Question

Determine the forces couple system at 0 which is equivalent to the two forces applied to the shaft AoB. Is R perpendicular to Mo?

Answer 1

Let use the following example

$F_1 = 710 \;N \\ F_2 = 770 \;N$

The resultant force $\vec{R}$ is given as

$\vec{R} = \vec{F_1} + \vec{F_2} \\ \vec{R} = (710 cos30 \hat{k} + 710 sin 30 \hat{j}) + (770 cos 53 \hat{k} -770 sin 53 \hat{j}) \\ = (614.87 \hat{k} + 355 \hat{j}) + (463.39 \hat{k} +614.949 \hat{j}) \\ = 1078.26 \hat{k} -259.949 \hat{j} \\ \vec{R} = 0 \hat{i} -259.949 \hat{j} + 1078.26 \hat{k}$

Taking moment about point O

$M_o = (-0.075 \hat{i}) \vec{F_1} + (0.18 \hat{i}) \vec{F_2} \\ = (-0.075) \hat{i} (614.8 \hat{k} + 355 \hat{j}) + (0.18 \hat{i})(463.39 \hat{k} -614.949 \hat{j}) \\ = (46.11 \hat{i} -26.625 \hat{k}) + (- 83.41 \hat{j} -110.689 \hat{k}) \\ M_0 = -37.3 \hat{j} -137.314 \hat{k}$

Check perpendicularity of R and M

$R.M_o = (-259.949 \hat{j} + 1078.26 \hat{k})(-37.3 \hat{j} -137.314 \hat{k}) \\ = 96961 -148060 \\ = -51099$

R.M_o ≠ 0

R is NOT perpendicular to M_o.