Chapter 15.4, Problem 15.128P

The elliptical exercise machine has fixed axes of rotation at points A and E. Knowing that at the instant shown the flywheel AB has a constant angular velocity of 6 rad/s clockwise, determine (a) the angular acceleration of bar DEF, (b) the acceleration of point F.

Fig. P15.127 and P15.128

To determine

(a)

Angular acceleration of bar DEF.

Answer to Problem 15.128P

The angular acceleration of bar DEF is $1.466rad/s^2$ counter clockwise.

Explanation of Solution

Given information:

Constant angular velocity of AB is $6rad/s$ clockwise.

Point A and E are fixed.

The absolute value of point A:

$v_A=v_B+v_{A/B}$

The relative velocity of A with respect to B is defined as:

$v_{A/B}=\omega k\times r_{A/B}$

$r_{A/B}$ - Position vector of A with respect to B

$\omega$ - Angular velocity

The absolute acceleration of point B is defined as:

$a_B=a_A+a_{B/A}$

$a_B=a_A+{\left(a_{B/A}\right)}_t+{\left(a_{B/A}\right)}_n$

The tangential acceleration is defined as:

${\left(a_{B/A}\right)}_t=\alpha k\times r_{B/A}$

The normal acceleration is defined as:

${\left(a_{B/A}\right)}_n=-r_{B/A}{\omega }_{AB}{}^2$

In above equations

$\alpha$ - Angular acceleration

$\omega$ - Angular velocity

$r$ - Distance from A to B

Calculation:

Position vector of B relative to A:

$r_{B/A}=0.2j$

Position vector of D relative to B:

$r_{D/B}=1.2i-0.2j$

Position vector of D relative to E:

$r_{D/E}=-0.12i-0.8j$

Absolute velocity of point B:

$v_B=v_A+{\omega }_{AB}k\times r_{B/A}\to \left(1\right)$

Absolute velocity of point D:

$\begin{array}{l}{v}_D=v_B+{\omega }_{BD}k\times r_{D/B}\to \left(2\right)\\ v_D=v_E+{\omega }_{DEF}k\times r_{D/E}\to \left(3\right)\end{array}$

We know that:

$\begin{array}{l}{v}_A=0\\ v_E=0\end{array}$

Substitute equation 1 into 2 and equate with 3:

${\omega }_{AB}k\times r_{B/A}+{\omega }_{BD}k\times r_{D/B}={\omega }_{DEF}k\times r_{D/E}$

Substitute:

$\begin{array}{l}-6k\times 0.2j+{\omega }_{BD}k\times \left(1.2i-0.2j\right)={\omega }_{DEF}k\times \left(-0.12i-0.8j\right)\\ 1.2i+1.2{\omega }_{BD}j+0.2{\omega }_{BD}i=-0.12{\omega }_{DEF}j+0.8{\omega }_{DEF}i\end{array}$

Equate components:

$\begin{array}{l}i:1.2+0.2{\omega }_{BD}=0.8{\omega }_{DEF}\\ j:1.2{\omega }_{BD}=-0.12{\omega }_{DEF}\end{array}$

Therefore:

$\begin{array}{l}{\omega }_{BD}=-0.1463rad/s\\ {\omega }_{DEF}=1.463rad/s\end{array}$

Similarly:

$\begin{array}{l}{a}_B=a_A+{\alpha }_{AB}k\times r_{B/A}-{\omega }_{AB}{}^2{r}_{B/A}\to \left(4\right)\\ a_D=a_B+{\alpha }_{BD}k\times r_{D/B}-{\omega }_{BD}{}^2{r}_{D/B}\to \left(5\right)\\ a_D=a_E+{\alpha }_{DEF}k\times r_{D/E}-{\omega }_{DEF}{}^2{r}_{D/E}\to \left(6\right)\end{array}$

But we know that:

$\begin{array}{l}{a}_A=0\\ a_E=0\\ {\alpha }_{AB}=0\end{array}$

Therefore, substitute equation 4 into 5 and equate with 6:

$-{\omega }_{AB}{}^2{r}_{B/A}+{\alpha }_{BD}k\times r_{D/B}-{\omega }_{BD}{}^2{r}_{D/B}={\alpha }_{DEF}k\times r_{D/E}-{\omega }_{DEF}{}^2{r}_{D/E}$

Substitute:

$\begin{array}{l}-{\left(6\right)}^2\left(0.2j\right)+{\alpha }_{BD}k\times \left(1.2i-0.2j\right)-{\left(-0.1463\right)}^2\left(1.2i-0.2j\right)=\\ {\alpha }_{DEF}k\times \left(-0.12i-0.8j\right)-{\left(1.463\right)}^2\left(-0.12i-0.8j\right)\end{array}$

Solve further:

$\begin{array}{l}-7.2j+1.2{\alpha }_{BD}j+0.2{\alpha }_{BD}i-0.0257i+0.00428j=\\ -0.12{\alpha }_{DEF}j+0.8{\alpha }_{DEF}i+0.257i+1.7123j\end{array}$

Equate components:

$\begin{array}{l}i:0.2{\alpha }_{BD}-0.0257=0.8{\alpha }_{DEF}+0.257\\ j:-7.2+1.2{\alpha }_{BD}+0.00428=-0.12{\alpha }_{DEF}+1.7123\end{array}$

Solve above equations:

$\begin{array}{l}{\alpha }_{BD}=7.278rad/s^2\\ {\alpha }_{DEF}=1.466rad/s^2\end{array}$

Conclusion:

The angular acceleration of bar DEF is $1.466rad/s^2$ counter clockwise.

To determine

(b)

Acceleration of point F

Answer to Problem 15.128P

The acceleration of point F is $1.574in/s^2$ at an angle $47.06°$ with horizontal downwards.

Explanation of Solution

Given information:

Constant angular velocity of AB is $6rad/s$ clockwise.

Point A and E are fixed.

The absolute acceleration of point B is defined as

$a_B=a_A+a_{B/A}$

$a_B=a_A+{\left(a_{B/A}\right)}_t+{\left(a_{B/A}\right)}_n$

The tangential acceleration is defined as

${\left(a_{B/A}\right)}_t=\alpha k\times r_{B/A}$

The normal acceleration is defined as

${\left(a_{B/A}\right)}_n=-r_{B/A}{\omega }_{AB}{}^2$

In above equations

$\alpha$ - Angular acceleration

$\omega$ - Angular velocity

$r$ - Distance from A to B

Calculation:

According to sub part a

$\begin{array}{l}{\alpha }_{DEF}=1.466rad/s^2\\ {\omega }_{DEF}=1.463rad/s\end{array}$

Absolute acceleration of point F

$a_F=a_E+{\alpha }_{DEF}k\times r_{F/E}-{\omega }_{DEF}{}^2{r}_{F/E}$

We know that

$a_E=0$

Therefore

$a_F={\alpha }_{DEF}k\times r_{F/E}-{\omega }_{DEF}{}^2{r}_{F/E}$

Substitute

$\begin{array}{l}{a}_F=1.466k\times \left(0.09i+0.6j\right)-{\left(1.463\right)}^2\left(0.09i+0.6j\right)\\ a_F=0.13194j-0.8796i-0.1926i-1.2842j\end{array}$

Then

$a_F=-1.0722i-1.1522j$

Find the magnitude and the angle

$a_F=1.574in/s^2\swarrow 47.06°$

Conclusion:

The acceleration of point F is $1.574in/s^2$ at an angle $47.06°$ with horizontal downwards.