Question 16.SP.6: The portion AOB of a mechanism consists of a 400-mm steel ro......
The portion AOB of a mechanism consists of a 400-mm steel rod OB welded to a gear E of radius 120 mm which can rotate about a horizontal shaft O. It is actuated by a gear D and, at the instant shown, has a clockwise angular velocity of 8 rad/s and a counterclockwise angular acceleration of 40 rad/s². Knowing that rod OB has a mass of 3 kg and gear E a mass of 4 kg and a radius of gyration of 85 mm, determine (a) the tangential force exerted by gear D on gear E, (b) the components of the reaction at shaft O.
In determining the effective forces of the rigid body AOB, gear E and rod OB will be considered separately. Therefore, the components of the acceleration of the mass center G_{O B} of the rod will be determined first:
\begin{aligned}& \left(\bar{a}_{O B}\right)_t=\bar{r} \alpha=(0.200 m )\left(40\text{ rad} / s ^2\right)=8 m / s ^2 \\& \left(\bar{a}_{O B}\right)_n=\bar{r} \omega^2=(0.200 m )(8 \text{ rad} / s )^2=12.8 m / s ^2\end{aligned}
Equations of Motion. Two sketches of the rigid body AOB have been drawn. The first shows the external forces consisting of the weight W _E of gear E, the weight W _{O B} of the rod OB, the force F exerted by gear D, and the components R _x \text { and } R _y of the reaction at O. The magnitudes of the weights are, respectively,
\begin{aligned}W_E & =m_E \text{g}=(4\text{ kg})\left(9.81 m / s ^2\right)=39.2 N \\W_{O B} & =m_{O B} \text{g}=(3 \text{ kg} )\left(9.81 m / s ^2\right)=29.4 N\end{aligned}
The second sketch shows the effective forces, which consist of a couple \bar{I}_E \alpha (since gear E is in centroidal rotation) and of a couple and two vector components at the mass center of OB. Since the accelerations are known, we compute the magnitudes of these components and couples:
\begin{aligned}\bar{I}_E \alpha=m_E \bar{k}_E^2 \alpha & =(4\text{ kg})(0.085 m )^2\left(40\text{ rad}/ s ^2\right)=1.156 N \cdot m \\m_{O B}\left(\bar{a}_{O B}\right)_t & =(3 \text{ kg} )\left(8 m / s ^2\right)=24.0 N \\m_{O B}\left(\bar{a}_{O B}\right)_n & =(3 \text{ kg} )\left(12.8 m / s ^2\right)=38.4 N\end{aligned}
\bar{I}_{O B} \alpha=\left(\frac{1}{12} m_{O B} L^2\right) \alpha=\frac{1}{12}(3 \text{ kg} )(0.400 m )^2\left(40 \text{ rad} / s ^2\right)=1.600 N \cdot m
Expressing that the system of the external forces is equivalent to the system of the effective forces, we write the following equations:
\begin{aligned}&+\uparrow\Sigma M_O=\Sigma\left(M_O\right)_{\text{eff}}:\\& F(0.120 m )=\bar{I}_E \alpha+m_{O B}\left(\bar{a}_{O B}\right)_t(0.200 m )+\bar{I}_{O B} \alpha \\& F(0.120 m )=1.156 N \cdot m +(24.0 N )(0.200 m )+1.600 N \cdot m\end{aligned}
F = 63.0 N F = 63.0 N↓
\stackrel{+}{\rightarrow} \Sigma F_x=\Sigma\left(F_x\right)_{ \text{eff}}: \quad R_x=m_{O B}\left(\bar{a}_{O B}\right)_t
R_x=24.0 N \quad R _x=24.0 N \rightarrow
\begin{gathered}+\uparrow \Sigma F_y=\Sigma\left(F_y\right)_{\text {eff }}: \quad R_y-F-W_E-W_{O B}=m_{O B}\left(\bar{a}_{O B}\right)_n \\R_y-63.0 N -39.2 N -29.4 N =38.4 N\end{gathered}
R_y=170.0 N \quad R _y=170.0 N \uparrow
