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:
Let us break the coupling moments being applied into components.
Mx=−35(0.175+0.175)−25(0.175+0.175)cos600
Mx=−16.625N⋅m
My=−25(0.175+0.175)sin600
My=−7.58N⋅m
The resultant couple moment is then these two components added together.
Mc={−16.625i−7.58j+0k}N⋅m
We can now determine the magnitude of this coupling moment:
magnitude of Mc=(−16.625)2+(−7.58)2+(0)2
magnitude of Mc=18.27N⋅m
The coordinate direction angles are:
α=cos−1(18.27−16.625)=151.770
β=cos−1(18.27−7.58)=114.50
γ=cos−1(18.270)=900
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