Question 4.10: A 450-lb load hangs from the corner C of a rigid piece of pi......
A 450-lb load hangs from the corner C of a rigid piece of pipe ABCD which has been bent as shown. The pipe is supported by the ball-and-socket joints A and D, which are fastened, respectively, to the floor and to a vertical wall, and by a cable attached at the midpoint E of the portion BC of the pipe and at a point G on the wall. Determine (a) where G should be located if the tension in the cable is to be minimum, (b) the corresponding minimum value of the tension.
Free-Body Diagram. The free-body diagram of the pipe includes the load W = (-450 lb)j, the reactions at A and D, and the force T exerted by the cable. To eliminate the reactions at A and D from the computations, we express that the sum of the moments of the forces about AD is zero. Denoting by λ the unit vector along AD, we write
\Sigma M_{A D}=0:\ \ \ \ \ \ L\cdot {(\overrightarrow{A E} }\times\mathrm{T)}\,+\,L \cdot(\overrightarrow{AC} \times\mathsf{W}\big)\,=0 (1)
The second term in Eq. (1) can be computed as follows:
{\overrightarrow{A C} }\times{{W}}=(12\mathrm{i}+\,12\mathrm{j})\,\times\,(-450\mathrm{j})\,=\,-5400\mathrm{k} \\ \\ L~=\frac{\overrightarrow{AD} }{AD} ~={\frac{12\mathrm{i}+12\mathrm{j}-\mathrm{6k}}{18}}={\frac{2}{3}}\mathrm{i}\,+\,{\frac{2}{3}}\mathrm{j}\,-\,{\frac{1}{3}}\mathrm{k} \\ \\ L\cdot({{{\overrightarrow{A C} }}}~\times~{ W})\,=\,(\textstyle{\frac{2}{3}}{\mathrm i}\,+\,\textstyle{\frac{2}{3}}\mathrm j\,-\,\textstyle{\frac{1}{3}}{\mathrm k})\,\cdot\,(-5400{\mathrm{k}})\,=\,+\,1800Substituting the value obtained into Eq. (1), we write
L\cdot({{\overrightarrow{{A E}} }}\times{{{T}}})=-1800\;\ \mathrm{lb}\cdot\mathrm{ft} (2)
Minimum Value of Tension. Recalling the commutative property for mixed triple products, we rewrite Eq. (2) in the form
{T}\cdot(L\times\overrightarrow{{A E}} )=-1800\,\mathrm{lb}\cdot\mathrm{ft} (3)
which shows that the projection of T on the vector L\times\overrightarrow{{A E}} is a constant. It follows that T is minimum when parallel to the vector
L\times\overrightarrow{{A E}}=(\textstyle{{\frac{2}{3}}}\mathrm i ~{{+}}\textstyle{\frac{2}{3}}\mathrm{{J}}-\textstyle{\frac{1}{3}}\mathrm{{k}})\times (\mathsf{6\mathrm i+12\mathrm j})=4{\mathrm i}-2\mathrm{j}+4{\mathrm{k}}Since the corresponding unit vector is {\textstyle\frac{2}{3}}\mathrm{i}\,-\,{\frac{1}{3}}\mathrm{j}\,+\,{\frac{2}{3}}\mathrm{k}, we write
{T}_{\mathrm{min}}=T(\textstyle{\frac{2}{3}}\mathrm{i}\,-\,\textstyle{\frac{1}{3}}\mathrm{j}\,+\,\textstyle{\frac{2}{3}}{\mathrm k}) (4)
Substituting for T and L\times\overrightarrow{{A E}} in Eq. (3) and computing the dot products, we obtain 6T = -1800 and, thus, T = -300. Carrying this value into (4), we obtain
{T}_{\mathrm{min}}=\,-200\mathrm{i}\,+\,100\mathrm{j}\,-\,200\mathrm{k}\qquad T_{\mathrm{min}}\,=\,300\ \mathrm{lb}Location of G. Since the vector \overrightarrow{EG} and the force {T}_{\mathrm{min}} have the same direction, their components must be proportional. Denoting the coordinates of G by x, y, 0, we write
{\frac{x-6}{-200}}={\frac{y-12}{+100}}={\frac{0-6}{-200}}\quad\quad x=0\quad\quad y=15~\mathrm{ft}
