Chapter 15.4, Problem 15.130P

(a)

To determine

The acceleration of point B.

Answer to Problem 15.130P

The acceleration of point B $\left({\text{a}}_B\right)$ is $\underset{\_}{138.1{\text{ft/s}}^2}$ with an angle of $\underset{\_}{78.7°\left(\text{II}-\text{quad}\right)}$.

Explanation of Solution

Given information:

The constant angular velocity of the bar DE $\left({\omega }_{DE}\right)$ is 18 rad/s.

Calculation:

Draw the free body diagram of the bar system as in Figure (1).

Determine the angular velocity at point D $\left(v_D\right)$ using the relation.

$v_D=l_{DE}{\omega }_{DE}$

Here, $l_{DE}$ is the length of bar DE.

Substitute 15.2 in. for $l_{DE}$ and 18 rad/s for ${\omega }_{DE}$.

$\begin{array}{c}{v}_D=15.2\times 18\\ =273.6\text{in}\text{./s}\end{array}$

The magnitude of the velocity at point D is ${\text{v}}_D=v_D\uparrow$ and the magnitude of the velocity at point B is ${\text{v}}_B=v_B\to$.

A point C is the instantaneous center of bar BD.

Determine the angular velocity of the bar BD using the relation.

${\omega }_{BD}=\frac{v_D}{l_{CD}}$

Substitute $273.6\text{in}\text{./s}$ for $v_B$ and 19.2 in. for $l_{CD}$.

$\begin{array}{c}{\omega }_{BD}=\frac{273.6}{19.2}\\ =14.25\text{rad/s}\left(\text{Anticlockwise}\right)\end{array}$

Determine the angular velocity at point B $\left(v_B\right)$ using the relation.

$v_B=l_{CB}{\omega }_{BD}$

Here, $l_{CB}$ is the length of bar CB.

Substitute 8 in. for $l_{CB}$ and 14.25 rad/s for ${\omega }_{BD}$.

$\begin{array}{c}{v}_B=8\times 14.25\\ =114\text{in}\text{./s}\end{array}$

Determine the angular velocity of the bar AB using the relation.

${\omega }_{AB}=\frac{v_B}{l_{AB}}$

Here, $l_{AB}$ is the length of bar AB.

Substitute $114\text{in}\text{./s}$ for $v_B$ and 8 in. for $l_{AB}$.

$\begin{array}{c}{\omega }_{AB}=\frac{114}{8}\\ =14.25\text{rad/s}\left(\text{Anticlockwise}\right)\end{array}$

The value of angular acceleration at bar DE is zero $\left({\alpha }_{DE}=0\right)$.

Determine the acceleration at point D using the relation.

${\text{a}}_D=l_{DE}{\omega }_{DE}^2$

Substitute 15.2 in. for $l_{DE}$ and 18 rad/s for ${\omega }_{DE}$.

$\begin{array}{c}{\text{a}}_D=15.2\times {18}^2\\ =4,924.8\text{in}{\text{./s}}^2(\to )\end{array}$

Determine the acceleration at point B using the relation.

${\text{a}}_B=\left[l_{AB}{\alpha }_{AB}\to \right]+\left[l_{AB}{\omega }_{AB}^2\uparrow \right]$

Substitute 8 in. for $l_{AB}$ and 14.25 rad/s for ${\omega }_{AB}$.

$\begin{array}{c}{\text{a}}_B=\left[8{\alpha }_{AB}\to \right]+\left[8\times {14.25}^2\uparrow \right]\\ =\left[8{\alpha }_{AB}\to \right]+\left[1,624.5\text{in}{\text{./s}}^{\text{2}}\uparrow \right]\end{array}$

Determine the tangential component of acceleration at point D with respect to B.

${\left({\text{a}}_{D/B}\right)}_t=\left[l_{DC}{\alpha }_{BD}\downarrow \right]+\left[l_{BC}{\alpha }_{BD}\to \right]$

Substitute 19.2 in. for $l_{DC}$ and 8 in. for $l_{BC}$.

${\left({\text{a}}_{D/B}\right)}_t=\left[19.2{\alpha }_{BD}\downarrow \right]+\left[8{\alpha }_{BD}\to \right]$

Determine the normal component of acceleration at point D with respect to B.

${\left({\text{a}}_{D/B}\right)}_n=\left[l_{DC}{\omega }_{BD}^2\to \right]+\left[l_{BC}{\omega }_{BD}^2\uparrow \right]$

Substitute 19.2 in. for $l_{DC}$, 14.25 rad/s for ${\omega }_{BD}$, and 8 in. for $l_{BC}$.

