Chapter 16.2, Problem 16.141P

Two rotating rods in the vertical plane are connected by a slider block P of negligible mass. The rod attached at A has a weight of 1.6 lb and a length of 8 in. Rod BP weighs 2 lb and is 10 in. long and the friction between block P and AE is negligible. The motion of the system is controlled by a couple M applied to rod BP. Knowing that rod BP has a constant angular velocity of 20 rad/s clockwise, determine (a) the couple M, (b) the components of the force exerted on AE by block P.

Fig. P16.141 and Fig. P16.142

To determine

Find the value of couple M.

Answer to Problem 16.141P

The value of couple M is $\underset{\_}{5.82\text{lb}\cdot \text{ft}}$.

Explanation of Solution

Given information:

The weight of the rod AE is $W_{AE}=1.6\text{lb}$.

The weight of the rod BP is $W_{BP}=2\text{lb}$.

The length of the rod AE is $L_{AE}=8\text{in}\text{.}$

The length of the rod BP is $L_{BP}=10\text{in}\text{.}$

The angular velocity is $\omega =20\text{rad}/\text{s}$.

Calculation:

Consider the acceleration due to gravity as $g=32.2\text{ft}/{\text{s}}^2$.

Calculate the position vector $\left(r\right)$ as shown below.

The position of P with respect to A.

$\begin{array}{c}{r}_{P/A}=\left(-10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\sin 30°\right)\text{j}\\ =-0.4167\text{j}\end{array}$

The position of P with respect to B.

$\begin{array}{c}{r}_{P/B}=-\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\cos 30°\right)\text{i}-\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\sin 30°\right)\text{j}\\ =-0.7217\text{i}-0.4167\text{j}\end{array}$

The position of P with respect to E.

$\begin{array}{c}{r}_{P/E}=-\left(8\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\text{j}\\ =-0.6667\text{j}\end{array}$

The angular velocity of rod BP in vector form is ${\omega }_{BP}=-\left(20\text{rad}/\text{s}\right)\text{k}$.

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

${\text{v}}_P={\omega }_{BP}\times r_{P/B}$

Substitute $-\left(20\text{rad}/\text{s}\right)\text{k}$ for ${\omega }_{BP}$, and $-0.7217\text{i}-\text{0}\text{.4167j}$ for $r_{P/B}$.

$\begin{array}{c}{\text{v}}_P=\left(-\left(20\text{rad}/\text{s}\right)\text{k}\right)\times \left(-0.7217\text{i}-\text{0}\text{.4167j}\right)\\ =14.434\text{j}-8.334\text{i}\\ =-8.334\text{i}+14.434\text{j}\end{array}$

Consider the relative angular velocity of rod AE as ${\text{ω}}_{AE}={\omega }_{AE}\text{k}$ and the collar P slides on the rod with relative velocity ${\text{v}}_{P/AE}=u\text{j}$.

Calculate the velocity of point P $\left({\text{v}}_P\right)$ as shown below.

${\text{v}}_P={\omega }_{AE}\times r_{P/A}+v_{P/AE}$

Substitute ${\omega }_{AE}\text{k}$ for ${\omega }_{AE}$, $-\text{0}\text{.4167j}$ for $r_{P/A}$, $-8.334\text{i}+14.434\text{j}$ for ${\text{v}}_p$, and $u\text{j}$ for $v_{P/AE}$.

$\begin{array}{c}-8.334\text{i}+14.434\text{j}={\omega }_{AE}\text{k}\times \left(-\text{0}\text{.4167j}\right)+u\text{j}\\ =0.4167{\omega }_{AE}\text{i}+u\text{j}\end{array}$

Resolving i and j components as shown below.

$\begin{array}{l}-8.334=0.4167{\omega }_{AE}\\ {\omega }_{AE}=-20\text{rad}/\text{s}\\ 14.434=u\\ u=14.434\text{ft}/\text{s}\end{array}$

Calculate the acceleration of rod BP $\left({\text{a}}_P\right)$ as shown below.

${\text{a}}_P={\alpha }_{BP}\times r_{P/B}-{\omega }_{BP}^2{r}_{P/B}$

Substitute 0 for ${\alpha }_{BP}$, $-0.7217\text{i}-0.4167\text{j}$ for $r_{P/B}$, and $20\text{rad}/\text{s}$ for ${\omega }_{BP}$.

$\begin{array}{c}{\text{a}}_P=0-{\left(20\right)}^2\left(-0.7217\text{i}-0.4167\text{j}\right)\\ =288.68\text{i}+166.68\text{j}\end{array}$

Calculate the acceleration of point P with respect to point E $\left({\text{a}}_{P^{\prime }}\right)$ as shown below.

${\text{a}}_{P^{\prime }}={\alpha }_{AE}\times r_{P/A}-{\omega }_{AE}^2{r}_{P/A}$

Substitute ${\alpha }_{AE}\text{k}$ for ${\alpha }_{AE}$, $-\text{0}\text{.4167j}$ for $r_{P/A}$, $\left(-20\text{rad}/\text{s}\right)\text{k}$ for ${\omega }_{AE}$.

