# The gear reducer is subjected to the couple moments

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

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

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

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

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

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

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

$M_c=M_1+M_2$

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

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

The magnitude of this resultant moment is:

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

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

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:

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

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

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

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

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

$\alpha=48.29^0$

$\beta=50.19^0$

$\gamma=67.41^0$