Question 4.SP.10: A 450-lb load hangs from the corner C of a rigid piece of pi......

STRATEGY: Draw the free-body diagram of the pipe showing the reactions at A and D. Isolate the unknown tension T and the known weight W by summing moments about the diagonal line AD, and compute values from the equilibrium equations.

MODELING and ANALYSIS:

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 (Fig. 1). To eliminate the reactions at A and D from the computations, take the sum of the moments of the forces about the line AD and set it equal to zero. Denote the unit vector along AD by λ, which enables you to write

ΣM_{AD} = 0:       \lambda\cdot(\overrightarrow{{{A E}}}\times\bf T)+\lambda\cdot(\overrightarrow{{{A C}}}\times{W})=0               (1)

You can compute the second term in Eq. (1) as follows:

\overrightarrow{A C} × W = (12i + 12j) × (−450j) = − 5400k

\lambda={\frac{{\overrightarrow{{A D}}}}{A D}}={\frac{12\mathbf{i}+12\mathbf{j}-6\mathbf{k}}{18}}={\frac{{2}}{3}}\mathbf{i}+{\frac{{2}}{3}}\mathbf{j}-{\frac{1}{3}}\mathbf{k}

\lambda\cdot(\overrightarrow{{{A C}}}\times{\bf W})=(\frac{2}{3}\mathbf{{i}}+\frac{2}{3}\mathbf{{j}}-{{\frac{1}{3}}\mathbf{k}})\cdot(-5400\mathrm{\bf{k}})=+1800

Substituting this value into Eq. (1) gives

\lambda\cdot(\overrightarrow{{A E}}\times\mathbf{T})=-1800\;\mathrm{lb}\cdot\mathrm{ft}           (2)

Minimum Value of Tension. Recalling the commutative property for mixed triple products, you can rewrite Eq. (2) in the form

\mathbf{T}\cdot(\lambda\times\overrightarrow{{A E}})=-1800\;\mathrm{lb}\cdot\mathrm{ft}            (3)

This shows that the projection of T on the vector λ × \overrightarrow{{A E}} is a constant. It follows that T is minimum when it is parallel to the vector

\lambda\times{\overrightarrow{{A E}}}=(\frac{2}{3}\mathbf{i}+{\frac{2}{3}}\mathbf{j}-{\frac{1}{3}}\mathbf{k})\times(6\mathbf{i}+12\mathbf{j})=4\mathbf{i}-2\mathbf{j}+4\mathbf{k}

The corresponding unit vector is \frac{2}{3}\mathbf{i}-{\frac{1}{3}}\mathbf{j}+{\frac{2}{3}}\mathbf{k}, which gives

\mathbf{T}_{\mathrm{min}}=T({\frac{2}{3}}\mathbf{i}-{\frac{1}{3}}\mathbf{j}+{\frac{2}{3}}\mathbf{k})         (4)

Substituting for T and λ × \overrightarrow{{A E}} in Eq. (3) and computing the dot products yields 6T = −1800 and, thus, T = −300. Carrying this value into Eq. (4) gives you

\mathbf{T}_{\mathrm{min}}=-200\mathbf{i}+100\mathbf{j}-200\mathbf{k}\qquad\qquad T_{\mathrm{min}}=300~\mathrm{lb}   ◂

Location of G. Because the vector \overrightarrow{{E G}} and the force \mathbf{T}_{\mathrm{min}} have the same direction, their components must be proportional. Denoting the coordinates of G by x, y, and 0 (Fig. 2), you get

{\frac{x-6}{-200}}={\frac{y-12}{+100}}={\frac{0-6}{-200}}\qquad x=0\qquad y=15{\mathrm{~ft}}   ◂

REFLECT and THINK: Sometimes you have to rely on the vector analysis presented in Chaps. 2 and 3 as much as on the conditions for equilibrium described in this chapter.