Chapter 15, Problem 15.257RP

(a)

To determine

Find the relative velocity of the block with respect to the each rod.

Answer to Problem 15.257RP

The relative velocity of the rod AE, BD, are $\underset{\_}{2\text{m/s}(\to )}$ and $\underset{\_}{0}$.

Explanation of Solution

Given information:

The value of the angle is $\theta =90°$.

Calculation:

Consider the value of the angle as $\theta =90°$.

Show the geometry of the assembly as shown in Figure 1.

Refer to Figure 1.

Show the relation between the distances $r_1$ and $r_2$ as follows:

$\frac{100\text{mm}}{\sin 60°}=\frac{r_1}{\sin 30°}=\frac{r_2}{\sin 90°}$

Calculate the distances $r_1$ and $r_2$ as follows:

$\begin{array}{l}\frac{100\text{mm}}{\sin 60°}=\frac{r_1}{\sin 30°}\\ r_1=\frac{100\text{mm}}{\sin 60°}\times \sin 30°\\ r_1=57.735\text{mm}\end{array}$

$\begin{array}{c}{r}_2=\frac{100\text{mm}}{\sin 60°}\times \sin 90°\\ =115.47\text{mm}\end{array}$

Consider the common angular velocity and angular acceleration of the rod AE and BD are $\omega =-20\text{rad/s}$ and $\alpha =4{\text{rad/s}}^2$.

Consider the double slider H as a particle sliding along the rotating rod AH with relative velocity $u_1\to$ and relative acceleration ${\dot{u}}_1\to$.

Consider $H^{\prime }$ is the point on the rod AE coincides with H.

Calculate the velocity at $H^{\prime }$ as follows:

$\begin{array}{c}{\text{v}}_{H^{\prime }}={\text{r}}_1\omega (\uparrow )\\ =\left(57.735\text{mm}\right)\left(-20\text{rad/s}\right)(\uparrow )\\ =1154.7{\text{mm/s}}^2(\downarrow )\end{array}$

Calculate the acceleration at $H^{\prime }$ as follows:

$\begin{array}{c}{\text{a}}_{H^{\prime }}={\text{r}}_1\alpha (\uparrow )+{\text{r}}_1{\omega }^2(\leftarrow )\\ =\left(57.735\text{mm}\right)\left(4{\text{rad/s}}^2\right)(\uparrow )+\left(57.735\text{mm}\right){\left(20\text{rad/s}\right)}^2\\ =230.9{\text{mm/s}}^2(\uparrow )+23094{\text{mm/s}}^2(\leftarrow )\end{array}$

Calculate the corresponding coriolis acceleration as follows:

$\begin{array}{c}{\text{a}}_1=\left(2\omega {\text{u}}_1\right)(\uparrow )\\ =2\times \left(-20\right)u_1(\uparrow )\\ =40u_1(\downarrow )\end{array}$

Show the velocity and acceleration at H as follows:

$\begin{array}{l}{\text{v}}_H={\text{v}}_{H^{\prime }}+u_1(\to )\\ {\text{v}}_H=1154.7(\downarrow )+u_1(\to )\end{array}$ (1)

$\begin{array}{l}{\text{a}}_H=a_{H^{\prime }}+{\dot{u}}_1(\to )+40u_1(\downarrow )\\ {\text{a}}_H=230.9(\uparrow )+23094(\leftarrow )+{\dot{u}}_1(\to )+40u_1(\downarrow )\end{array}$ (2)

Consider the double slider H is a particle sliding along BD with relative velocity $u_2\left(\text{Acting at 60}°\text{above horizontal}\right)$ and relative acceleratio ${\dot{u}}_2\left(\text{Acting at 60}°\text{above horizontal}\right)$

Consider the point $H^{″}$ on the rod BD coincides with H.

