Chapter 15.4, Problem 15.135P

(a)

To determine

 Find the angular acceleration of the bar DE.

Answer to Problem 15.135P

The angular acceleration of bar DE is $\underset{\_}{16.53{\text{rad/s}}^2\left(\text{Counterclockwise}\right)}$.

Explanation of Solution

Given information:

The constant angular velocity of AB is ${\text{ω}}_{AB}=4\text{rad/s}\left(\text{Clockwise}\right)=-4\text{k}$.

Calculation:

Consider the bar AB.

Show the position $\left({\text{r}}_{B/A}\right)$ of the point B with respect to A as follows:

$\begin{array}{c}{\text{r}}_{B/A}=3\text{i}+\sqrt{{12}^2-3^2}\text{j}\\ =3\text{i}+\sqrt{135}\text{j}\end{array}$

Calculate the velocity at B using the relation:

${\text{v}}_B={\text{ω}}_{AB}\times {\text{r}}_{B/A}$

Substitute $-4\text{k}$ for ${\text{ω}}_{AB}$ and $3\text{i}+\sqrt{135}\text{j}$ for ${\text{r}}_{B/A}$.

$\begin{array}{c}{\text{v}}_B=-4\text{k}\times \left(3\text{i}+\sqrt{135}\text{j}\right)\\ =-12\text{j}+4\sqrt{135}\text{i}\\ =4\sqrt{135}\text{i}-12\text{j}\end{array}$

Consider the object BDF.

Consider the position of the point D with respect to B is denoted by ${\text{r}}_{D/B}=\left(\text{6in}\text{.}\right)\text{i}$.

Consider the position of the point F with respect to B is denoted by ${\text{r}}_{F/B}=\left(3\text{i}-\sqrt{135}\text{j}\right)\text{in}\text{.}$.

Consider the angular velocity of the bar BD is denoted by ${\text{ω}}_{BD}={\omega }_{BD}\text{k}$.

Calculate the velocity at D using the relation:

$\begin{array}{c}{\text{v}}_D={\text{v}}_B+{\text{v}}_{B/D}\\ ={\text{v}}_B+{\omega }_{BD}\text{k}\times {\text{r}}_{D/B}\end{array}$

Substitute $4\sqrt{135}\text{i}-12\text{j}$ for ${\text{v}}_B$ and $\left(\text{6in}\text{.}\right)\text{i}$ for ${\text{r}}_{D/B}$.

$\begin{array}{c}{\text{v}}_D=\left(4\sqrt{135}\text{i}-12\text{j}\right)+{\omega }_{BD}\text{k}\times \left(\text{6in}\text{.}\right)\text{i}\\ =\left(4\sqrt{135}\text{i}-12\text{j}\right)+6{\omega }_{BD}\text{j}\end{array}$ (1)

Consider the bar DE.

Consider the position of the point D with respect to E is denoted by ${\text{r}}_{D/E}=\left(-3\text{i}+\sqrt{135}\text{j}\right)\text{in}\text{.}$.

The velocity at E is ${\text{v}}_E=0$, as point E is fixed.

Calculate the velocity at D using the relation:

${\text{v}}_D={\text{v}}_E+{\omega }_{DE}\text{k}\times {\text{r}}_{D/E}$

Substitute $0$ for ${\text{v}}_E$ and $\left(-3\text{i}+\sqrt{135}\text{j}\right)\text{in}\text{.}$ for ${\text{r}}_{D/E}$.

$\begin{array}{c}{\text{v}}_D=0+{\omega }_{DE}\text{k}\times \left(-3\text{i}+\sqrt{135}\text{j}\right)\text{in}\\ =-3{\omega }_{DE}\text{j}-\sqrt{135}{\omega }_{DE}\text{i}\\ \text{=}-\sqrt{135}{\omega }_{DE}\text{i}+\left(-3{\omega }_{DE}\right)\text{j}\end{array}$ (2)

Equate Equation (1) and (2).

