3 - 65. if the tension developed in either cable ab or ac cannot exceed 1000 lb, determine the maximum…

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?\n\nsolution\nforce vectors: we can express each of the forces on the free - body diagram shown in fig. a in cartesian vector form as\n mathbf{f}_{ab}=f_{ab}left\frac{(10 - 0)mathbf{i}+(-15 - 0)mathbf{j}+(-30 - 0)mathbf{k}}{sqrt{(10 - 0)^{2}+(-15 - 0)^{2}+(-30 - 0)^{2}}}\right=\frac{2}{7}f_{ab}mathbf{i}-\frac{3}{7}f_{ab}mathbf{j}-\frac{6}{7}f_{ab}mathbf{k} \n mathbf{f}_{ac}=f_{ac}left\frac{(-15 - 0)mathbf{i}+(-10 - 0)mathbf{j}+(-30 - 0)mathbf{k}}{sqrt{(-15 - 0)^{2}+(-10 - 0)^{2}+(-30 - 0)^{2}}}\right=-\frac{3}{7}f_{ac}mathbf{i}-\frac{2}{7}f_{ac}mathbf{j}-\frac{6}{7}f_{ac}mathbf{k} \n mathbf{f}_{ad}=fleft\frac{(0 - 0)mathbf{i}+(12.5 - 0)mathbf{j}+(-30 - 0)mathbf{k}}{sqrt{(0 - 0)^{2}+(12.5 - 0)^{2}+(-30 - 0)^{2}}}\right=\frac{5}{13}fmathbf{j}-\frac{12}{13}fmathbf{k} \n mathbf{f}_{ae}=f_{ae}mathbf{k}

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?\n\nsolution\nforce vectors: we can express each of the forces on the free - body diagram shown in fig. a in cartesian vector form as\n mathbf{f}_{ab}=f_{ab}left\frac{(10 - 0)mathbf{i}+(-15 - 0)mathbf{j}+(-30 - 0)mathbf{k}}{sqrt{(10 - 0)^{2}+(-15 - 0)^{2}+(-30 - 0)^{2}}}\right=\frac{2}{7}f_{ab}mathbf{i}-\frac{3}{7}f_{ab}mathbf{j}-\frac{6}{7}f_{ab}mathbf{k} \n mathbf{f}_{ac}=f_{ac}left\frac{(-15 - 0)mathbf{i}+(-10 - 0)mathbf{j}+(-30 - 0)mathbf{k}}{sqrt{(-15 - 0)^{2}+(-10 - 0)^{2}+(-30 - 0)^{2}}}\right=-\frac{3}{7}f_{ac}mathbf{i}-\frac{2}{7}f_{ac}mathbf{j}-\frac{6}{7}f_{ac}mathbf{k} \n mathbf{f}_{ad}=fleft\frac{(0 - 0)mathbf{i}+(12.5 - 0)mathbf{j}+(-30 - 0)mathbf{k}}{sqrt{(0 - 0)^{2}+(12.5 - 0)^{2}+(-30 - 0)^{2}}}\right=\frac{5}{13}fmathbf{j}-\frac{12}{13}fmathbf{k} \n mathbf{f}_{ae}=f_{ae}mathbf{k}

Answer

Explanation:

Step1: Set up equilibrium equations

At point A, $\sum F_x = 0$, $\sum F_y=0$, $\sum F_z = 0$. From $\sum F_x = 0$, we have $\frac{2}{7}F_{AB}-\frac{3}{7}F_{AC}=0$. Since the maximum of $F_{AB}$ and $F_{AC}$ is 1000 lb. Let's assume $F_{AB} = 1000$ lb, then from $\frac{2}{7}F_{AB}-\frac{3}{7}F_{AC}=0$, we get $F_{AC}=\frac{2}{3}\times1000=\frac{2000}{3}$ lb (less than 1000 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 $-\frac{3}{7}\times1000-\frac{2}{7}\times\frac{2000}{3}=-\frac{9000 + 4000}{21}=-\frac{13000}{21}$. Then, $\frac{5}{13}F_{AD}=\frac{13000}{21}$, and $F_{AD}=\frac{13000}{21}\times\frac{13}{5}=\frac{33800}{21}\approx1609.52$ 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 $-\frac{6}{7}(1000+\frac{2000}{3})=-\frac{6}{7}\times\frac{3000 + 2000}{3}=-\frac{6}{7}\times\frac{5000}{3}=-\frac{10000}{7}$. And $-\frac{12}{13}F_{AD}=-\frac{12}{13}\times\frac{33800}{21}=-\frac{31200}{21}$. Then $F_{AE}=\frac{10000}{7}+\frac{31200}{21}=\frac{30000 + 31200}{21}=\frac{61200}{21}\approx2914.29$ lb.

Answer:

The maximum tension in cable $AD$ is $\frac{33800}{21}\approx1609.52$ lb, and the force developed along the antenna - tower at point $A$ (i.e., $F_{AE}$) is $\frac{61200}{21}\approx2914.29$ lb.