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:
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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