# A couple acts on each of the handles

A couple acts on each of the handles of the minidual valve. Determine the magnitude and coordinate direction angles of the resultant couple moment.

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

#### Solution:

Let us break the coupling moments being applied into components.

$M_x=-35(0.175+0.175)-25(0.175+0.175)\cos60^0$

$M_x=-16.625\,\text{N}\cdot\text{m}$

$M_y=-25(0.175+0.175)\sin60^0$

$M_y=-7.58\,\text{N}\cdot\text{m}$

The resultant couple moment is then these two components added together.

$M_c=\left\{-16.625i-7.58j+0k\right\}\,\text{N}\cdot\text{m}$

We can now determine the magnitude of this coupling moment:

magnitude of $M_{c}\,=\,\sqrt{(-16.625)^2+(-7.58)^2+(0)^2}$

magnitude of $M_{c}=18.27\,\text{N}\cdot\text{m}$

The coordinate direction angles are:

$\alpha=\cos^{-1}\left(\dfrac{-16.625}{18.27}\right)=151.77^0$

$\beta=\cos^{-1}\left(\dfrac{-7.58}{18.27}\right)=114.5^0$

$\gamma=\cos^{-1}\left(\dfrac{0}{18.27}\right)=90^0$

magnitude of $M_{c}=18.27\,\text{N}\cdot\text{m}$

$\alpha=151.77^0$

$\beta=114.5^0$

$\gamma=90^0$