Chapter 15.2, Problem 15.69P

In the position shown, bar DE has a constant angular velocity of 10 rad/s clockwise. Determine (a) the distance h for which the velocity of point F is vertical, (b) the corresponding velocity of point F.

    

To determine

(a)

The distance $h$ for which velocity of point F is vertical.

Answer to Problem 15.69P

The distance $h$ for which velocity of point F is vertical is $1.501801\text{m}$.

Explanation of Solution

Given information:

The angular velocity of link DE is $10\text{rad}/\text{s}$.

Figure-(1) represents the geometry of links attached in mechanism.

Write the expression for position vector of point $D$ with respect to point $E$.

${\overrightarrow{r}}_{D/E}=-\left(100\text{mm}\right)\text{i}+\left(200\text{mm}\right)\text{j}$ ...... (I)

Write the expression for velocity component of point $D$.

${\overrightarrow{v}}_D={\overrightarrow{\omega }}_{DE}\times {\overrightarrow{r}}_{D/E}$ ...... (II)

Here, angular velocity of link $DE$ is ${\overrightarrow{\omega }}_{DE}$.

Write the expression for position vector of point $B$ with respect to point $D$.

${\overrightarrow{r}}_{B/D}=-\left(300\text{mm}\right)\text{i}+\left(100\text{mm}\right)\text{j}$ ...... (III)

Write the expression for velocity component of $B$ with respect to point $D$.

${\overrightarrow{v}}_{B/D}={\overrightarrow{\omega }}_{BD}\times {\overrightarrow{r}}_{B/D}$ ...... (IV)

Here, angular velocity of link $BD$ is ${\overrightarrow{\omega }}_{BD}$.

Write the expression for velocity component of point $B$.

${\overrightarrow{v}}_B={\overrightarrow{v}}_D+{\overrightarrow{v}}_{B/D}$ ...... (V)

Write the expression for position vector of point $B$ with respect to point $A$.

${\overrightarrow{r}}_{B/A}=\left(420\text{mm}\right)\text{j}$ ...... (VI)

Write the expression for velocity component of point $B$.

${\overrightarrow{v}}_B={\overrightarrow{\omega }}_{AB}\times {\overrightarrow{r}}_{B/A}$ ...... (VII)

Here, angular velocity of link $AB$ is ${\overrightarrow{\omega }}_{AB}$.

Write the expression for the proportion of rod $FBD$.

$\frac{{\overrightarrow{r}}_{F/D}}{{\overrightarrow{r}}_{B/D}}=\frac{h+300\text{mm}}{300\text{mm}}$ ...... (VIII)

Here, horizontal distance of point F from point A is $h$ and position vector of $F$ with respect to $D$ is ${\overrightarrow{r}}_{F/D}$.

Write the expression for velocity component $v_{F/D}$.

${\overrightarrow{v}}_{F/D}={\overrightarrow{\omega }}_{BD}\times {\overrightarrow{r}}_{F/D}$ ...... (IX)

Write the expression for velocity of point $F$.

${\overrightarrow{v}}_F={\overrightarrow{v}}_D+{\overrightarrow{v}}_{F/D}$ ...... (X)

Calculation:

Consider the unit vector $\overrightarrow{k}$ along the angular velocity direction in clockwise as negative and counter clockwise as positive.

Substitute $-10\text{rad}/\text{s}$ for ${\overrightarrow{\omega }}_{DE}$ and $-\left(100\text{mm}\right)\text{i}+\left(200\text{mm}\right)\text{j}$ for ${\overrightarrow{r}}_{D/E}$ in Equation (II).

$\begin{array}{c}{\overrightarrow{v}}_D=\left(-10\text{rad}/\text{s}\right)\text{k}\times \left(-\left(100\text{mm}\right)\text{i}+\left(200\text{mm}\right)\text{j}\right)\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& -10\text{rad}/\text{s}\\ -100\text{mm}& 200\text{mm}& 0\end{array}\right|\\ =\left(0+2000\text{mm}/\text{s}\right)\text{i}-\left(0-1000\text{mm}/\text{s}\right)\text{j}\\ =\left(2000\text{mm}/\text{s}\right)\text{i}+\left(1000\text{mm}/\text{s}\right)\text{j}\end{array}$

Substitute $-\left(300\text{mm}\right)\text{i}+\left(100\text{mm}\right)\text{j}$ for ${\overrightarrow{r}}_{B/D}$ in Equation (IV).

