) at supports A and B:

y=\frac{\omega x^{2}}{2T_{O}}                (11.3A)
At A :            y_{A}=10  ft = \frac{\left(30  lb/ft\right) \left(x_{B} – 150  ft\right)^{2} }{2T_{O}}                 (1)
At B :            y_{B}=22  ft = \frac{\left(30  lb/ft\right) \left(x_{B} \right)^{2} }{2T_{O}}                             (2)

We eliminate T_{O} by dividing (1) by (2), which gives

\frac{10 ft}{22 ft} = \frac{\left(x_{B} – 150 ft\right)^{2}}{x^{2}_{B}}

Rearranging and simplifying produces an equation that can be solved using the quadratic formula:

x^{2}_{B} – 550 x_{B}  + 41, 250 =0
x_{B} =\frac{550\pm \sqrt{\left(- 550\right)^{2}- 4(1)(41, 250)} }{2(1)} =275  ft \pm 185.4  ft \Rightarrow  x_{B} =460.4  ft  or  89.6  ft

We discard x_{B}= 460.4  ft  because it is not a realistic solution when the constraints of this problem are considered. Therefore x_{B}= 89.6  ft . Substituting this value of x_{B} into (2), we solve for the minimum cable tension T_{O}

y_{B}=\frac{\omega x^{2}_{B}}{2T_{O}} =22  ft = \frac{\left(30  lb/ft\right)\left(89.6  ft\right)^{2} }{2T_{O}} \Rightarrow  T_{O}=5475  lb

T_{max} occurs at support B, because the slope of the cable is steepest there.
Using (11.4), we find T_{max} :

T^{2}=\omega ^{2} x^{2} +T_{O}^{2}       or       T =\sqrt{\omega ^{2} x^{2} + T_{O}^{2}}             (11.4)
T_{max} =T_{@ x=89.6  ft}=\sqrt{\left\lgroup30\frac{lb}{ft} \right\rgroup^{2} \left(89.6  ft\right)^{2} + \left(5474  lb\right)^{2} }   \Rightarrow  T_{max} =6098  lb

The tensile force is tangent to the cable at any point along the cable. Therefore, the angle the cable force makes with the horizontal and the angle the cable makes with the horizontal are the same. We call this angle \theta _{B} at support B. Figure 3 shows the relationship between the cable force at B and the horizontal component of that force.

T_{Bx}=T_{B}\cos \theta _{B} \Rightarrow \theta _{B} =\cos ^{-1} \left\lgroup\frac{5474 lb}{6098 lb} \right\rgroup \Rightarrow \theta _{B} =26.2^{\circ }

Finally, to find the length of the cable L, substitute x_{A}= 89.6  ft−150  ft = −60.4  ft , x_{B}= 89.6  ft , T_{O}= 5474  lb  , and ω = 30 lb/ft into (IV.5A):

L = – s_{OA} + s_{OB}=- \left[x_{A}+ \frac{\omega ^{2} x^{3}_{A}}{ 6T_{O}^{2}} – \frac{\omega ^{4} x^{5}_{A}}{ 40T_{O}^{4}}\right] + \left[x_{B}+ \frac{\omega ^{2} x^{3}_{B}}{ 6T_{O}^{2}} – \frac{\omega ^{4} x^{5}_{B}}{ 40T_{O}^{4}}\right]
L=154.6 ft

Check Aside from rechecking calculations, there are no alternative calculations that serve as a check for this problem.