$\begin{array}{c}{\left({\text{a}}_{D/B}\right)}_n=\left[19.2\times {14.25}^2\to \right]+\left[8\times {14.25}^2\uparrow \right]\\ =\left[3,898.8\text{in}{\text{./s}}^{\text{2}}\to \right]+\left[1,624.5\text{in}{\text{./s}}^{\text{2}}\uparrow \right]\end{array}$

Determine the acceleration at point D using the relation.

${\text{a}}_D={\text{a}}_B+{\left({\text{a}}_{D/B}\right)}_t+{\left({\text{a}}_{D/B}\right)}_n$

Substitute $4,924.8\text{in}{\text{./s}}^2(\to )$ for ${\text{a}}_D$, $\left[8{\alpha }_{AB}\to \right]+\left[1,624.5\text{in}{\text{./s}}^{\text{2}}\uparrow \right]$ for ${\text{a}}_B$, $\left[19.2{\alpha }_{BD}\downarrow \right]+\left[8{\alpha }_{BD}\to \right]$ for ${\left({\text{a}}_{D/B}\right)}_t$, and $\left[3,898.8\text{in}{\text{./s}}^{\text{2}}\to \right]+\left[1,624.5\text{in}{\text{./s}}^{\text{2}}\uparrow \right]$ for ${\left({\text{a}}_{D/B}\right)}_n$.

$4,924.8\text{in}{\text{./s}}^2(\to )=\left(\begin{array}{l}\left[8{\alpha }_{AB}\to \right]+\left[1,624.5\text{in}{\text{./s}}^{\text{2}}\uparrow \right]+\left[19.2{\alpha }_{BD}\downarrow \right]+\left[8{\alpha }_{BD}\to \right]+\\ \left[3,898.8\text{in}{\text{./s}}^{\text{2}}\to \right]+\left[1,624.5\text{in}{\text{./s}}^{\text{2}}\uparrow \right]\end{array}\right)$ (1)

Equate the vertical components in Equation (1).

$\begin{array}{l}0=1,624.5-19.2{\alpha }_{BD}+1,624.5\\ 19.2{\alpha }_{BD}=3,249\\ {\alpha }_{BD}=\frac{3,249}{19.2}\\ {\alpha }_{BD}=169.21{\text{rad/s}}^{\text{2}}\end{array}$

Equate the horizontal components in Equation (1).

$4,928=8{\alpha }_{AB}+8{\alpha }_{BD}+3,898.8$

Substitute $169.21{\text{rad/s}}^{\text{2}}$ for ${\alpha }_{BD}$.

$\begin{array}{l}4,928=8{\alpha }_{AB}+8\times 169.21+3,898.8\\ 4,928-5,252.48=8{\alpha }_{AB}\\ {\alpha }_{AB}=\frac{-324.48}{8}\\ {\alpha }_{AB}=-40.56{\text{rad/s}}^{\text{2}}\end{array}$

Determine the acceleration at point B.

${\text{a}}_B=\left[l_{AB}{\alpha }_{AB}\to \right]+\left[l_{AB}{\omega }_{AB}^2\uparrow \right]$

Substitute 8 in. for $l_{AB}$, $-40.56{\text{rad/s}}^{\text{2}}$ for ${\alpha }_{AB}$, and 14.25 rad/s for ${\omega }_{AB}$.

$\begin{array}{c}{\text{a}}_B=\left[8\times -40.56\to \right]+\left[8\times {14.25}^2\uparrow \right]\\ =\left[324.48\text{in}{\text{./s}}^{\text{2}}\leftarrow \right]+\left[1,624.5\text{in}{\text{./s}}^{\text{2}}\uparrow \right]\end{array}$

Determine the magnitude of the acceleration at point B.

${\text{a}}_B=\sqrt{{\left({\text{a}}_B\right)}_x^2+{\left({\text{a}}_B\right)}_y^2}$

Substitute $-\text{324}\text{.48}\text{in}{\text{./s}}^{\text{2}}$ in ${\left({\text{a}}_B\right)}_x$ and $\text{1,624}\text{.5}\text{in}{\text{./s}}^{\text{2}}$ for ${\left({\text{a}}_B\right)}_y$.