$\begin{array}{c}{\text{a}}_{P^{\prime }}=\left({\alpha }_{AE}\text{k}\right)\times \left(-\text{0}\text{.4167j}\right)-{\left(-20\right)}^2\left(-0.4167\text{j}\right)\\ =0.4166{\alpha }_{AE}\text{i}-\left(400\right)\left(-\text{0}\text{.4167j}\right)\\ =0.4167{\alpha }_{AE}\text{i}+166.68\text{j}\end{array}$

Calculate the acceleration of point P $\left(a_P\right)$ as shown below.

${\text{a}}_P={\text{a}}_{P^{\prime }}+{\text{a}}_{P/AE}+2{\omega }_{BP}\times v_{P/AE}$

Substitute $288.68\text{i}+166.68\text{j}$ for ${\text{a}}_P$, $0.4167{\alpha }_{AE}\text{i}+166.68\text{j}$ for ${\text{a}}_{P^{\prime }}$, 0 for $a_{P/AE}$, $-\left(20\text{rad}/\text{s}\right)\text{k}$, for ${\omega }_{BP}$, and $14.434\text{j}$ for $v_{P/AE}$.

$\begin{array}{c}288.68\text{i}+166.68\text{j}=0.4167{\alpha }_{AE}\text{i}+166.68\text{j}+0+2\left(-20\text{k}\right)\times \left(14.434\text{j}\right)\\ =0.4167{\alpha }_{AE}\text{i}+166.68\text{j}+577.36\text{i}\\ =\left(0.4167{\alpha }_{AE}+577.36\right)\text{i}+166.68\text{j}\end{array}$

Resolving i and j components as shown below.

$\begin{array}{l}288.68=0.4167{\alpha }_{AE}+577.36\\ 0.4167{\alpha }_{AE}=-288.68\\ {\alpha }_{AE}=-692.8\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Calculate the mass of $\left(m\right)$ as shown below.

$m=\frac{W}{g}$ (1)

For rod AE.

Substitute $1.6\text{ lb}$ for $W$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_{AE}=\frac{1.6}{32.2}\\ =0.049689\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

For rod BP.

Substitute $2\text{ lb}$ for $W$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_{BP}=\frac{2}{32.2}\\ =0.062112\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

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

$\overline{I}=\frac{m{L}^2}{12}$ (2)

For rod AE.

Substitute $0.049689\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m$ and $8\text{ in}\text{.}$ for $L$ in Equation (2).

$\begin{array}{c}{I}_{AE}=\frac{\left(0.049689\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\times {\left(8\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2}{12}\\ =0.00184\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

For rod BP.

Substitute $0.062112\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m$ and $\text{10 in}\text{.}$ for $L$ in Equation (2).

$\begin{array}{c}{I}_{BP}=\frac{\left(0.062112\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\times {\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2}{12}\\ =0.00359\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

Calculate the position vector $\left(r\right)$ as shown below.

The position of mass center G with respect to the rod AE.

$\begin{array}{c}{r}_{G/A}=\left(-4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\text{j}\\ =-0.3333\text{j}\end{array}$

The position of mass center H with respect to the rod BP.

$\begin{array}{c}{r}_{H/B}=-\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\cos 30°\right)\text{i}-\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\sin 30°\right)\text{j}\\ =-0.7217\text{i}-0.4167\text{j}\end{array}$

Calculate the acceleration of point G $\left({\text{a}}_G\right)$ as shown below.

${\text{a}}_G={\alpha }_{AE}{r}_{G/A}-{\omega }_{AE}^2{r}_{G/A}$

Substitute $\left(-692.8\text{rad}/{\text{s}}^2\right)\text{k}$ for ${\alpha }_{AE}$, $-\text{0}\text{.3333j}$ for $r_{G/A}$, and $\left(-20\text{rad}/\text{s}\right)\text{k}$ for ${\omega }_{AE}$.

$\begin{array}{c}{\text{a}}_G=\left(-692.8\text{k}\right)\times \left(-\text{0}\text{.3333j}\right)-{\left(-20\right)}^2\left(-\text{0}\text{.3333j}\right)\\ =-230.91\text{i}+133.32\text{j}\end{array}$

Calculate the acceleration of point H $\left({\text{a}}_H\right)$ as shown below.

${\text{a}}_H={\alpha }_{BP}{r}_{H/B}-{\omega }_{BP}^2{r}_{H/B}$

Substitute 0 for ${\alpha }_{BP}$, $-0.7217\text{i}-0.4167\text{j}$ for $r_{H/B}$, and $\left(-20\text{rad}/\text{s}\right)\text{k}$ for ${\omega }_{BP}$.

