The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the...

Write the equation of the position vector for the line \(D G\),

$$ \begin{aligned} \overrightarrow{D G} &=21 \mathbf{i}-(15+23) \mathbf{j}+18 \mathbf{k} \\ &=21 \mathbf{i}-38 \mathbf{j}+18 \mathbf{k} \end{aligned} $$

Calculate the unit vector of \(\overrightarrow{D G}\),

$$ \begin{aligned} \lambda_{D G} &=\frac{\overrightarrow{D G}}{|\overrightarrow{D G}|} \\ &=\frac{21 \mathbf{i}-38 \mathbf{j}+18 \mathbf{k}}{|2 \mathbf{i}-38 \mathbf{j}+18 \mathbf{k}|} \\ &=\frac{21 \mathbf{i}-38 \mathbf{j}+18 \mathbf{k}}{\sqrt{21^{2}+(-38)^{2}+18^{2}}} \\ &=0.447 \mathbf{i}-0.81 \mathbf{j}+0.383 \mathbf{k} \end{aligned} $$

Calculate the force vector for the force \(P\),

$$ \begin{aligned} \mathbf{P} &=(P)(\text { unit vecor } \overrightarrow{D G}) \\ &=(P)\left(\boldsymbol{\lambda}_{D G}\right) \end{aligned} $$

Here,

Magnitude of the force is \(P\),

Substitute \(235 \mathrm{lb}\) for \(P\) and the respective values in the above equation,

$$ \begin{aligned} \mathbf{P} &=235(0.447 \mathbf{i}-0.81 \mathbf{j}+0.383 \mathbf{k}) \\ &=105.045 \mathbf{i}-190.35 \mathbf{j}+90.005 \mathbf{k} \end{aligned} $$

Write the equation of the position vector for the line \(A B\),

$$ \overrightarrow{A B}=32 \mathbf{i}-30 \mathbf{j}-24 \mathbf{k} $$

Calculate the unit vector of \(\overrightarrow{A B}\),

$$ \begin{aligned} \lambda_{A B} &=\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|} \\ &=\frac{32 \mathbf{i}-30 \mathbf{j}-24 \mathbf{k}}{|32 \mathbf{i}-30 \mathbf{j}-24 \mathbf{k}|} \\ &=\frac{32 \mathbf{i}-30 \mathbf{j}-24 \mathbf{k}}{\sqrt{32^{2}+30^{2}+24^{2}}} \\ &=0.64 \mathbf{i}-0.6 \mathbf{j}-0.48 \mathbf{k} \end{aligned} $$

Write the equation of the position vector for the line \(A D\),

$$ \overrightarrow{A D}=16 \mathbf{i}+8 \mathbf{j}-12 \mathbf{k} $$

Calculate the vector product for the line \(A D\) and force vector \(P\),

$$ \begin{aligned} (\overrightarrow{A D}) \times(\mathbf{P}) &=(16 \mathbf{i}+8 \mathbf{j}-12 \mathbf{k}) \times(105.045 \mathbf{i}-190.35 \mathbf{j}+90.005 \mathbf{k}) \\ &=-3045.6 \mathbf{k}-1440.08 \mathbf{j}-840.36 \mathbf{k}+720.04 \mathbf{i}-1260.54 \mathbf{j}-2284.2 \mathbf{i} \\ &=-1564.16 \mathbf{i}-2700.62 \mathbf{j}-3885.96 \mathbf{k} \end{aligned} $$

Calculate the moment about of the given force \(P\) about line \(A B\),

$$ \begin{aligned} \mathbf{M}_{A B} &=\lambda_{A B} \cdot((\overrightarrow{A D}) \times(\mathbf{P})) \\ &=(0.64 \mathbf{i}-0.6 \mathbf{j}-0.48 \mathbf{k}) \cdot(-1564.16 \mathbf{i}-2700.62 \mathbf{j}-3885.96 \mathbf{k}) \\ &=-1001.0624+1620.372+1865.261 \\ &=2484.57 \mathrm{lb} \cdot \mathrm{in} \\ &=207.047 \mathrm{lb} \cdot \mathrm{ft} \end{aligned} $$

Therefore, the moment of \(P\) about the \(\operatorname{rod} A B\) is \(207.047 \mathrm{lb} \cdot \mathrm{ft}\)