# The meshed gears are subjected

The meshed gears are subjected to the couple moments shown. Determine the magnitude of the resultant couple moment and specify its 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:

We will first express each moment in Cartesian vector form. (Don’t remember how?)

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

To express $M_2$ in Cartesian vector form, we first need to find M’, which is the component which lies on the x-y plane.

$M'=20\cos20^0=18.79$

(Using M’ we can figure out $M_{2x}$ and $M_{2y}$)

$M_x=18.79\sin30^0=-9.4$

$M_y=18.79\cos30^0=-16.27$

(both the x and y components are negative because both components lie in the negative x and y axes)

$M_z=20\sin20^0=6.84$

We can now express $M_2$ in Cartesian vector form:

$M_2=\left\{-9.4i-16.27j+6.84k\right\}\,\text{N}\cdot\text{m}$

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

$M_c=M_1+M_2$

$M_c=\left\{0i+0j+50k\right\}+\left\{-9.4i-16.27j+6.84k\right\}$

$M_c=\left\{-9.4i-16.27j+56.84k\right\}\,\text{N}\cdot\text{m}$

The magnitude of this couple moment is:

magnitude of $M_{c}\,=\,\sqrt{(-9.4)^2+(-16.27)^2+(56.84)^2}$

magnitude of $M_{c}=59.86\,\text{N}\cdot\text{m}$

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.

The coordinate direction angles can be found by:

$\alpha=\cos^{-1}\left(\dfrac{-9.4}{59.86}\right)=99.03^0$

$\beta=\cos^{-1}\left(\dfrac{-16.27}{59.86}\right)=105.77^0$

$\gamma=\cos^{-1}\left(\dfrac{56.84}{59.86}\right)=18.28^0$

$M_c=\left\{-9.4i-16.27j+56.84k\right\}\,\text{N}\cdot\text{m}$

magnitude of $M_{c}=59.86\,\text{N}\cdot\text{m}$

$\alpha=99.03^0$

$\beta=105.77^0$

$\gamma=18.28^0$