The gear reducer is subjected to the couple moments shown. Determine the resultant couple moment and specify its magnitude and coordinate direction angles.

#### Solution:

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Let us first express each moment acting upon the gear reducer in Cartesian vector form.

M_1=\left\{0i+50j+0k\right\}\,\text{N}\cdot\text{m}

M_2=\left\{60\cos30^0i+0j+60\sin30^0k\right\}\,\text{N}\cdot\text{m}

M_2=\left\{51.96i+0j+30k\right\}\,\text{N}\cdot\text{m}

M_2=\left\{60\cos30^0i+0j+60\sin30^0k\right\}\,\text{N}\cdot\text{m}

M_2=\left\{51.96i+0j+30k\right\}\,\text{N}\cdot\text{m}

The resultant couple moment can be found by adding the two moments together.

M_c=M_1+M_2

M_c=\left\{0i+50j+0k\right\}+\left\{51.96i+0j+30k\right\}

M_c=\left\{51.96i+50j+30k\right\}\,\text{N}\cdot\text{m}

M_c=\left\{0i+50j+0k\right\}+\left\{51.96i+0j+30k\right\}

M_c=\left\{51.96i+50j+30k\right\}\,\text{N}\cdot\text{m}

The magnitude of this resultant moment is:

M_c=\sqrt{(51.96)^2+(50)^2+(30)^2}

M_c=78.1\,\text{N}\cdot\text{m}

M_c=78.1\,\text{N}\cdot\text{m}

The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was r\,=\,ai+bj+ck, then the magnitude would be, r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.

We can also calculate the coordinate direction angles as follows:

\alpha=\cos^{-1}\left(\dfrac{51.96}{78.1}\right)=48.29^0

\beta=\cos^{-1}\left(\dfrac{50}{78.1}\right)=50.19^0

\gamma=\cos^{-1}\left(\dfrac{30}{78.1}\right)=67.41^0

\beta=\cos^{-1}\left(\dfrac{50}{78.1}\right)=50.19^0

\gamma=\cos^{-1}\left(\dfrac{30}{78.1}\right)=67.41^0

#### Final Answers:

M_c=\left\{51.96i+50j+30k\right\}\,\text{N}\cdot\text{m}

Magnitude of M_c=78.1\,\text{N}\cdot\text{m}

Magnitude of M_c=78.1\,\text{N}\cdot\text{m}

\alpha=48.29^0

\beta=50.19^0

\gamma=67.41^0