Chapter 15.2, Problem 15.68P

In the position shown, bar DE has a constant angular velocity of 10 rad/s clockwise. Knowing that $h=500$ mm, determine (a) the angular velocity of bar FBD, (b) the velocity of point F.

    

To determine

(a)

The angular velocity of rod $FBD$.

Answer to Problem 15.68P

The angular velocity of rod $FBD$ is $3.33\text{rad}/\text{s}$.

Explanation of Solution

Given Information:

The constant angular velocity of bar $DE$ is $10\text{rad}/\text{s}$ and the height is $500\text{mm}$.

Draw the schematic diagram for the system.

Figure-(1)

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

$v_D={\omega }_{DE}\left(r_{D/E}\right)$ ...... (I)

Here, the angular velocity at point $D$ is ${\omega }_{DE}$ and the position vector at point $D$ with respect to $E$ is $r_{D/E}$.

Write the expression for the relative velocity component at point $B$.

$v_{B/D}={\omega }_{BD}\left(r_{B/D}\right)$ ...... (II)

Here, the angular velocity at point $B$ is ${\omega }_{BD}$ and the position vector at point $B$ with respect to $D$ is $r_{B/D}$.

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

$v_B=v_D+v_{B/D}$ ...... (III)

Here, the velocity at point $A$ is $v_A$.

Write the expression for the velocity component $V_B$.

$V_B={\omega }_{AB}{r}_{B/A}$ ...... (IV)

Calculation:

Consider the unit vector along $X$ and $Y$ directions as $i$ and $j$ respectively and angular direction as $k$.

The coordinate of the point $D$ with point $E$ as reference are $\left(-100\text{mm},200\text{mm}\right)$. Therefore, the position vector is $r_{D/E}=-\left(\left(100\text{mm}\right)i-\left(200\text{mm}\right)j\right)$.

The coordinate of the point $B$ with point $A$ as reference are $\left(0,420\text{mm}\right)$. Therefore, the position vector is $r_{B/A}=\left(420\text{mm}\right)j$.

The coordinate of the point $B$ with point $D$ as reference are $\left(-300\text{mm,100}\text{mm}\right)$. Therefore, the position vector is $r_{B/D}=-\left(300\text{mm}\right)i+\left(100\text{mm}\right)j$.

Substitute $-\left(\left(100\text{mm}\right)i-\left(200\text{mm}\right)j\right)$ for $r_{D/E}$ and $\left({\omega }_{DE}\right)k$ for ${\omega }_{DE}$ in equation (I).

$\begin{array}{c}{v}_D=\left({\omega }_{DE}\right)k\left(-\left(\left(100\text{mm}\right)i-\left(200\text{mm}\right)j\right)\right)\\ =\left|\begin{array}{ccc}i& j& k\\ 0& 0& -10\\ -100& 200& 0\end{array}\right|\\ =i\left(0+2000\right)-j\left(0-\left(1000\right)\right)+k\left(0-0\right)\\ =\left(2000\right)i-\left(1000\right)j\end{array}$

Substitute $-\left(\left(300\text{mm}\right)i+\left(100\text{mm}\right)j\right)$ for $r_{B/D}$ and $\left({\omega }_{BD}\right)k$ for ${\omega }_{BD}$ in equation (II).

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

Substitute $\left(-100{\omega }_{BD}\right)i-\left(300{\omega }_{BD}\right)j$ for $v_{B/D}$ and $\left(2000\right)i-\left(1000\right)j$ for $v_D$ in Equation (III).

$\left(v_B\right)=\left(2000\right)i-\left(1000\right)j+\left(-100{\omega }_{BD}\right)i-\left(300{\omega }_{BD}\right)j$ ...... (V)

Substitute $\left(420\text{mm}\right)j$ for $r_{B/A}$ and $\left({\omega }_{AB}\right)k$ for ${\omega }_{AB}$ in equation (IV).

$\begin{array}{c}{v}_B=\left({\omega }_{AB}\right)k\left(\left(420\text{mm}\right)j\right)\\ =\left|\begin{array}{ccc}i& j& k\\ 0& 0& {\omega }_{AB}\\ 0& 420& 0\end{array}\right|\\ =i\left(0-420{\omega }_{AB}\right)-j\left(0-0\right)+k\left(0-0\right)\\ =\left(-420{\omega }_{AB}\right)i\end{array}$ ...... (VI)

Substitute $\left(2000\right)i-\left(1000\right)j+\left(-100{\omega }_{BD}\right)i-\left(300{\omega }_{BD}\right)j$ for $v_B$ in Equation (VI).

$\left(2000\right)i-\left(1000\right)j+\left(-100{\omega }_{BD}\right)i-\left(300{\omega }_{BD}\right)j=\left(-420{\omega }_{AB}\right)i$ ...... (VII)

Compare the component along $y$ -direction.

$\begin{array}{l}\left(1000\text{mm}/\text{s}\right)j-\left(300\text{mm}{\omega }_{BD}\right)j=0\\ {\omega }_{BD}=\frac{1000\text{mm}/\text{s}}{300\text{mm}}\\ {\omega }_{BD}=3.33\text{rad}/\text{s}\end{array}$

Compare the component along $x$ -direction.