$\begin{array}{c}{\overrightarrow{v}}_{B/D}=\left({\omega }_{BD}\right)\text{k}\times \left(-\left(300\text{mm}\right)\text{i}+\left(100\text{mm}\right)\text{j}\right)\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& {\omega }_{BD}\\ -300\text{mm}& 100\text{mm}& 0\end{array}\right|\\ =\left(0-100\text{mm}\times {\omega }_{BD}\right)\text{i}-\left(0+300\text{mm}\times {\omega }_{BD}\right)\text{j}=\left(-100\text{mm}\times {\omega }_{BD}\right)\text{i}-\left(300\text{mm}\times {\omega }_{BD}\right)\text{j}\end{array}$

Substitute $\left(2000\text{mm}/\text{s}\right)\text{i}+\left(1000\text{mm}/\text{s}\right)\text{j}$ for ${\overrightarrow{v}}_D$ and $\left(-100\text{mm}\times {\omega }_{BD}\right)\text{i}-\left(300\text{mm}\times {\omega }_{BD}\right)\text{j}$ for ${\overrightarrow{v}}_{B/D}$ in Equation (V).

$\begin{array}{c}{\overrightarrow{v}}_B=\left[\begin{array}{c}\left(2000\text{mm}/\text{s}\right)\text{i}+\left(1000\text{mm}/\text{s}\right)\text{j}+\\ \left(-100\text{mm}\times {\omega }_{BD}\right)\text{i}-\left(300\text{mm}\times {\omega }_{BD}\right)\text{j}\end{array}\right]\\ =\left[\begin{array}{l}\left(2000\text{mm}/\text{s}-100\text{mm}\times {\omega }_{BD}\right)\text{i}+\\ \left(1000\text{mm}/\text{s-}300\text{mm}\times {\omega }_{BD}\right)\text{j}\end{array}\right]\end{array}$ ...... (XI)

Substitute $\left(420\text{mm}\right)\overrightarrow{j}$ for ${\overrightarrow{r}}_{B/A}$ in Equation (VII).

$\begin{array}{r}{\overrightarrow{v}}_B=\left({\omega }_{AB}\right)\text{k}\times \left(420\text{mm}\right)\text{j}\\ =\left(-420\text{mm}\times {\omega }_{AB}\right)\text{i}\end{array}$ ...... (XII)

Substitute $\left(-420\text{mm}\times {\omega }_{AB}\right)\text{i}$ for ${\overrightarrow{v}}_B$ in Equation (XI).

$\left[\begin{array}{l}\left(2000\text{mm}/\text{s}-100\text{mm}\times {\omega }_{BD}\right)\text{i}+\\ \left(1000\text{mm}/\text{s}-300\text{mm}\times {\omega }_{BD}\right)\text{j}\end{array}\right]=\left(-420\text{mm}\times {\omega }_{AB}\right)\text{i}$ ...... (XIII)

Compare the coefficient of $\text{j}$ in Equation (XIII).

$\begin{array}{l}1000\text{mm}/\text{s}-300\text{mm}\times {\omega }_{BD}=0\\ {\omega }_{BD}=\frac{1000\text{mm}/\text{s}}{300\text{mm}}\\ {\omega }_{BD}=3.33\text{rad}/\text{s}\end{array}$

Substitute $3.33\text{rad}/\text{s}$ for ${\omega }_{BD}$ and compare the coefficient of $\text{i}$ in Equation (XIII).

$\begin{array}{l}2000\text{mm}/\text{s}-100\text{mm}\times \left(3.33\text{rad}/\text{s}\right)=-420\text{mm}\times {\omega }_{AB}\\ {\omega }_{AB}=-\frac{2000\text{mm}/\text{s}-\text{333}\text{mm}/\text{s}}{420\text{mm}}\\ {\omega }_{AB}=-3.9683\text{rad}/\text{s}\end{array}$

Substitute $-\left(300\text{mm}\right)\text{i}+\left(100\text{mm}\right)\text{j}$ for ${\overrightarrow{r}}_{B/D}$ in Equation (VIII).

$\begin{array}{l}\frac{{\overrightarrow{r}}_{F/D}}{-\left(300\text{mm}\right)\text{i}+\left(100\text{mm}\right)\text{j}}=\frac{h+300\text{mm}}{300\text{mm}}\\ {\overrightarrow{r}}_{F/D}=\left(\frac{h+300\text{mm}}{300\text{mm}}\right)\left[-\left(300\text{mm}\right)\text{i}+\left(100\text{mm}\right)\text{j}\right]\end{array}$

Substitute $3.33\text{rad}/\text{s}$ for ${\omega }_{BD}$ and $\left(\frac{h+300\text{mm}}{300\text{mm}}\right)\left[-\left(300\text{mm}\right)\text{i}+\left(100\text{mm}\right)\text{j}\right]$ for ${\overrightarrow{r}}_{F/D}$ in Equation (IX)

