The meshed gears are subjected to the couple moments shown. Determine the magnitude of the resultant couple moment and specify its coordinate direction angles.
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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We will first express each moment in Cartesian vector form. (Don’t remember how?)
To express M_2 in Cartesian vector form, we first need to find M’, which is the component which lies on the x-y plane.
(Using M’ we can figure out M_{2x} and M_{2y})
M_x=18.79\sin30^0=-9.4
M_y=18.79\cos30^0=-16.27
(both the x and y components are negative because both components lie in the negative x and y axes)
M_z=20\sin20^0=6.84
We can now express M_2 in Cartesian vector form:
The resultant couple moment can be found by adding the two moments together.
M_c=\left\{0i+0j+50k\right\}+\left\{-9.4i-16.27j+6.84k\right\}
M_c=\left\{-9.4i-16.27j+56.84k\right\}\,\text{N}\cdot\text{m}
The magnitude of this couple moment is:
magnitude of M_{c}=59.86\,\text{N}\cdot\text{m}
The coordinate direction angles can be found by:
\beta=\cos^{-1}\left(\dfrac{-16.27}{59.86}\right)=105.77^0
\gamma=\cos^{-1}\left(\dfrac{56.84}{59.86}\right)=18.28^0
Final Answers:
magnitude of M_{c}=59.86\,\text{N}\cdot\text{m}
\alpha=99.03^0
\beta=105.77^0
\gamma=18.28^0