$\left(2000\right)i-\left(100{\omega }_{BD}\right)i=\left(-420{\omega }_{AB}\right)i$ ...... (VIII)

Substitute $3.33\text{rad}/\text{s}$ for ${\omega }_{BD}$ in Equation (VIII).

$\begin{array}{l}\left(2000\right)i-\left(100\left(3.33\text{rad}/\text{s}\right)\right)i=\left(-420{\omega }_{AB}\right)i\\ \left(2000\right)i-\left(333\text{rad}/\text{s}\right)i=\left(-420{\omega }_{AB}\right)i\\ {\omega }_{AB}=-\frac{1667}{420}\text{rad}/\text{s}\\ {\omega }_{AB}=-3.968\text{rad}/\text{s}\end{array}$

Conclusion:

The angular velocity of rod $FBD$ is $3.33\text{rad}/\text{s}$.

To determine

(b)

The velocity of point $F$.

Answer to Problem 15.68P

The velocity of point $F$ is $2\text{m}/\text{s}$.

Explanation of Solution

Given Information:

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

$R=\frac{r_{F/D}}{r_{B/D}}$ ...... (IX)

Here, the distance between the point $F$ and $D$ is $r_{F/D}$ and the distance between the point $B$ and $D$ is $r_{B/D}$.

Write the expression for the position vector at point $F$ with respect to $D$.

$r_{F/D}=R\left(r_{B/D}\right)$ ...... (X)

Write the expression for the velocity component at $F$ with respect to $D$.

$v_{F/D}={\omega }_{BD}{r}_{F/D}$ ...... (XI)

Write the expression for the velocity at point $F$.

$v_F=v_D+v_{F/D}$ ...... (XII)

Write the expression for the resultant velocity at point $F$.

$v_F=\sqrt{x^2+y^2}$ ...... (XIII)

Here, the coefficient of $i$ vector is $x$ and the coefficient of $j$ vector is $y$.

Write the expression for the angle of the velocity at point $F$.

$\varphi ={\tan }^{-1}\left(\frac{y}{x}\right)$ ...... (XIV)

Calculation:

Substitute $\left(500\text{mm}+300\text{mm}\right)$ for $r_{F/D}$ and $300\text{mm}$ for $r_{B/D}$ in equation (IX).

$\begin{array}{c}R=\frac{\left(500\text{mm}+300\text{mm}\right)}{300\text{mm}}\\ =\frac{800\text{mm}}{300\text{mm}}\\ =2.666\end{array}$

Substitute $2.666$ for $R$ and $-\left(300\text{mm}\right)i+\left(100\text{mm}\right)j$ for $r_{B/D}$ in Equation (X).

$\begin{array}{c}{r}_{F/D}=2.666\left(-\left(300\text{mm}\right)i+\left(100\text{mm}\right)j\right)\\ =-\left(800\text{mm}\right)i+\left(266.67\text{mm}\right)j\end{array}$

Substitute $-\left(800\text{mm}\right)i+\left(266.67\text{mm}\right)j$ for $r_{F/D}$ and $\left(3.33\text{rad}/\text{s}\right)k$ for ${\omega }_{BD}$ in Equation (XI).

$\begin{array}{c}{v}_{F/D}=\left(\left(3.33\text{rad}/\text{s}\right)k\right)\left(-\left(800\text{mm}\right)i+\left(266.67\text{mm}\right)j\right)\\ =\left|\begin{array}{ccc}i& j& k\\ 0& 0& 3.33\\ -800& 266.67& 0\end{array}\right|\\ =i\left(0-266.67\left(3.33\right)\right)-j\left(0-\left(-800\left(3.33\right)\right)\right)+k\left(0-0\right)\\ =-888.01i-2664j\end{array}$

Substitute $-888.01i-2664j$ for $v_{F/D}$ and $2000i+1000j$ for $v_D$ in Equation (XII).

$\begin{array}{c}\left(v_B\right)=\left(2000i+1000j\right)+\left(-888.01i-2664j\right)\\ =\left(1111.99\text{mm}/\text{s}\right)i-\left(1664\text{mm}/\text{s}\right)j\end{array}$

Substitute $1111.99\text{mm}/\text{s}$ for $x$ and $1664\text{mm}/\text{s}$ for $y$ in Equation (XIII).

$\begin{array}{c}{v}_F=\sqrt{{\left(1111.99\text{mm}/\text{s}\right)}^2+{\left(1664\text{mm}/\text{s}\right)}^2}\\ =\sqrt{\left(4005417.76{\text{mm}}^2/{\text{s}}^2\right)}\\ =2001.35\text{mm}/\text{s}\left(\frac{\text{1}\text{m}}{{\text{10}}^3\text{mm}}\right)\\ =2\text{m}/\text{s}\end{array}$

Substitute $1111.99\text{mm}/\text{s}$ for $x$ and $1664\text{mm}/\text{s}$ for $y$ in Equation (XIV).

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{1664\text{mm}/\text{s}}{1111.99\text{mm}/\text{s}}\right)\\ ={\tan }^{-1}\left(1.4964\right)\\ =56.24°\end{array}$

Conclusion:

The velocity of point $F$ is $2\text{m}/\text{s}$.