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

#### Solution:

Show me the final answerâ†“

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_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}

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

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

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

#### Final Answer:

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

\alpha=151.77^0

\beta=114.5^0

\gamma=90^0

\alpha=151.77^0

\beta=114.5^0

\gamma=90^0