Question 3.6: Determine the components of the single couple equivalent to ......
Determine the components of the single couple equivalent to the two couples shown.
Our computations will be simplified if we attach two equal and opposite 20-lb forces at A . This enables us to replace the original 20-lb-force couple by two new 20-lb-force couples, one of which lies in the zx plane and the other in a plane parallel to the xy plane. The three couples shown in the adjoining sketch can be represented by three couple vectors M_ x \ , \ M _y \ , \ and \ M_ z directed along the coordinate axes. The corresponding moments are
\begin{array}{l l}{{M_{x}=\,-(30 \ \mathrm lb)(18\ \mathrm{in.})\,=\,-{5}40 \ \mathrm l{b~~\cdot\ \mathrm {in}.}}}\\ {{M_{y}=\,+(20 \ \mathrm lb)(12\ \mathrm{in.})\,=\,+240\ \mathrm lb\cdot\ \mathrm{in.}}}\\ {{M_{z}\,=\,+(20\ \mathrm lb)(9\ \mathrm{in.})\,=\,+180\ \mathrm lb \ {\mathrm{\cdot \ in.}}}}\end{array}
These three moments represent the components of the single couple M equivalent to the two given couples. We write
{M}\,=\,-(540 \ \mathrm { lb}\,\cdot\,\mathrm{in.})\mathrm{i}\,+\,(240 \ \mathrm{l b}\,\cdot\,\mathrm{in.})\mathrm{j}\,+\,(180 \ \mathrm{l b}\,\cdot\,\mathrm{in.})\mathrm{k}
Alternative Solution. The components of the equivalent single couple M can also be obtained by computing the sum of the moments of the four given forces about an arbitrary point. Selecting point D , we write
{M}={M}_{D}=(18\ {\mathrm{in}}.){{\mathrm{j}}}\times (-30 \ \mathrm{l b}){\mathrm k}\:+\:[(9\mathrm{~in.}){\mathrm j}\:-\:(12\mathrm{~in.}){\mathrm k}]{\times}\:(-20 \ \mathrm {lb}){\mathrm i}and, after computing the various cross products,
{ M}=\,-(540 \ \mathrm{l b}\cdot\mathrm{in}.){\mathrm i}\,+\,(240 \ \mathrm {l b}\cdot\mathrm{in}.){\mathrm j}\,+\,(180 \ \mathrm{l b}\cdot\mathrm{in}.){\mathrm k}\qquad \quad
