Chapter 16.2, Problem 16.89P

To determine

The component of reaction at point C at $\theta =0°$.

Answer to Problem 16.89P

The reaction at point C is $30.9918\text{ lb}$ in the direction of $83.18°$ measured form positive $x$ -axis.

Explanation of Solution

Given:

Weight of rod AB is $2\text{lb}$.

Weight of rod BC is $4\text{lb}$.

Angular velocity is $10\text{rad/s}$.

Angle $\theta$ is $0°$.

Concept used:

Write the expression for the moment of inertia of rod AB.

$I_{AB}=\frac{W_{AB}{L}_{AB}^2}{12g}$ ...... (1)

Here, $I_{AB}$ is the moment of inertia of rod AB, $g$ is acceleration due to gravity, $W_{AB}$ is the weight of bar AB and $L_{AB}$ is the length of bar AB.

Write the expression for the moment of inertia of rod BC.

$I_{BC}=\frac{W_{BC}{L}_{BC}^2}{12g}$ ...... (2)

Here, $I_{BC}$ is the moment of inertia of rod BC, $W_{BC}$ is the weight of bar BC and $L_{BC}$ is the length of bar BC.

Draw the diagram to illustrate the geometry of system as shown below:

From above figure calculate the distance between Point C and D,

$r_{CD}=\sqrt{L_{BC}^2+L_{BD}^2}$ ...... (3)

Here, $r_{CD}$ is the distance between C and D and $L_{BD}$ is the distance between point B and D.

Write the expression for distance between B and E.

$r_{BE}=\frac{L_{BC}}{2}$ ...... (4)

Here, $r_{BE}$ is the distance between points B and E.

Write the expression for angle $\beta$.

$\tan \beta =\frac{L_{BD}}{L_{BC}}$ ...... (5)

Write the expression for tangential acceleration of rod AB.

${\left(a_{AB}\right)}_t=\alpha r_{CD}$

Here, ${\left(a_{AB}\right)}_t$ is the tangential component of acceleration of rod AB and $\alpha$ is the angular acceleration of rod.

Write the expression for centripetal acceleration of rod AB.

${\left(a_{AB}\right)}_n={\omega }^2{r}_{CD}$

Here, ${\left(a_{AB}\right)}_n$ is the centripetal acceleration and $\omega$ is the angular velocity of rod.

Write the expression for tangential acceleration of rod BC.

${\left(a_{BC}\right)}_t=\alpha r_{CE}$

Here, ${\left(a_{BC}\right)}_t$ is the tangential component of acceleration of rod BC.

Write the expression for centripetal acceleration of rod BC.

${\left(a_{BC}\right)}_n={\omega }^2{r}_{CE}$

Here, ${\left(a_{BC}\right)}_n$ is the centripetal acceleration or rod BC.

Draw the free body diagram of the system.

Write the expression for moment about point C.

$\frac{W_{AB}{L}_{AB}}{2}=I_{AB}\alpha +I_{BC}\alpha +r_{CD}\left(\frac{W_{CD}}{g}\right){\left(a_{AB}\right)}_t+r_{CE}\left(\frac{W_{BC}}{g}\right){\left(a_{BC}\right)}_t$ ...... (6)

Write the expression for the equilibrium of forces in horizontal direction.

$C_x=\left(\frac{W_{AB}}{g}\right){\left(a_{AB}\right)}_t\cos \beta +\left(\frac{W_{AB}}{g}\right){\left(a_{AB}\right)}_n\sin \beta +\left(\frac{W_{BC}}{g}\right){\left(a_{BC}\right)}_t$ ...... (7)

Here, $C_x$ is the horizontal reaction at point C.

Write the expression for equilibrium of forces in vertical direction.

$C_y-W_{AB}-W_{Bc}=-\left(\frac{W_{AB}}{g}\right){\left(a_{AB}\right)}_t\sin \beta +\left(\frac{W_{AB}}{g}\right){\left(a_{AB}\right)}_n\cos \beta +\left(\frac{W_{BC}}{g}\right){\left(a_{BC}\right)}_n$ ...... (8)

Here, $C_y$ is the vertical reaction at point C.

