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.

A couple acts on each of the handles

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_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

 

Final Answer:

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

\alpha=151.77^0

\beta=114.5^0

\gamma=90^0

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 4, question 4-88.

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