Question 3.8: A 4.80-m-long beam is subjected to the forces shown. Reduce ......
A 4.80-m-long beam is subjected to the forces shown. Reduce the given system of forces to ( a ) an equivalent force-couple system at A, (b) an equivalent force-couple system at B, (c) a single force or resultant.
Note . Since the reactions at the supports are not included in the given system of forces, the given system will not maintain the beam in equilibrium.
a. Force-Couple System at A . The force-couple system at A equivalent to the given system of forces consists of a force R and a couple {M}_{A}^{R} defined as follows:
R \ = \ \Sigma{F} \\ \\ ~~~~=(150~ N)\mathrm j – (600 ~N)\mathrm j + (100 ~N)\mathrm j – (250 ~N)\mathrm j = -(600 ~N)\mathrm j{M}_{A}^{R} \ = \ \Sigma{(r \ \times \ F)} \\ \\ ~~~~=(1.6\mathrm i)\times (-600\mathrm j) + (2.8\mathrm i) \ \times (100\mathrm j) + (4.8\mathrm i) \ \times (-250\mathrm j) \\ \\ ~~~~~~~~~~~~~~~= -(1880~ N \cdot~ \mathrm m)\mathrm k
The equivalent force-couple system at A is thus
{ R}=600\;{ N}{\mathrm w}\qquad{ M}_{A}^{R}=1880\;{ N}\cdot{\mathrm m}\;{\mathrm i}b. Force-Couple System at B . We propose to find a force-couple system at B equivalent to the force-couple system at A determined in part a . The force R is unchanged, but a new couple {M}_{B}^{R} must be determined, the moment of which is equal to the moment about B of the force-couple system determined in part a . Thus, we have
{M}_{B}^{R}={M}_{\mathrm{A}}^{R}+\overrightarrow{BA}~ \times~{R} \\ \\ ~~~~=-(1880\mathrm{N}\cdot\mathrm{m})\mathrm{k}+(-4.8\mathrm{m})\mathrm{i}\times(-600\mathrm{N})\mathrm{j} \\ \\ ~~~~=-(1880\mathrm{N}\cdot\mathrm{m})\mathrm{k}+(2880\mathrm{N}\cdot\mathrm{m})\mathrm{k}=\ +({1}000\mathrm{N}\cdot\mathrm{m})\mathrm{k}The equivalent force-couple system at B is thus
{\mathrm R}\,=\,600\;{\mathrm N}~{\mathrm w}\qquad{\ \mathrm M}_{B}^{R}\,=\,1000\;{\mathrm N}\,\cdot\,{\mathrm m}\;{\mathrm l}c. Single Force or Resultant. The resultant of the given system of forces is equal to R , and its point of application must be such that the moment of R about A is equal to {M}_{\mathrm{A}}^{R} . We write
{\mathrm{r}}\times{{\mathrm R}}={M}_{\mathrm{A}}^{R} \\ \\ {x\mathrm{i}}\times(-600\;\mathrm{N})\mathrm {j}=-{{(1880~N}~\cdot\mathrm{m})}\mathrm{k} \\ \\ -x(600\ {\mathrm{N}}){\mathrm{k}}\,=\,-(1880\ {\mathrm{N}}\,\cdot\,{\mathrm{m}}){\mathrm{k}}and conclude that x = 3.13 m. Thus, the single force equivalent to the given system is defined as
{\mathrm{R}}\,=\,600\,{\mathrm{N}}~\mathrm W\qquad x\,=\,3.13\,{\mathrm{m}}
