Question 3.5: A cube of side a is acted upon by a force P as shown. Determ......
A cube of side a is acted upon by a force P as shown. Determine the moment of P ( a ) about A , ( b ) about the edge AB , ( c ) about the diagonal AG of the cube, ( d ). Using the result of part c , determine the perpendicular distance between AG and FC.
a. Moment about A. Choosing x , y , and z axes as shown, we resolve into rectangular components the force P and the vector { r}_{F/A}\,=\,{{\overrightarrow{{A F}} }} drawn from A to the point of application F of P.
{ r}_{F/A}=a{\mathrm i}-a{\mathrm j}=a({\mathrm i}-{\mathrm j}) \\ \\ { P}\,=\,(P/\ \!1\overline{{{2}}}){\mathrm j}\,-\,(P/1\overline{{{2}}}){\mathrm k}\,=\,(P/\ 1\overline{{{2}}})({\mathrm j}\,-\,{\mathrm k})
The moment of P about A is
{ M}_{\scriptscriptstyle A}={ r}_{F/{ A}}\times\mathrm{P}=\,a(\mathrm{i}\,-\,\mathrm{j})\;{\times}\;(P/\mathrm{1}\,\overline{2})(\mathrm{j}\,-\,\mathrm{k})
{ M}_{\scriptscriptstyle A}\,=\,(a P/\,\displaystyle{ 1}\;\overline{{{2}}})({\mathrm i}\,+\,\ \!\mathrm{j}\,+\,\mathrm{k})\qquad
b. Moment about AB. Projecting {M}_{A} on AB , we write
M_{A B}=\mathrm{i}\cdot\mathrm{M}_{A}=\mathrm{i}\cdot(a P/1\,\overline{{{2}}})(\mathrm{i}\ +\mathrm{j}\ +\mathrm{k})
M_{A B}=a P/\ 1\ \overline{2} \qquad
We verify that, since AB is parallel to the x axis, M_{A B} is also the x component of the moment M_{A }.
c. Moment about Diagonal AG. The moment of P about AG is obtained by projecting M_{A } on AG . Denoting by L the unit vector along AG , we have
L=\frac{\overrightarrow{AG} }{AG}={\frac{a\mathrm{i}\,-\,a\mathrm{j}\,-\,a\mathrm{k}}{a\,1\,\overline{{{{3}}}}}}=\,(1/\,1\,\overline{{{{3}}}})(\mathrm{i}\,-\,\mathrm{j}\, – \,\mathrm{k}) \\ \\ M_{A G}=L\cdot{ M}_{A}=(1/1\,\bar{3})(\mathrm{i}\,-\,\mathrm{j}\,-\,\mathrm{k})\cdot\,(a P/1\,\bar{2})(\mathrm{i}\,+\,\mathrm{j}\,+\,\mathrm{k}) \\ \\ M_{A G}=\,(a P/\, 1 \ \bar{6})(1\,-\,1\,-\,1\,)\,\,\,\,\,\,M_{A G}\,=\,-a P/\,1 \ \bar{6}
Alternative Method. The moment of P about AG can also be expressed in the form of a determinant:
\qquad M_{A G}=\left|\begin{array}{l l l}{{I_{x}}}&{{I_{y}}}&{{I_{z}}}\\ {{x_{F/A}}}&{{y_{F/A}}}&{{{{z}}_{F/A}}}\\ {{{{ F}_{x}}}}&{{{ F}_{y}}}&{{{ F}_{z}}}\end{array}\right|=\left|\begin{array}{c c c}{{1/ \ 1 \ \mathrm{~}\bar{3}}}&{{-1/ \ 1 \ \mathrm{~}\bar{3}}}&{{-1/ \ 1 \ \mathrm{~}\bar{3}}}\\ {{a}}&{{-a}}&{{0}}\\ {{0}}&{{P/ \ 1 \ \mathrm{~}\bar{2}}}& – {{P/ \ 1 \ \mathrm{~}\bar{2}}}\end{array}\right|=\,-a P/\,1 \ \bar{6}d. Perpendicular Distance between AG and FC. We first observe that P is perpendicular to the diagonal AG . This can be checked by forming the scalar product {P}\cdot L and verifying that it is zero:
\qquad {P}\cdot L = (P/{\ 1 }\ {\overline{{\ 2}}})(\mathrm{j}-\mathrm{k})\ . \ (\,1/\,{1}\,{\overline{{{3}}}})(\mathrm{i}\,-\,\mathrm{j\,-\,k})\,=(P\ 1 \ \overline{{{6}}})(0-1+1)=0The moment M_{A G} can then be expressed as – Pd , where d is the perpendicular distance from AG to FC . (The negative sign is used since the rotation imparted to the cube by P appears as clockwise to an observer at G.) Recalling the value found for M_{A G} in part c,
M_{A G}=-P d=\,-a P/\,1\,\overline{{{6}}}\qquad\qquad\qquad d={a}/\,1\,\overline{{{6}}}