$\begin{array}{c}{\text{a}}_B=\sqrt{{\left(-324.48\right)}^2+{\left(1,624.5\right)}^2}\\ =1,656.6\text{in}{\text{./s}}^{\text{2}}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =138.1{\text{ft/s}}^{\text{2}}\end{array}$

Determine the direction of the acceleration at point B.

$\tan \alpha =\frac{{\left({\text{a}}_B\right)}_y}{{\left({\text{a}}_B\right)}_x}$

Substitute $\text{1,624}\text{.5}\text{in}{\text{./s}}^{\text{2}}$ for ${\left({\text{a}}_B\right)}_y$ and $\text{324}\text{.48}\text{in}{\text{./s}}^{\text{2}}$ in ${\left({\text{a}}_B\right)}_x$.

$\begin{array}{l}\tan \alpha =\frac{1,624.5}{324.48}\\ \alpha ={\tan }^{-1}\left(0.1997\right)\\ \alpha =78.7°\end{array}$

Therefore, the acceleration of point B $\left({\text{a}}_B\right)$ is $\underset{\_}{138.1{\text{ft/s}}^2}$ with an angle of $\underset{\_}{78.7°\left(\text{II}-\text{quad}\right)}$.

(b)

To determine

The acceleration of point G.

Answer to Problem 15.130P

The acceleration of point G is $\underset{\_}{203{\text{ft/s}}^2}$ with an angle of $\underset{\_}{19.5°\left(\text{I}-\text{quad}\right)}$.

Explanation of Solution

Given information:

The constant angular velocity of the bar DE $\left({\omega }_{DE}\right)$ is 18 rad/s.

Calculation:

Determine the acceleration of point G using the relation.

$\begin{array}{c}{\text{a}}_G={\text{a}}_B+\left({\text{a}}_{G/B}\right)\\ ={\text{a}}_B+\frac{1}{2}\left({\text{a}}_{D/B}\right)\\ ={\text{a}}_B+\frac{1}{2}\left({\text{a}}_D-{\text{a}}_B\right)\\ =\frac{{\text{a}}_B+{\text{a}}_D}{2}\end{array}$ (2)

Substitute $\left[324.48\text{in}{\text{./s}}^{\text{2}}\leftarrow \right]+\left[1,624.5\text{in}{\text{./s}}^{\text{2}}\uparrow \right]$ for ${\text{a}}_B$ and $4,924.8\text{in}{\text{./s}}^2(\to )$ for ${\text{a}}_D$.

$\begin{array}{c}{\text{a}}_G=\frac{\left[324.48\leftarrow \right]+\left[1,624.5\uparrow \right]+4,924.8(\to )}{2}\\ =\left[2,300.16\text{in}{\text{./s}}^{\text{2}}\right]\to +\left[812.25\text{in}{\text{./s}}^{\text{2}}\uparrow \right]\end{array}$

Determine the magnitude of the acceleration at point G.

${\text{a}}_G=\sqrt{{\left({\text{a}}_G\right)}_x^2+{\left({\text{a}}_G\right)}_y^2}$

Substitute $2,300.16\text{in}{\text{./s}}^{\text{2}}$ in ${\left({\text{a}}_G\right)}_x$ and $\text{812}\text{.25}\text{in}{\text{./s}}^{\text{2}}$ for ${\left({\text{a}}_G\right)}_y$.

$\begin{array}{c}{\text{a}}_G=\sqrt{{\left(2,300.16\right)}^2+{\left(812.25\right)}^2}\\ =2,439.36\text{in}{\text{./s}}^{\text{2}}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =203{\text{ft/s}}^{\text{2}}\end{array}$

Determine the direction of the acceleration at point B.

$\tan \alpha =\frac{{\left({\text{a}}_B\right)}_y}{{\left({\text{a}}_B\right)}_x}$

Substitute $\text{812}\text{.25}\text{in}{\text{./s}}^{\text{2}}$ for ${\left({\text{a}}_G\right)}_y$ and $2,300.16\text{in}{\text{./s}}^{\text{2}}$ in ${\left({\text{a}}_G\right)}_x$.

$\begin{array}{l}\tan \alpha =\frac{812.25}{2,300.16}\\ \alpha ={\tan }^{-1}\left(0.353\right)\\ \alpha =19.5°\end{array}$

Therefore, the acceleration of point G is $\underset{\_}{203{\text{ft/s}}^2}$ with an angle of $\underset{\_}{19.5°\left(\text{I}-\text{quad}\right)}$.