Chapter 15.4, Problem 15.135P

Roberts linkage is named after Richard Roberts (1789-1864) and can be used to draw a close approximation to a straight line by locating a pen at point F. The distance AB is the same as BF, DF, and DE. Knowing that at the instant shown bar AB has a constant angular velocity of 4 rad/s clockwise, determine (a) the angular acceleration of bar DE, (b) the acceleration of point F.

Fig. P 15.135

To determine

(a)

Angular acceleration of bar DE.

Answer to Problem 15.135P

The angular acceleration ${\alpha }_{DE}$ of bar DE is $16.52rad/s^2$ counter clockwise.

Explanation of Solution

Given information:

Constant angular velocity of link AB is $4rad/s$ clockwise.

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:

The position vector $r_{B/A}$ of B relative to A:

$r_{B/A}=3i+\sqrt{{12}^2-3^2}j=3i+\sqrt{135}j$

The velocity $v_B$ of point B:

$v_B=v_A+{\omega }_{AB}k\times r_{B/A}$

Substitute:

$\begin{array}{l}{v}_B=-4k\times \left[3i+\sqrt{135}j\right]\\ v_B=-12j+4\sqrt{135}i\end{array}$

The position vector $r_{D/B}$ of D relative to B:

$r_{D/B}=6i$

The velocity $v_D$ of point D:

$v_D=v_B+{\omega }_{BD}k\times r_{D/B}$

Substitute

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

For bar DE

The position vector $r_{D/E}$ of D relative to E:

$r_{D/E}=-3i+\sqrt{135}j$

The velocity $v_D$ of point D:

$\begin{array}{l}{v}_D={\omega }_{DE}k\times \left[-3i+\sqrt{135}j\right]\\ v_D=-3{\omega }_{DE}j-\sqrt{135}{\omega }_{DE}i\to \left(2\right)\end{array}$

Equate components in equation 1 and 2:

$\begin{array}{l}i:4\sqrt{135}=-\sqrt{135}{\omega }_{DE}\\ j:6{\omega }_{BD}-12=-3{\omega }_{DE}\end{array}$

Therefore

$\begin{array}{l}{\omega }_{DE}=-4rad/s\\ {\omega }_{BD}=4rad/s\end{array}$

Bar AB has constant angular velocity. Therefore angular acceleration ${\alpha }_{AB}$ is zero.

Acceleration $a_B$ of point B:

$\begin{array}{l}{a}_B=-{\omega }_{AB}{}^2{r}_{B/A}\\ a_B=-{\left(4rad/s\right)}^2\left(3i+\sqrt{135}j\right)\\ a_B=-48i-16\sqrt{135}j\end{array}$

For object BDF:

Acceleration $a_D$ of point D:

$a_D=a_B+{\alpha }_{BD}k\times r_{D/B}-{\omega }_{BD}{}^2{r}_{D/B}$

Substitute:

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

For bar DE

Acceleration $a_D$ of point D:

$a_D=a_E+{\alpha }_{DE}k\times r_{D/E}-{\omega }_{DE}{}^2{r}_{D/E}$

Substitute

$\begin{array}{l}{a}_D=0+{\alpha }_{DE}k\times \left(-3i+\sqrt{135}j\right)-{\left(-4\right)}^2\left(-3i+\sqrt{135}j\right)\\ a_D=-3{\alpha }_{DE}j-\sqrt{135}{\alpha }_{DE}i+48i-16\sqrt{135}j\end{array}$

Equate components

$\begin{array}{l}i:-144=-\sqrt{135}{\alpha }_{DE}+48\\ j:-16\sqrt{135}+6{\alpha }_{BD}=-3{\alpha }_{DE}-16\sqrt{135}\end{array}$

Therefore

$\begin{array}{l}{\alpha }_{DE}=\frac{192}{\sqrt{135}}=16.52rad/s^2\\ {\alpha }_{BD}=-\frac{1}{2}{\alpha }_{DE}=-\frac{96}{\sqrt{135}}\end{array}$

Conclusion:

The angular acceleration ${\alpha }_{DE}$ of bar DE is $16.52rad/s^2$ counter clockwise.

To determine

(b)

Acceleration of point F.

Answer to Problem 15.135P

The acceleration $a_F$ of point F is $193.6in/s^2$ an angle $7.35°$ with horizontal downwards.

Explanation of Solution

Given information:

Constant angular velocity of link AB is $4rad/s$ clockwise.

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:

The acceleration $a_B$ of point B:

$a_B=-48i-16\sqrt{135}j$

Angular acceleration ${\alpha }_{BD}$ of object BD:

${\alpha }_{BD}=-\frac{96}{\sqrt{135}}rad/s^2$

Angular velocity ${\omega }_{BD}$ of object BD:

${\omega }_{BD}=4rad/s$

Position vector $r_{F/B}$ of point F relative to point B:

$r_{F/B}=3i-\sqrt{135}j$

Acceleration $a_F$ of point F:

$a_F=a_B+{\alpha }_{BD}k\times r_{F/B}-{\omega }_{BD}{}^2{r}_{F/B}$

Substitute

$\begin{array}{l}{a}_F=-48i-16\sqrt{135}j+\left(-\frac{96}{\sqrt{135}}k\right)\times \left(3i-\sqrt{135}j\right)-{\left(4\right)}^2\left(3i-\sqrt{135}j\right)\\ a_F=-48i-16\sqrt{135}j-\frac{288}{\sqrt{135}}j-96i-48i+16\sqrt{135}j\end{array}$

Solve

$\begin{array}{l}{a}_F=-192i-\frac{288}{\sqrt{135}}j\\ a_F=-192i-24.78j\end{array}$

The magnitude and angle of acceleration $a_F$ of point F:

$a_F=193.6in/s^2\swarrow 7.35°$

Conclusion:

The acceleration $a_F$ of point F is $193.6in/s^2$ an angle $7.35°$ with horizontal downwards.