$\begin{array}{c}{\text{a}}_H=0-{\left(-20\right)}^2\left(-0.7217\text{i}-0.4167\text{j}\right)\\ =288.68\text{i}+166.68\text{j}\end{array}$

Calculate the inertial terms of the mass center $\left(ma\right)$ as shown below.

For rod AE.

$\begin{array}{c}{m}_{AE}{\text{a}}_G=\left(0.049689\right)\left(-230.91\text{i}+133.32\text{j}\right)\\ =-11.474\text{i}+6.625\text{j}\end{array}$

For rod BP.

$\begin{array}{c}{m}_{BP}{\text{a}}_H=\left(0.062112\right)\left(288.68\text{i}+166.68\text{j}\right)\\ =17.93\text{i}+10.353\text{j}\end{array}$

Calculate the effective couples at mass center $\left(\overline{I}\alpha \right)$ as shown below.

For rod AE.

$\begin{array}{c}{I}_{AE}{\alpha }_{AE}=\left(0.00184\right)\left(-692.8\text{k}\right)\\ =-1.275\text{k}\end{array}$

For rod BP.

$\begin{array}{c}{I}_{BP}{\alpha }_{BP}=\left(0.00359\right)\left(0\right)\\ =0\end{array}$

Sketch the Free Body Diagram of rod AE as shown in Figure 1.

Refer to Figure 1.

Apply the Equilibrium of moment about A as shown below.

$\begin{array}{l}{\displaystyle \sum M_A}=I_G\alpha +mad\\ r_{P/A}\times \left(-P\text{i}\right)=I_{AE}{\alpha }_{AE}+r_{G/A}\times \left(m_{AE}{a}_G\right)\end{array}$

Substitute $-0.4167\text{j}$ for $r_{P/A}$, $-1.275\text{k}$ for $I_{AE}{\alpha }_{AE}$, $-0.3333\text{j}$ for $r_{G/A}$, and $-11.474\text{i}+6.625\text{j}$ for $m_{AE}{a}_G$.

$\begin{array}{l}\left(-0.4167\text{j}\right)\times \left(-P\text{i}\right)=-1.275\text{k}+\left(-0.3333\text{j}\right)\times \left(-11.474\text{i}+6.625\text{j}\right)\\ -0.4167P\text{k}=-1.275\text{k}-3.8243\text{k}\\ 0.4167P=5.0993\\ P=12.24\text{lb}\end{array}$

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

Refer to Figure 2.

Apply the Equilibrium of moment about A as shown below.

$\begin{array}{l}{\displaystyle \sum M_B}=I_G\alpha +mad\\ r_{P/B}\times P\text{i}+\frac{r_{H/B}}{2}\times \left(-W_{BP}\text{j}\right)+M\text{k}=r_{H/B}\times \left(m_{BP}{a}_H\right)+I_{BP}{\alpha }_{BP}\end{array}$

Substitute $-0.7217\text{i}-0.4167\text{j}$ for $r_{P/B}$, $12.24\text{lb}$ for P, $-0.7217\text{i}-0.4167\text{j}$ for $r_{H/B}$, $2\text{lb}$ for $W_{BP}$, $17.93\text{i}+10.353\text{j}$ for $m_{BP}{a}_H$, and $0$ for $I_{BP}{\alpha }_{BP}$.

$\begin{array}{l}\left[\begin{array}{l}\left(-0.7217\text{i}-0.4167\text{j}\right)\times 12.24\text{i}+\\ \frac{1}{2}\left(-0.7217\text{i}-0.4167\text{j}\right)\times \left(-2\text{j}\right)+M\text{k}\end{array}\right]=\left[\begin{array}{l}\left(-0.7217\text{i}-0.4167\text{j}\right)\times \\ \left(17.93\text{i}+10.353\text{j}\right)+0\end{array}\right]\\ 5.1004\text{k}+0.7217\text{k}+M\text{k}=-7.4718\text{k}+7.4714\text{k}\\ M\text{k}=-5.8225\text{k}\\ M=-5.82\text{lb}\cdot \text{ft}\end{array}$

Hence, the couple M is $\underset{\_}{5.82\text{lb}\cdot \text{ft}}$.

To determine

The components of the force exerted on AE by block.

Answer to Problem 16.141P

The components of the force exerted on AE by block is $\underset{\_}{P=12.24\text{lb}}$.

Explanation of Solution

Given information:

The weight of the rod AE is $W_{AE}=1.6\text{lb}$.

The weight of the rod BP is $W_{BP}=2\text{lb}$.

The length of the rod AE is $L_{AE}=8\text{in}\text{.}$

The length of the rod BP is $L_{BP}=10\text{in}\text{.}$

The angular velocity is $\omega =20\text{rad}/\text{s}$.

Calculation:

Refer to part (a).

The components of the force exerted on AE by block $P=12.24\text{lb}$.

Therefore, the components of the force exerted on AE by block is $\underset{\_}{P=12.24\text{lb}}$.