Question

3 - 65. if the tension developed in either cable ab or ac cannot exceed 1000 lb, determine the maximum tension that can be developed in cable ad when it is tightened by the turnbuckle. also, what is the force developed along the antenna tower at point a? solution force vectors: we can express each of the forces on the free - body diagram shown in fig. a in cartesian vector form as f_ab = f_ab (10 - 0)i+(-15 - 0)j+(-30 - 0)k/sqrt((10 - 0)^2+(-15 - 0)^2+(-30 - 0)^2) = 2/7 f_ab i - 3/7 f_ab j - 6/7 f_ab k f_ac = f_ac (-15 - 0)i+(-10 - 0)j+(-30 - 0)k/sqrt((-15 - 0)^2+(-10 - 0)^2+(-30 - 0)^2) = - 3/7 f_ac i - 2/7 f_ac j - 6/7 f_ac k f_ad = f * (0 - 0)i+(12.5 - 0)j+(-30 - 0)k/sqrt((0 - 0)^2+(12.5 - 0)^2+(-30 - 0)^2) = 5/13 f j - 12/13 f k f_ae = f_ae k

Explanation:

Step1: Establish equilibrium equations

At point A, $\sum F_x = 0$, $\sum F_y=0$, $\sum F_z = 0$. From $\sum F_x=0$: $\frac{2}{7}F_{AB}-\frac{3}{7}F_{AC}=0$. Since the maximum of $F_{AB}$ and $F_{AC}$ is 1000 lb, assume $F_{AB} = 1000$ lb (we can also assume $F_{AC}=1000$ lb, the result will be the same), then from $\frac{2}{7}F_{AB}-\frac{3}{7}F_{AC}=0$, we have $\frac{2}{7}\times1000-\frac{3}{7}F_{AC}=0$, so $F_{AC}=\frac{2000}{3}\text{ lb}< 1000\text{ lb}$.

Step2: Use $\sum F_y = 0$

$\sum F_y=-\frac{3}{7}F_{AB}-\frac{2}{7}F_{AC}+\frac{5}{13}F_{AD}=0$. Substitute $F_{AB} = 1000$ lb and $F_{AC}=\frac{2000}{3}$ lb into it: $-\frac{3}{7}\times1000-\frac{2}{7}\times\frac{2000}{3}+\frac{5}{13}F_{AD}=0$. First, calculate the left - hand side: $-\frac{3000}{7}-\frac{4000}{21}=-\frac{9000 + 4000}{21}=-\frac{13000}{21}$. Then, from $-\frac{13000}{21}+\frac{5}{13}F_{AD}=0$, we get $F_{AD}=\frac{13000}{21}\times\frac{13}{5}=\frac{33800}{21}\approx1609.52\text{ lb}$.

Step3: Use $\sum F_z = 0$

$\sum F_z=-\frac{6}{7}F_{AB}-\frac{6}{7}F_{AC}-\frac{12}{13}F_{AD}+F_{AE}=0$. Substitute $F_{AB} = 1000$ lb, $F_{AC}=\frac{2000}{3}$ lb and $F_{AD}=\frac{33800}{21}$ lb into it. First, calculate $-\frac{6}{7}(F_{AB}+F_{AC})=-\frac{6}{7}(1000+\frac{2000}{3})=-\frac{6}{7}\times\frac{3000 + 2000}{3}=-\frac{10000}{7}$. And $-\frac{12}{13}F_{AD}=-\frac{12}{13}\times\frac{33800}{21}=-\frac{31200}{21}$. Then $F_{AE}=\frac{10000}{7}+\frac{31200}{21}=\frac{30000 + 31200}{21}=\frac{61200}{21}\approx2914.29\text{ lb}$.

Answer:

The maximum tension in cable $AD$ is approximately $1609.52$ lb and the force developed along the antenna tower at point $A$ ($F_{AE}$) is approximately $2914.29$ lb.