Question 3.2: A force of 800 N acts on a bracket as shown. Determine the m......
A force of 800 N acts on a bracket as shown. Determine the moment of the force about B.
The moment {M}_{B} of the force F about B is obtained by forming the vector product
{M}_{B}\,=\,{r}_{A/B}\,\times\,{F}
where {r}_{A/B} is the vector drawn from B to A . Resolving {r}_{A/B} and F into rectangular components, we have
{r}_{A/B}\,=\,-(0.2\,\mathrm{m})\mathrm{i}\,+\,(0.16\,\mathrm{m})\mathrm{j} \\ \\ ~~~~~~~~~{{F}}=({\mathrm{800~N}})\cos\,60^{\circ}{\mathrm{i}}\,+\,({\mathrm{800~N}})\sin\,60^{\circ}{\mathrm{j}}\, \\ \\~~~~~~~~~=\,(400\ N)\mathrm{i}\,+\,(693\ \mathrm{N})\mathrm{j}
Recalling the relations (3.7) for the cross products of unit vectors (Sec. 3.5), we obtain
The moment {M}_{B} is a vector perpendicular to the plane of the figure and pointing into the paper.
