Replace the force-couple system shown in Fig. (a) with an equivalent force-couple system, with the force acting at point A, given that F =100 N and C =3600 N · mm.
Replace the force-couple system shown in Fig. (a) with an equivalent force-couple system, with the force acting at point A, given that F =100 N and C =3600 N · mm.
Moving the given force F from point B to point A requires the introduction of a couple of transfer C^{T}. This couple is then added to the given couple-vector C, thereby obtaining the resultant couple-vector, which we label C^{R}. The couplevector C^{R} and the force F located at point A will then be the required force-couple system. Owing to the three-dimensional nature of this problem, it is convenient to use vector methods in the solution. Writing F in vector form, we obtain
F=100\lambda _{BE}=100\frac{\overrightarrow{BE}}{\left|\overrightarrow{BE} \right|}=100\left(\frac{−120i + 60k}{134.164} \right)=−89.44i + 44.72k NThe position vector from A to B is r_{AB} =120j − 60k m. The couple of transfer is equal to the moment of the given force F about point A, so we have
C^{T}=M_{A}=r_{AB}\times F=\left|\begin{matrix} i & j & k \\ 0 & 120 & -60 \\ −89.44 & 0 & 44.72 \end{matrix} \right|= 5366.4i + 5366.4j + 10732.8k N · mm
Expressing the given couple-vector C shown in Fig. (a) in vector form,
C=3600\lambda _{DB}=3600\frac{\overrightarrow{DB}}{\left|\overrightarrow{DB} \right|}=3600\left(\frac{120i + 120j − 60k}{180} \right)=2400i + 2400j − 1200k N · mmAdding C^{T} and C (remember that couple-vectors are free vectors), the resultant couple vector is
C^{R} =C^{T} + C=7766.4i + 7766.4j + 9532.8k N · mm
The magnitude of C^{R} is given by
C^{R}=\sqrt{(7766.4)^{2}+(7766.4)^{2}+(9532.8)^{2}}=14543.3 N · mmThe equivalent force-couple system is shown in Fig. (b). Note that the force acts at point A. For convenience of representation, C^{R} is shown at point O, but being a free vector, it could be placed anywhere.