The gear reducer is subjected to the couple moments shown. Determine the resultant couple...

Write the moment \(M_{1}\).

$$ M_{1}=50 \mathrm{j} \mathrm{N} \cdot \mathrm{m} $$

Write the moment \(M_{2}\).

$$ M_{2}=60 \cos 30^{\circ} \mathbf{i}+60 \sin 30^{\circ} \mathbf{k} $$

Calculate the resultant moment \(M_{R}\) as follows.

$$ \begin{aligned} M_{R} &=M_{1}+M_{2} \\ &=50 \mathbf{i}+60 \cos 30^{\circ} \mathbf{i}+60 \sin 30^{\circ} \mathbf{k} \\ &=51.96 \mathbf{i}+50 \mathbf{j}+30 \mathbf{k} \end{aligned} $$

Calculate the magnitude of the resultant moment as follows.

$$ \begin{aligned} M &=\left|M_{R}\right| \\ &=\sqrt{(51.96)^{2}+(50)^{2}+(30)^{2}} \\ &=78.10 \mathrm{~N} \cdot \mathrm{m} \end{aligned} $$

Thus, the resultant couple moment is \(78.10 \mathrm{~N} \cdot \mathrm{m}\)

Calculate the unit vector of the resultant moment is

\(\begin{aligned} u &=\frac{M_{R}}{M} \\ &=\frac{51.96 \mathbf{i}+50 \mathbf{j}+30 \mathbf{k}}{78.10} \\ &=0.665 \mathbf{i}+0.64 \mathbf{j}+0.384 \mathbf{k} \end{aligned}\)

Calculate the general unit vector of force is as follow

\(u=\operatorname{Cos}\left(\alpha_{1}\right)+\operatorname{Cos}\left(\beta_{1}\right)+\operatorname{Cos}(\gamma)\)

The direction angles of the resultant moments can be determined as follow

Compare unit vector of resultant moment and general unit vector.

\(\cos \alpha_{1}=0.665\)

\(\alpha_{1}=\cos ^{-1}(0.1665)\)

\(\quad=48.31^{*}\)

And

\(\cos \beta_{1}=0.64\)

\(\begin{aligned} \beta_{1} &=\cos ^{-1}(0.64) \\ &=50.2^{*} \end{aligned}\)

Also,

\(\cos \gamma_{1}=0.384\)

\(\begin{aligned} \gamma_{1} &=\cos ^{-1}(0.384) \\ &=67.4 \mathrm{l}^{*} \end{aligned}\)

Thus, the direction angles of the resultant force are \(\alpha_{1}=48.31^{\circ}, \beta_{1}=50.2^{\circ}\), and

\(\gamma_{1}=67.41^{\circ}\)