Chapter 16.2, Problem 16.136P

(a)

To determine

Find the couple M.

Answer to Problem 16.136P

The couple M is $\underset{\_}{78.3\text{N}\cdot \text{m}}$.

Explanation of Solution

Given information:

The mass of the rod BC is $m_{BC}=6\text{kg}$.

The mass of the disk is $m_d=10\text{kg}$.

The mass of the rod CD is $m_{CD}=5\text{kg}$.

The angular velocity is ${\omega }_{AB}=36\text{rad}/\text{s}$.

The angular acceleration is $\alpha =150\text{rad}/{\text{s}}^{\text{2}}$.

Calculation:

Consider the acceleration due to gravity as $g=9.81\text{m}/{\text{s}}^2$.

Calculate the velocity of disk AB $\left({\text{v}}_B\right)$ as shown below.

$v_B=\overline{AB}{\omega }_{AB}$

Substitute $200\text{mm}$ for $\overline{AB}$ and $36\text{rad}/\text{s}$ for ${\omega }_{AB}$.

$\begin{array}{c}{v}_B=200\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\times 36\text{rad}/\text{s}\\ =7.2\text{m}/\text{s}\end{array}$

Calculate the velocity of rod BC $\left({\text{v}}_C\right)$ as shown below.

The velocity of disk AB is equal to the velocity of rod BC.

${\text{v}}_B={\text{v}}_C$

Substitute $7.2\text{m}/\text{s}$ for ${\text{v}}_B$.

${\text{v}}_C=7.2\text{m}/\text{s}$

Calculate the angular velocity of rod CD $\left({\omega }_{CD}\right)$ as shown below.

${\omega }_{CD}=\frac{v_C}{l_{CD}}$

Substitute $7.2\text{m}/\text{s}$ for $v_C$ and $250\text{mm}$ for $l_{CD}$.

$\begin{array}{c}{\omega }_{CD}=\frac{7.2\text{m}/\text{s}}{\left(250\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}\\ =28.8\text{rad}/\text{s}\end{array}$

Apply the acceleration analysis as shown below.

Calculate the acceleration for disk AB $\left({\text{a}}_B\right)$ as shown below.

${\text{a}}_B={\alpha }_{AB}\times {\text{r}}_{B/A}-{\omega }_{AB}^2{\text{r}}_{B/A}$

Substitute $150\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_{AB}$, $0.2\text{m}$ for ${\text{r}}_{B/A}$, and $36\text{m}/{\text{s}}^{\text{2}}$ for ${\omega }_{AB}$.

$\begin{array}{c}{\text{a}}_B=150\times 0.2-{36}^2\times 0.2\\ =\left[30\nwarrow +259.2\swarrow \right]\text{m}/{\text{s}}^{\text{2}}\end{array}$

Calculate the acceleration for rod BC $\left({\text{a}}_C\right)$ as shown below.

${\text{a}}_C={\text{a}}_B+{\left({\text{a}}_{C/B}\right)}_t-{\omega }_{BC}^2{\text{r}}_{C/B}$

Substitute $\left[30\nwarrow +259.2\swarrow \right]\text{m}/{\text{s}}^{\text{2}}$ for ${\text{a}}_B$, $0.4{\alpha }_{BC}$ for ${\left({\text{a}}_{C/B}\right)}_t$, and $0$ for ${\omega }_{BC}$.

${\text{a}}_C=30\nwarrow +259.2\swarrow +0.4{\alpha }_{BC}\uparrow$ (1)

Calculate the acceleration for rod CD $\left({\text{a}}_C\right)$ as shown below.

${\text{a}}_C={\left({\text{a}}_{C/D}\right)}_t-{\omega }_{CD}^2{\text{r}}_{C/D}$

Substitute $0.25{\alpha }_{CD}$ for ${\left({\text{a}}_{C/D}\right)}_t$, $0.25\text{m}$ for ${\text{r}}_{C/D}$, and $28.8\text{rad}/\text{s}$ for ${\omega }_{CD}$.