$\left(4\sqrt{135}\text{i}-12\text{j}\right)+6{\omega }_{BD}\text{j}=-\sqrt{135}{\omega }_{DE}\text{i}+\left(-3{\omega }_{DE}\right)\text{j}$ (3)

Equate i component of Equation (3) as follows:

$\begin{array}{l}4\sqrt{135}=-\sqrt{135}{\omega }_{DE}\\ {\omega }_{DE}=-\frac{4\sqrt{135}}{\sqrt{135}}\\ {\omega }_{DE}=-4\text{rad/s}\\ {\omega }_{DE}=4\text{rad/s}\left(\text{Clockwise}\right)\end{array}$

Equate j component of the Equation (3).

$\begin{array}{l}-12+6{\omega }_{BD}=-3{\omega }_{DE}\\ 6{\omega }_{BD}=-3{\omega }_{DE}+12\\ {\omega }_{BD}=\left(\frac{-3{\omega }_{DE}+12}{6}\right)\end{array}$

Substitute $-4\text{rad/s}$ for ${\omega }_{DE}$.

$\begin{array}{c}{\omega }_{BD}=\left(\frac{-3\times \left(-4\right)+12}{6}\right)\\ =\left(\frac{24}{6}\right)\\ =4\text{rad/s}\left(\text{Counterclockwise}\right)\end{array}$

The angular acceleration of AB is ${\alpha }_{AB}=0$.

Consider the bar AB.

Calculate the acceleration at B using the relation:

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

Substitute 0 for ${\text{α}}_{AB}$, $3\text{i}+\sqrt{135}\text{j}$ for ${\text{r}}_{B/A}$ and $-4\text{rad/s}$ for ${\omega }_{AB}$.

$\begin{array}{c}{\text{a}}_B=0-{\left(-4\right)}^2\times \left(3\text{i}+\sqrt{135}\text{j}\right)\\ =-48\text{i}-16\sqrt{135}\text{j}\end{array}$

Consider the object BDF.

Calculate the acceleration at D using the relation:

$\begin{array}{c}{\text{a}}_D={\text{a}}_B+{\text{a}}_{D/B}\\ ={\text{a}}_B+{\text{α}}_{BD}\times {\text{r}}_{D/B}-{\omega }_{BD}^2{\text{r}}_{D/B}\\ ={\text{a}}_B+\left({\alpha }_{BD}\text{k}\right)\times {\text{r}}_{D/B}-{\omega }_{BD}^2{\text{r}}_{D/B}\end{array}$

Substitute $-48\text{i}-16\sqrt{135}\text{j}$ for ${\text{a}}_B$, $\left(\text{6in}\text{.}\right)\text{i}$ for ${\text{r}}_{D/B}$, and $4\text{rad/s}$ for ${\omega }_{BD}$.

$\begin{array}{l}{\text{a}}_D=-48\text{i}-16\sqrt{135}\text{j}+\left({\alpha }_{BD}\text{k}\right)\times 6\text{i}-4^2\times 6\text{i}\\ {\text{a}}_D=-48\text{i}-16\sqrt{135}\text{j}+6{\alpha }_{BD}\text{j}-96\text{i}\\ {\text{a}}_D=-144\text{i+}\left(-16\sqrt{135}+6{\alpha }_{BD}\right)\text{j}\end{array}$ (4)

Consider the bar DE.

Calculate the acceleration at D using the relation:

$\begin{array}{c}{\text{a}}_D={\text{α}}_{DE}\times {\text{r}}_{D/E}-{\omega }_{DE}^2{\text{r}}_{D/E}\\ =\left({\alpha }_{DE}\text{k}\right)\times {\text{r}}_{D/E}-{\omega }_{DE}^2{\text{r}}_{D/E}\end{array}$

Substitute $\left(-3\text{i}+\sqrt{135}\text{j}\right)\text{in}\text{.}$ for ${\text{r}}_{D/E}$ and $-4\text{rad/s}$ for ${\omega }_{DE}$.

