Question 4.SP.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 that has been bent, as shown. The pipe is supported by 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.
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
\Sigma M_{AD}=0: \quad \pmb{\lambda} \cdot(\overrightarrow{A E} \times \pmb{T})+\pmb{\lambda} \cdot(\overrightarrow{A C} \times \pmb{W})=0 (1)
You can compute the second term in Eq. (1) as follows:
\begin{aligned}&\overrightarrow{AC} \times \pmb{W}=(12 \pmb{i}+12 \pmb{j}) \times(-450 \pmb{j})=-5400 \pmb{k} \\&\pmb{\lambda}=\frac{\overrightarrow{A D}}{A D}=\frac{12 \pmb{i}+12 \pmb{j}-6 \pmb{k}}{18}=\frac{2}{3} \pmb{i}+\frac{2}{3} \pmb{j}-\frac{1}{3} \pmb{k}\end{aligned}Substituting this value into Eq. (1) \quad \pmb{\lambda} \cdot(\overrightarrow{A C} \times \pmb{W})=\left(\frac{2}{3} \pmb{i}+\frac{2}{3} \pmb{j}-\frac{1}{3} \pmb{k}\right)(-5400 \pmb{k})=+1800 gives
\pmb{\lambda} \cdot(\overrightarrow{AE} \times \pmb{T})=-1800 lb \cdot ft (2)
Minimun Value of Tension. Recalling the commutative property for mixed triple products, you can rewrite Eq. (2) in the form
\pmb{T} \cdot(\pmb{\lambda} \times \overrightarrow{A E})=-1800 lb \cdot ft (3)
This shows that the projection of T on the vector \pmb{\lambda} \times \overrightarrow{AE} is a constant. It follows that T is minimum when it is parallel to the vector
\pmb{\lambda} \times \overrightarrow{A E}=\left(\frac{2}{3} \pmb{i}+\frac{2}{3} \pmb{j}-\frac{1}{3} \pmb{k}\right) \times(6 \pmb{i}+12 \pmb{j})=4 \pmb{i}-2 \pmb{j}+4 \pmb{k}The corresponding unit vector is \frac{2}{3} \pmb{i}-\frac{1}{3} \pmb{j}+\frac{2}{3} \pmb{k}, which gives
\pmb{T}_{\min }=T\left(\frac{2}{3} \pmb{i}-\frac{1}{3} \pmb{j}+\frac{2}{3} \pmb{k}\right) (4)
Substituting for T and \pmb{\lambda} \times \overrightarrow{AE} in Eq. (3) and computing the dot products yields 6T = -1800 and, thus, T = −300. Carrying this value into Eq. (4) gives you
\pmb{T}_{\min }= −200i + 100j − 200k
T_{\min }= 300 lb
Location of G.
Because the vector \overrightarrow{EG} and the force \pmb{T}_{\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}x = 0 y = 15ft
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.