$\begin{array}{c}{\text{a}}_C=0.25{\alpha }_{CD}\swarrow -{28.8}^2\times 0.25\swarrow \\ =0.25{\alpha }_{CD}\swarrow -207.36\swarrow \end{array}$ (2)

Equating the components of Equations (1) and (2) as shown below.

Along x component.

$\begin{array}{l}-30\cos 30°-259.2\cos 60°=-0.25{\alpha }_{CD}\cos 30°-207.36\cos 60°\\ 0.2165{\alpha }_{CD}=51.9\\ {\alpha }_{CD}=239.72\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Along y component.

$\begin{array}{l}30\cos 30°-259.2\sin 60°+0.4{\alpha }_{BC}=0.25{\alpha }_{CD}\sin 30°-207.36\sin 60°\\ 0.4{\alpha }_{BC}=0.125{\alpha }_{CD}+18.914\end{array}$

Substitute $239.72\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_{CD}$.

$\begin{array}{l}0.4{\alpha }_{BC}=0.125\times 239.72+18.914\\ {\alpha }_{BC}=122.198\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Calculate the acceleration of the mass centers as shown below.

Calculate the acceleration of mass center for disk AB $\left({\text{a}}_A\right)$ as shown below.

${\text{a}}_{AB}={\text{a}}_A=0$

Calculate the acceleration of the mass center at P for rod BC $\left({\text{a}}_P\right)$ as shown below.

${\text{a}}_P={\text{a}}_B+{\alpha }_{BC}\times {\text{r}}_{P/B}-{\omega }_{BC}^2{\text{r}}_{P/B}$

Substitute $\left[30\nwarrow +259.2\swarrow \right]\text{m}/{\text{s}}^{\text{2}}$ for ${\text{a}}_B$, $0.2\text{m}$ for ${\text{r}}_{P/B}$, and $0$ for ${\omega }_{BC}$.

$\begin{array}{c}{\text{a}}_P=30\nwarrow +259.2\swarrow +0.2{\alpha }_{BC}\uparrow +0\\ =30\nwarrow +259.2\swarrow +0.2{\alpha }_{BC}\uparrow \end{array}$

Substitute $122.198\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_{BC}$.

$\begin{array}{c}{\text{a}}_P=30\nwarrow +259.2\swarrow +0.2\times 122.198\uparrow \\ =30\nwarrow +259.2\swarrow +24.44\uparrow \end{array}$

Calculate the acceleration of the mass center at Q for rod CD $\left({\text{a}}_Q\right)$ as shown below.

${\text{a}}_Q={\alpha }_{CD}\times r_{Q/D}-{\omega }_{CD}^2{\text{r}}_{Q/D}$

Substitute $239.72\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_{CD}$, $0.125\text{m}$ for ${\text{r}}_{Q/D}$, and $28.8\text{rad}/\text{s}$ for ${\omega }_{CD}$.

$\begin{array}{c}{\text{a}}_Q=0.125\times 239.72\swarrow -{28.8}^2\times 0.125\swarrow \\ =29.965\swarrow -103.68\swarrow \end{array}$

Calculate the inertial terms at mass centers as shown below.

The inertia terms at centers are $ma$.

For disk AB.

$m_{AB}{a}_A=0$

For rod BC.

$m_{BC}{\text{a}}_P$

Substitute $6\text{kg}$ for $m_{BC}$ and $30\nwarrow +259.2\swarrow +24.44\uparrow$ for ${\text{a}}_P$.

$\begin{array}{c}{m}_{BC}{\text{a}}_P=6\times \left(30\nwarrow +259.2\swarrow +24.44\uparrow \right)\\ =180\nwarrow +1,555.2\swarrow +179.5656\uparrow \end{array}$

For rod CD.

$m_{CD}{\text{a}}_Q$

Substitute $5\text{kg}$ for $m_{BC}$ and $29.965\swarrow -103.68\swarrow$ for ${\text{a}}_P$.