$\begin{array}{l}{\overrightarrow{v}}_{F/D}=\left(3.33\text{rad}/\text{s}\right)\text{k}\times \left(\left(\frac{h+300\text{mm}}{300\text{mm}}\right)\left[-\left(300\text{mm}\right)\text{i}+\left(100\text{mm}\right)\text{j}\right]\right)\\ =\left[\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& 3.33\text{rad}/\text{s}\\ -\left(h+300\text{mm}\right)& \frac{h+300\text{mm}}{3\text{mm}}& 0\end{array}\right]\\ =\left(0-\frac{h+300\text{mm}}{3\text{mm}}\times 3.33\text{rad}/\text{s}\right)\text{i}-\left(0+\left(h+300\text{mm}\right)\times 3.33\text{rad}/\text{s}\right)\text{j}\\ =-\left(\frac{h+300\text{mm}}{3\text{mm}}\times 3.33\text{rad}/\text{s}\right)\text{i}-\left(\left(h+300\text{mm}\right)\times 3.33\text{rad}/\text{s}\right)\text{j}\end{array}$

Substitute $-\left(\frac{h+300\text{mm}}{3\text{mm}}\times 3.33\text{rad}/\text{s}\right)\text{i}-\left(\left(h+300\text{mm}\right)\times 3.33\text{rad}/\text{s}\right)\text{j}$ for ${\overrightarrow{v}}_{F/D}$ and $\left(2000\text{mm}/\text{s}\right)\text{i}+\left(1000\text{mm}/\text{s}\right)\text{j}$ for ${\overrightarrow{v}}_D$ in Equation (x).

$\begin{array}{c}{\overrightarrow{v}}_F=\left[\begin{array}{l}\left(2000\text{mm}/\text{s}\right)\text{i}+\left(1000\text{mm}/\text{s}\right)\text{j}+\\ -\left(\frac{h+300\text{mm}}{3\text{mm}}\times 3.33\text{rad}/\text{s}\right)\text{i}-\left(\left(h+300\text{mm}\right)\times 3.33\text{rad}/\text{s}\right)\text{j}\end{array}\right]\\ =\left[\begin{array}{l}\left(2000\text{mm}/\text{s}-\frac{h+300\text{mm}}{3\text{mm}}\times 3.33\text{rad}/\text{s}\right)\text{i}+\\ \left(1000\text{mm}/\text{s}-\left(h+300\text{mm}\right)\times 3.33\text{rad}/\text{s}\right)\text{j}\end{array}\right]\end{array}$ ...... (XIV)

Substitute $0$ for $\text{i}$ coefficient as velocity of point $F$ is in vertical direction only.

$\begin{array}{l}2000\text{mm}/\text{s}-\frac{h+300\text{mm}}{3\text{mm}}\times 3.33\text{rad}/\text{s}=0\\ \frac{h+300\text{mm}}{3\text{mm}}\times 3.33\text{rad}/\text{s}=2000\text{mm}/\text{s}\\ h+300\text{mm}=1801.801801\text{mm}\\ h=1501.801801\text{mm}\end{array}$

$\begin{array}{c}h=1501.801801\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\\ =1.501801\text{m}\end{array}$

Conclusion:

The distance $h$ for which velocity of point F is vertical is $1.501801\text{m}$.

To determine

(b)

The velocity of point $F$

Answer to Problem 15.69P

The velocity of point $F$ is $5\text{m}/\text{s}$downward direction.

Explanation of Solution

Calculation:

Substitute $1.501\text{m}$ for $h$ in Equation (XIV).

$\begin{array}{c}{\overrightarrow{v}}_F=\left[\begin{array}{l}\left(2000\text{mm}/\text{s}-\frac{1501.801+300\text{mm}}{3\text{mm}}\times 3.33\text{rad}/\text{s}\right)\text{i}+\\ \left(1000\text{mm}/\text{s}-\left(1501.801\text{mm}+300\text{mm}\right)\times 3.33\text{rad}/\text{s}\right)\text{j}\end{array}\right]\\ =0\text{i}-\left(1000\text{mm}/\text{s}+6000\text{mm}/\text{s}\right)\text{j}\\ =\left(-5000\text{mm}/\text{s}\times \frac{1\text{m}/\text{s}}{1000\text{mm}/\text{s}}\right)\text{j}\\ =\left(-5\text{m}/\text{s}\right)\text{j}\end{array}$

Conclusion:

The velocity of point $F$ is $5\text{m}/\text{s}$downward direction.