Write the expression for resultant reaction.

$C_R=\sqrt{C_x^2+C_y^2}$ ...... (9)

Here $C_R$ is the resultant reaction at point C.

Write the expression for the direction of resultant reaction.

$\varphi ={\tan }^{-1}\left(\frac{C_y}{C_x}\right)$ ...... (10)

Here, $\varphi$ is the direction of resultant reaction at point C.

Calculation:

Substitute $2\text{ lb}$ for $W_{AB}$, $32.2{\text{ft/s}}^2$ for $g$ and $1\text{ ft}$ for $L_{AB}$ in equation (1).

$\begin{array}{c}{I}_{AB}=\frac{\left(2\right){\left(1\right)}^2}{12\left(32.2\right)}\\ =5.175\times {10}^{-3}\text{ slug}\cdot {\text{ft}}^2\end{array}$

Substitute $\text{4 lb}$ for $W_{BC}$, $32.2{\text{ft/s}}^2$ for $g$ and $2\text{ ft}$ for $L_{BC}$ in equation (2).

$\begin{array}{c}{I}_{BC}=\frac{\left(4\right){\left(2\right)}^2}{12\left(32.2\right)}\\ =41.4\times {10}^{-3}\text{ slug}\cdot {\text{ft}}^2\end{array}$

Substitute $2\text{ ft}$ for $L_{BC}$ and $0.5\text{ ft}$ for $L_{BD}$ in equation (3).

$\begin{array}{c}{r}_{CD}=\sqrt{{\left(2\right)}^2+{\left(0.5\text{}\right)}^2}\\ =2.0615\text{ft}\end{array}$

Substitute $2\text{}\text{ft}$ for $L_{BC}$ in equation (4).

$\begin{array}{c}{r}_{CE}=\frac{2}{2}\\ =1\text{ ft}\end{array}$

Substitute $0.5\text{ ft}$ for $L_{BD}$ and $2\text{ ft}$ for $L_{BC}$ in equation (5).

$\begin{array}{c}\tan \beta =\frac{0.5}{2}\\ \beta ={\tan }^{-1}\left(\frac{1}{4}\right)\\ \beta =14.0362°\end{array}$

Substitute $2\text{ lb}$ for $W_{AB}$, $1\text{ ft}$ for $L_{AB}$, $5.175\times {10}^{-3}\text{ slug}\cdot {\text{ft}}^2$ for $I_{AB}$, $41.4\times {10}^{-3}\text{ slug}\cdot {\text{ft}}^2$ for $I_{BC}$, $1\text{ft}$ for $r_{CE}$, $2.0615\text{ft}$ for $r_{CD}$, $32.2{\text{ ft/s}}^2$ for $g$, $\alpha r_{CD}$ for ${\left(a_{AB}\right)}_t$ and $\alpha r_{CE}$ for ${\left(a_{BC}\right)}_t$ in equation (6).

$\frac{2\left(1\right)}{2}=\left[\begin{array}{l}\left(5.175\times {10}^{-3}\right)\alpha +\left(41.4\times {10}^{-3}\right)\alpha \\ +\left(2.0615\right)\left(\frac{2}{32.2}\right)\left(\alpha r_{CD}\right)+\left(1\right)\left(\frac{4}{32.2}\right)\left(\alpha r_{CE}\right)\end{array}\right]$

Substitute $1\text{ft}$ for $r_{CE}$ and $2.0615\text{ft}$ for $r_{CD}$ in above expression and simplify for $\alpha$.