$\begin{array}{c}{m}_{CD}{\text{a}}_Q=5\times \left(29.965\swarrow -103.68\swarrow \right)\\ =149.825\swarrow +518.4\swarrow \end{array}$

Calculate the mass moment of inertia $\left(\overline{I}\right)$ as shown below.

For disk AB.

${\overline{I}}_{AB}=\frac{1}{2}{m}_{AB}{r}_{AB}^2$

Substitute $10\text{kg}$ for $m_{AB}$ and $0.2\text{m}$ for $r_{AB}$.

$\begin{array}{c}{\overline{I}}_{AB}=\frac{1}{2}\times 10\times {0.2}^2\\ =0.2\text{kg}\cdot {\text{m}}^2\end{array}$

For rod BC.

${\overline{I}}_{BC}=\frac{1}{12}{m}_{BC}{l}_{BC}^2$

Substitute $6\text{kg}$ for $m_{BC}$ and $0.4\text{m}$ for $l_{BC}$.

$\begin{array}{c}{\overline{I}}_{BC}=\frac{1}{12}\times 6\times {0.4}^2\\ =0.08\text{kg}\cdot {\text{m}}^2\end{array}$

For rod CD.

${\overline{I}}_{CD}=\frac{1}{12}{m}_{CD}{l}_{CD}^2$

Substitute $5\text{kg}$ for $m_{CD}$ and $0.25\text{m}$ for $l_{CD}$.

$\begin{array}{c}{\overline{I}}_{CD}=\frac{1}{12}\times 5\times {0.25}^2\\ =0.0260417\text{kg}\cdot {\text{m}}^2\end{array}$

Calculate the effective couples at mass centers as shown below.

The inertia terms at centers are $\overline{I}\alpha$.

For disk AB.

${\overline{I}}_{AB}{\alpha }_{AB}=0$

For rod BC.

${\overline{I}}_{BC}{\alpha }_{BC}$

Substitute $0.08\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_{BC}$ and $122.198\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_{BC}$.

$\begin{array}{c}{\overline{I}}_{BC}{\alpha }_{BC}=0.08\times 122.198\\ =9.776\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\times \frac{\text{1}\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ =9.776\text{N}\cdot \text{m}\end{array}$

For rod CD.

${\overline{I}}_{CD}{\alpha }_{CD}$

Substitute $0.0260417\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_{CD}$ and $239.72\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_{CD}$.

$\begin{array}{c}{\overline{I}}_{CD}{\alpha }_{CD}=0.0260417\times 239.72\\ =6.243\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\times \frac{\text{1}\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ =6.243\text{N}\cdot \text{m}\end{array}$

Sketch the effective force and couples on the system as shown in Figure 1.

Sketch the Free Body Diagram of the rod BC as shown in Figure 2.

Refer to Figure 2.

Apply the Equilibrium of moment about B as shown below.

$\begin{array}{l}{\displaystyle \sum M_B}={\overline{I}}_G\alpha +mad\\ l_{BC}{C}_y-\frac{1}{2}{l}_{BC}{m}_{BC}g={\overline{I}}_{BC}{\alpha }_{BC}+mad\end{array}$

Substitute $0.4\text{m}$ for $l_{BC}$, $6\text{kg}$ for $m_{BC}$, $9.81\text{m}/{\text{s}}^2$ for g, $9.776\text{N}\cdot \text{m}$ for ${\overline{I}}_{BC}{\alpha }_{BC}$, $180\nwarrow +1,555.2\swarrow +179.5656\uparrow$ for ma, and $0.2\text{m}$ for d.

$\begin{array}{l}0.4C_y-\frac{1}{2}\times 0.4\times 6\times 9.81=9.776+\left(180\sin 30°-1,555.2\sin 60°+179.5656\right)\times 0.2\\ 0.4C_y=-193.9\\ C_y=-484.75\text{N}\end{array}$

Sketch the Free Body Diagram of the rod CD as shown in Figure 3.

Refer to Figure 3.

Apply the Equilibrium of moment about D as shown below.

