Replace the couple and force shown by an equivalent single force applied to the lever. Determine the distance from the shaft to the point of application of this equivalent force.
First the given force and couple are replaced by an equivalent force-couple system at O . We move the force F = -(400 N) j to O and at the same time add a couple of moment {M}_{O} equal to the moment about O of the force in its original position.
\qquad {M}_{O}={\overrightarrow{{{O B}}} }\times{F}=\left[(0.150 \ \mathrm m)\mathrm i \ + \ (0.260 \ \mathrm m)\mathrm j\right]~\times ~(-400 \ \mathrm N)\mathrm j \\ \\ ~~~~~~~~~~~~~~~~~~=-(60 \ \mathrm N \ \cdot \ \mathrm m ) \mathrm kThis couple is added to the couple of moment – (24 N · m) k formed by the two 200-N forces, and a couple of moment – (84 N · m) k is obtained. This last couple can be eliminated by applying F at a point C chosen in such a way that
\qquad -(84 ~{N}\cdot\mathrm {m})\mathrm{k}={{{\overrightarrow{{OC}} }}}\times F \\ \\~~~~~~~~~~~~~~~~~~~~~~=\left[\left(O C\right)\cos60^{\circ}\mathrm{i}+\left(O C\right)\sin60^{\circ}\mathrm{j}\right]\times(-400\mathrm{N})\mathrm{j} \\ \\~~~~~~~~~~~~~~~~~~~~~~=\;-(O C)\cos\ 60^{\circ}(400\,\mathrm{N})\mathrm{k}We conclude that
({O C})\,\cos\,60^{\circ}=\,0.210\,\,\mathrm{m}\,=\,210\,\,\mathrm{m}{\mathrm{m}}\qquad{ O}C\,=\,420\,\,\mathrm{m}{\mathrm{m}}
Alternative Solution. Since the effect of a couple does not depend on its location, the couple of moment – (24 N · m) k can be moved to B ; we thus obtain a force-couple system at B . The couple can now be eliminated by applying F at a point C chosen in such a way that
-(24 ~{N}\cdot\mathrm {m})\mathrm{k}={{{\overrightarrow{{BC}} }}}\times F \\ \\~~~~~~~~~~~~~~~~~~~~~~=-(B C)\cos60^{\circ}(400\,\mathrm{N})\mathrm{k}
We conclude that
\left(B C\right)\,\cos\,60^{\circ}=0.060\,\,{\mathrm{m}}\,=\,60\,\,{\mathrm{mm}}\quad\quad B C\,=\,120~{\mathrm{mm}} \\ \\ O C\,=\,O B\,+\,B C\,=\,300\,\mathrm{{mm}}\,+\,120\,\mathrm{{mm}}\qquad O C\,=\,420\,\mathrm{{mm}}