$\begin{array}{l}{\text{a}}_D={\alpha }_{DE}\text{k}\times \left(-3\text{i}+\sqrt{135}\text{j}\right)-{\left(-4\right)}^2\times \left(-3\text{i}+\sqrt{135}\text{j}\right)\\ {\text{a}}_D=-3{\alpha }_{DE}\text{j}-\sqrt{135}{\alpha }_{DE}\text{i}+48\text{i}-16\sqrt{135}\text{j}\\ {\text{a}}_D=\left(-\sqrt{135}{\alpha }_{DE}+48\right)\text{i}+\left(-3{\alpha }_{DE}-16\sqrt{135}\right)\text{j}\end{array}$ (5)

Equate Equation (4) and (5).

$-144\text{i+}\left(-16\sqrt{135}+6{\alpha }_{BD}\right)\text{j}=\left(-\sqrt{135}{\alpha }_{DE}+48\right)\text{i}+\left(-3{\alpha }_{DE}-16\sqrt{135}\right)\text{j}$ (6)

Equate i component of Equation (6).

$\begin{array}{c}-144=-\sqrt{135}{\alpha }_{DE}+48\\ {\alpha }_{DE}=\frac{192}{\sqrt{135}}\\ =16.53{\text{rad/s}}^2\left(\text{Counterclockwise}\right)\end{array}$

Thus, the angular acceleration of DE is $\underset{\_}{16.53{\text{rad/s}}^2\left(\text{Counterclockwise}\right)}$.

(b)

To determine

Find the acceleration of point F.

Answer to Problem 15.135P

The acceleration at F is $\underset{\_}{193.6{\text{rad/s}}^2}$ making angle $7.36°$ clockwise below horizontal

Explanation of Solution

Given information:

Calculation:

Refer Part (a).

Consider the position of the point F with respect to point B is denoted by ${\text{r}}_{F/B}=3\text{i}-\sqrt{135}\text{j}$.

Calculate the velocity $\left({\text{a}}_F\right)$ at F using the relation:

${\text{a}}_F={\text{a}}_B+{\alpha }_{BD}\times {\text{r}}_{F/B}-{\omega }_{BD}^2{\text{r}}_{F/B}$

Substitute $\left(-48\text{i}-16\sqrt{135}\text{j}\right)$ for ${\text{a}}_B$, $3\text{i}-\sqrt{135}\text{j}$  for ${\text{r}}_{F/B}$, $-8.265\text{k}$ for ${\alpha }_{BD}$, and $-4\text{rad/s}$ for ${\omega }_{BD}$

$\begin{array}{l}{\text{a}}_F=\left(-48\text{i}-16\sqrt{135}\text{j}\right)-8.265\text{k}\times \left(3\text{i}-\sqrt{135}\text{j}\right)-{\left(-4\right)}^2\times \left(3\text{i}-\sqrt{135}\text{j}\right)\\ {\text{a}}_F=\left(-48\text{i}-16\sqrt{135}\text{j}\right)-\left(\text{24}\text{.795j}+96.03\text{i}\right)-\left(\text{48i}-185.903\text{j}\right)\\ {\text{a}}_F=\left(-48\text{i}-16\sqrt{135}\text{j}\right)+\left(\text{-24}\text{.795j}-96.03\text{i}\right)-\left(\text{48i}-185.903\text{j}\right)\\ {\text{a}}_F=-192\text{i}-24.795\text{j}\end{array}$

Calculate the magnitude of acceleration at F using the relation:

$\begin{array}{c}{\text{a}}_F=\sqrt{{\left(-192\right)}^2+{\left(-24.795\right)}^2}\\ =193.6{\text{rad/s}}^2\end{array}$

Find the direction of the acceleration at F as follows:

$\begin{array}{c}\tan \theta =\left(\frac{24.795}{192}\right)\\ \theta ={\tan }^{-1}\left(\frac{24.795}{192}\right)\\ =7.36°\end{array}$

Thus, the acceleration at F is $\underset{\_}{193.6{\text{rad/s}}^2}$ making angle $7.36°$ clockwise below horizontal.