$\begin{array}{l}1=\left[\begin{array}{l}\left(5.175\times {10}^{-3}\right)\alpha +\left(41.4\times {10}^{-3}\right)\alpha \\ +\left(2.0615\right)\left(\frac{2}{32.2}\right)\left(\alpha \left(2.0615\right)\right)+\left(1\right)\left(\frac{4}{32.2}\right)\left(\alpha \left(1\right)\right)\end{array}\right]\\ 1=0.434\alpha \\ \alpha =2.3{\text{ rad/s}}^2\end{array}$

Substitute $2\text{ lb}$ for $W_{AB}$, $4\text{ lb}$ for $W_{BC}$, $32.2{\text{ ft/s}}^2$ for $g$, $\alpha r_{CD}$ for ${\left(a_{AB}\right)}_t$, ${\omega }^2{r}_{CD}$ for ${\left(a_{AB}\right)}_n$, $\alpha r_{CE}$ for ${\left(a_{BC}\right)}_t$ and $14.0362°$ for $\beta$ in equation (7).

$C_x=\left(\frac{2}{32.2}\right)\alpha r_{CD}\left(\cos 14.0326°\right)+\left(\frac{2}{32.2}\right)\left(r_{CD}{\omega }^2\right)\sin 14.0326°+\left(\frac{4}{32.2}\right)\alpha r_{CE}$

Substitute $1\text{ft}$ for $r_{CE}$, $2.0615\text{ft}$ for $r_{CD}$, $2.3{\text{ rad/s}}^2$ for $\alpha$ and $10\text{rad/s}$ for $\omega$ in above expression.

$\begin{array}{c}{C}_x=\left[\begin{array}{c}\left(\frac{2}{32.2}\right)\left(2.3\right)\left(2.0615\right)\left(\cos 14.0326°\right)\\ +\left(\frac{2}{32.2}\right)\left(\left(2.0615\right){\left(10\right)}^2\right)\sin 14.0326°\\ +\left(\frac{4}{32.2}\right)\left(2.3\right){\left(1\right)}^2\end{array}\right]\\ =3.677\text{ lb}\end{array}$

Substitute $2\text{ lb}$ for $W_{AB}$, $4\text{ lb}$ for $W_{BC}$, $32.2{\text{ ft/s}}^2$ for $g$, $\alpha r_{CD}$ for ${\left(a_{AB}\right)}_t$, ${\omega }^2{r}_{CD}$ for ${\left(a_{AB}\right)}_n$, $\alpha r_{CE}$ for ${\left(a_{BC}\right)}_t$

$1\text{ft}$ for $r_{CE}$, $2.0615\text{ft}$ for $r_{CD}$, $2.3{\text{ rad/s}}^2$ for $\alpha$, $10\text{rad/s}$ for $\omega$ and $14.0362°$ for $\beta$ in equation (8).

$\begin{array}{c}{C}_y-2-4=\left[\begin{array}{l}-\left(32.2\right)\left(2.0615\right)\left(2.3\right)\sin \left(14.0362°\right)\\ +\left(\frac{2}{32.2}\right)\left(2.0615\right){\left(10\right)}^2\cos 14.0362°+\left(\frac{4}{32.2}\right)\left(1\right){\left(10\right)}^2\end{array}\right]\\ C_y-6=24.773\\ C_y=30.773\text{lb}\end{array}$

Substitute $30.773\text{lb}$ for $C_y$ and $3.677\text{lb}$ for $C_x$ in equation (9).

$\begin{array}{l}{C}_R=\sqrt{{\left(3.677\right)}^2+{\left(30.773\right)}^2}\\ C_R=30.9918\text{lb}\end{array}$

Substitute $30.773\text{lb}$ for $C_y$ and $3.677\text{lb}$ for $C_x$ in equation (10).

$\begin{array}{l}\varphi ={\tan }^{-1}\left(\frac{30.773}{3.677}\right)\\ \varphi =83.18°\end{array}$

The reaction at point C is $30.9918\text{ lb}$ in the direction of $83.18°$ measured form positive $x$ -axis.

Conclusion:

Thus, the reaction at point C is $30.9918\text{ lb}$ in the direction of $83.18°$ measured form positive $x$ -axis.