Calculate the velocity and acceleration at $H^{″}$ as follows:

$\begin{array}{c}{\text{v}}_{H^{″}}={\text{r}}_2\omega \left(\text{Acting at 60}°\text{above horizontal}\right)\\ =\left(115.47\text{mm}\right)\left(-20\text{rad/s}\right)\left(\text{Acting at 60}°\text{above horizontal}\right)\\ =2309.4{\text{mm/s}}^2\left(\text{Acting at 60}°\text{below horizontal}\right)\end{array}$

Calculate the corresponding coriolis acceleration as follows:

$\begin{array}{c}{\text{a}}_2=\left(2\omega {\text{u}}_2\right)\left(\text{Acting at 30}°\text{above horizontal}\right)\\ =2\times \left(-20\right)u_2\\ =40u_2\left(\text{Acting at 30}°\text{below horizontal}\right)\end{array}$

Show the velocity and acceleration at H as follows:

$\begin{array}{l}{\text{v}}_H={\text{v}}_{H^{″}}+u_2\left(\text{Acting at 60}°\text{above horizontal}\right)\\ {\text{v}}_H=2309.4\left(\text{Acting at 30}°\text{below horizontal}\right)+u_2\left(\text{Acting at 60}°\text{above horizontal}\right)\end{array}$ (3)

$\begin{array}{l}{\text{a}}_H=a_{H^{″}}+{\dot{u}}_2\left(\text{Acting at 60}°\text{above horizontal}\right)+40u_2\left(\text{Acting at 30}°\text{below horizontal}\right)\\ {\text{a}}_H=\left[\begin{array}{l}461.9\left(\text{Acting at 30}°\text{below horizontal}\right)+\\ 46188\left(\text{Acting at 60}°\text{below horizontal}\right)+\\ {\dot{u}}_2\left(\text{Acting at 60}°\text{above horizontal}\right)+\\ 40u_2\left(\text{Acting at 30}°\text{below horizontal}\right)\end{array}\right]\end{array}$ (4)

Equate vertical components of Equation (1) and (3).

$\begin{array}{c}1154.7=-2309.4\sin 30°+u_2\sin 60°\\ u_2=\left(\frac{2309.4\sin 30°-1154.7}{\sin 60°}\right)\\ =0\end{array}$

Equate horizontal components of Equation (1) and (3).

$\begin{array}{l}{u}_1=2309.4\sin \left(30°\right)+0\\ u_1=2000\text{mm/s}\end{array}$

Substitute $2000\text{mm/s}$ for $u_1$ and 0 for $u_2$ in Equation (2) and (4).

${\text{a}}_H=230.9(\uparrow )+23094(\leftarrow )+{\dot{u}}_1(\to )+40\times \left(2000\right)(\downarrow )$ (5)

${\text{a}}_H=\left[\begin{array}{l}461.9\left(\text{Acting at 30}°\text{below horizontal}\right)+\\ 46188\left(\text{Acting at 60}°\text{below horizontal}\right)+\\ {\dot{u}}_2\left(\text{Acting at 60}°\text{above horizontal}\right)+\\ 40u_2\left(\text{Acting at 30}°\text{below horizontal}\right)\end{array}\right]$ (6)

Equate the vertical component of Equation (5) and (6)

$\begin{array}{l}230.9-40\times 2000=461.9\sin 30°+46188\times \sin 60°+{\dot{u}}_2\sin \left(60°\right)-40\times 0\times \sin 30°\\ {\dot{u}}_2=\frac{230.9-80000-230.9+40000}{\sin 60°}\\ {\dot{u}}_2=-46188{\text{mm/s}}^2\end{array}$

Equate the horizontal component of Equation (5) and (6)

$\begin{array}{l}{\dot{u}}_1-23094=-\left(461.4\right)\cos \left(30°\right)-\left(46188\right)\cos \left(60°\right)\\ {\dot{u}}_1=-23494{\text{mm/s}}^2\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\\ {\dot{u}}_1=-23.5{\text{m/s}}^2\end{array}$

Thus, the relative velocity of the rod AE, BD, are $2\text{m/s}(\to )$ and $0$.

(b)

To determine

Find the relative acceleration of each rod.

Answer to Problem 15.257RP

Thus, The relative acceleration of the rod AE and BD are $\underset{\_}{23.5{\text{m/s}}^2(\leftarrow )}$ and $\underset{\_}{46.2{\text{m/s}}^2}$.

Explanation of Solution

Given information:

Calculation:

Refer Part (a).

The relative acceleration of the rod AE and BD are $\underset{\_}{23.5{\text{m/s}}^2(\leftarrow )}$ and $\underset{\_}{46.2{\text{m/s}}^2}$.