$\begin{array}{l}{\displaystyle \sum M_D}={\overline{I}}_G\alpha +mad\\ C_x{l}_{CD}\cos 30°+484.75\times 0.125-m_{CD}g\times \frac{0.125}{2}={\overline{I}}_{CD}{\alpha }_{CD}+mad\\ 0.866C_x{l}_{CD}+60.594-0.0625m_{CD}g={\overline{I}}_{CD}{\alpha }_{CD}+mad\end{array}$

Substitute $0.25\text{m}$ for $l_{CD}$, $5\text{kg}$ for $m_{CD}$, $9.81\text{m}/{\text{s}}^2$ for g, $6.243\text{N}\cdot \text{m}$ for ${\overline{I}}_{CD}{\alpha }_{CD}$, $149.825\text{N}$ for ma, and $0.125\text{m}$ for d.

$\begin{array}{l}0.866C_x\times 0.25+65.535-0.0625\times 5\times 9.81=6.243+149.825\times 0.125\\ 0.2165C_x=-37.49825\\ C_x=-173.2\text{N}\end{array}$

Sketch the Free Body Diagram of the rod AB and BC as shown in Figure 4.

Refer to Figure 4.

Apply the Equilibrium of moment about A as shown below.

$\begin{array}{l}{\displaystyle \sum M_A}={\overline{I}}_G\alpha +mad\\ \left[\begin{array}{l}-M-484.75\times \left(0.4+0.2\sin 30°\right)\\ +173.2\times \left(0.2\cos 30°\right)-m_{BC}g\times \left(0.2+0.2\sin 30°\right)\end{array}\right]={\overline{I}}_{BC}{\alpha }_{BC}+mad\\ -M-212.375-0.3m_{BC}g={\overline{I}}_{BC}{\alpha }_{BC}+mad\end{array}$

Substitute $6\text{kg}$ for $m_{BC}$, $9.81\text{m}/{\text{s}}^2$ for g, $9.776\text{N}\cdot \text{m}$ for ${\overline{I}}_{BC}{\alpha }_{BC}$, $180\nwarrow +1,555.2\swarrow +179.5656\uparrow$ for ma, and $0.2\text{m}$ for d.

$\begin{array}{l}-M-212.375-0.3\times 6\times 9.81=\left[\begin{array}{l}9.776+180\left(0.2+0.2\sin 30°\right)\\ -1,555.2\cos 30°\times 0.2\\ +179.5656\times \left(0.2+0.2\sin 30°\right)\end{array}\right]\\ -M-230.033=-151.722\\ -M=78.3\text{N}\cdot \text{m}\\ M=-78.3\text{N}\cdot \text{m}\end{array}$

Therefore, the couple M is $\underset{\_}{78.3\text{N}\cdot \text{m}\left(\text{Counter-clockwise}\right)}$.

(b)

To determine

The components of force exerted at C on rod BC.

Answer to Problem 16.136P

The components of force exerted at C on rod BC is $\underset{\_}{C_x=173\text{N}\leftarrow \text{ and}{C}_y=485\text{N}\downarrow }$.

Explanation of Solution

Given information:

The mass of the rod BC is $m_{BC}=6\text{kg}$.

The mass of the disk is $m_d=10\text{kg}$.

The mass of the rod CD is $m_{CD}=5\text{kg}$.

The angular velocity is ${\omega }_{AB}=36\text{rad}/\text{s}$.

The angular acceleration is $\alpha =150\text{rad}/{\text{s}}^2$.

Calculation:

Refer to part (a).

The components of force exerted at C on rod BC along x direction is $\begin{array}{c}{C}_x=-173.2\text{N}\\ \approx 173\text{N}\leftarrow \end{array}$

The components of force exerted at C on rod BC along y direction is $\begin{array}{c}{C}_y=-484.75\text{N}\\ \approx 485\text{N}\downarrow \end{array}$

Therefore, the components of force exerted at C on rod BC is $\underset{\_}{C_x=173\text{N}\leftarrow \text{ and}{C}_y=485\text{N}\downarrow }$.