Chapter 15.2, Problem 15.68P

(a)

To determine

The angular velocity of the rod FBD.

Answer to Problem 15.68P

The angular velocity of the rod FBD $\left({\omega }_{BD}\right)$ is $\underset{\_}{3.33\text{rad/s}\left(\text{Anticlockwise}\right)}$.

Explanation of Solution

Given information:

The constant angular velocity of the bar DE ${\omega }_{DE}$ is $10\text{rad/s}$.

The value of h is 500 mm.

Calculation:

Write the value of relative position vector of D with respect to E.

$\begin{array}{c}{\text{r}}_{D/E}=\left[\left(-100\right)\text{i}-\left(200\right)\text{j}\right]\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\\ =\left(-0.1\text{m}\right)\text{i}+\left(0.2\text{m}\right)\text{j}\end{array}$

Determine the velocity vector of D $\left({\text{v}}_D\right)$.

${\text{v}}_D={\text{ω}}_{DE}\times {\text{r}}_{D/E}$

Substitute $\left(-10\text{k}\right)\text{rad/s}$ for ${\text{ω}}_{DE}$ and $\left(-0.1\text{m}\right)\text{i}+\left(0.2\text{m}\right)\text{j}$ for ${\text{r}}_{D/E}$.

$\begin{array}{c}{\text{v}}_D=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& -10\\ -0.1& 0.2& 0\end{array}\right|\\ =\text{i}\left(0+2\right)-\text{j}\left(0-1\right)+\text{k}\left(0\right)\\ =\left(2\text{i}+\text{j}\right)\text{m/s}\end{array}$

Bar FBD:

Write the value of relative position vector of B with respect to D.

$\begin{array}{c}{\text{r}}_{B/D}=\left[\left(-300\right)\text{i}+\left(100\right)\text{j}\right]\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\\ =\left(-0.3\text{m}\right)\text{i}+\left(0.1\text{m}\right)\text{j}\end{array}$

Determine the velocity vector of B $\left({\text{v}}_B\right)$.

$\begin{array}{l}\text{Plane}\text{motion}=\text{Translation}\text{with}D+\text{Rotation}\text{about}D\\ {\text{v}}_B={\text{v}}_D+{\text{ω}}_{BD}\times {\text{r}}_{B/D}\end{array}$

Substitute $\left(2\text{i}+\text{j}\right)\text{m/s}$ for ${\text{v}}_D$, ${\omega }_{BD}\text{k}$ for ${\text{ω}}_{BD}$, and $\left(-0.3\text{m}\right)\text{i}+\left(0.1\text{m}\right)\text{j}$ for ${\text{r}}_{B/D}$.

$\begin{array}{c}{\text{v}}_B=\left(2\text{i}+\text{j}\right)+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& {\omega }_{BD}\\ -0.3& 0.1& 0\end{array}\right|\\ =\left(2\text{i}+\text{j}\right)+\text{i}\left(0-0.1{\omega }_{BD}\right)-\text{j}\left(0+0.3{\omega }_{BD}\right)+\text{k}\left(0\right)\\ =\left(-0.1{\omega }_{BD}+\text{2}\right)\text{i}+\left(1-0.3{\omega }_{BD}\right)\text{j}\end{array}$ (1)

BAR AB:

Determine the velocity vector of point B $\left({\text{v}}_B\right)$ with respect to rotation about A.

${\text{v}}_B={\text{ω}}_{AB}\times {\text{r}}_{B/A}$

Here, ${\text{r}}_{B/A}$ is the relative position vector of B with respect to A.

Substitute ${\omega }_{AB}\text{k}$ for ${\text{ω}}_{AB}$ and (0.42 m) j for ${\text{r}}_{B/A}$.

$\begin{array}{c}{\text{v}}_B={\omega }_{AB}\text{k}\times \left(0.42\text{m}\right)\text{ j}\\ =-0.42{\omega }_{AB}\text{i}\end{array}$ (2)

Equate the j components in Equation (1) and (2).

$\begin{array}{l}\left(1-0.3{\omega }_{BD}\right)=0\\ 1=0.3{\omega }_{BD}\\ {\omega }_{BD}=\frac{1}{0.3}\\ {\omega }_{BD}=3.33\text{rad/s}\left(\text{Anticlockwise}\right)\end{array}$

Therefore, the angular velocity of the rod FBD $\left({\omega }_{BD}\right)$ is $\underset{\_}{3.33\text{rad/s}\left(\text{Anticlockwise}\right)}$.

Equate the i components in Equation (1) and (2).

$\left(-0.1{\omega }_{BD}+\text{2}\right)=-0.42{\omega }_{AB}$

Substitute $3.33\text{rad/s}$ for ${\omega }_{BD}$.

$\begin{array}{l}\left(-0.1\times 3.33+\text{2}\right)=-0.42{\omega }_{AB}\\ {\omega }_{AB}=\frac{-1.667}{0.42}\\ {\omega }_{AB}=-3.97\text{rad/s}\\ {\omega }_{AB}=3.97\text{rad/s}\left(\text{clockwise}\right)\end{array}$

(b)

To determine

The velocity of point F.

Answer to Problem 15.68P

The velocity of point F is $\underset{\_}{2\text{m/s}}$ with the angle of $\underset{\_}{56.3°\left(\text{IV}-\text{Quadrant}\right)}$.

Explanation of Solution

Given information:

The constant angular velocity of the bar DE ${\omega }_{DE}$ is $10\text{rad/s}$.

The value of h is 500 mm.

Calculation:

BAR FBD:

Determine the value of relative position vector of B with respect to A using proportions.

${\text{r}}_{F/D}=\left(\frac{h+300}{300}\right){\text{r}}_{B/D}$

Substitute 500 mm for h and $\left(-0.3\text{m}\right)\text{i}+\left(0.1\text{m}\right)\text{j}$ for ${\text{r}}_{B/D}$.

$\begin{array}{c}{\text{r}}_{F/D}=\left(\frac{500+300}{300}\right)\left[\left(-0.3\text{m}\right)\text{i}+\left(0.1\text{m}\right)\text{j}\right]\\ =\left[\left(-0.8\text{m}\right)\text{i}+\left(0.267\text{m}\right)\text{j}\right]\end{array}$

Determine the velocity vector of point F.

${\text{v}}_F={\text{v}}_D+{\text{ω}}_{BD}\times {\text{r}}_{F/D}$

Substitute $\left(2\text{i}+\text{j}\right)\text{m/s}$ for ${\text{v}}_D$, $3.33\text{rad/s}$ for ${\omega }_{BD}$, and $\left[\left(-0.8\text{m}\right)\text{i}+\left(0.267\text{m}\right)\text{j}\right]$ for ${\text{r}}_{F/D}$.

$\begin{array}{c}{\text{v}}_F=2\text{i}+\text{j}+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& 3.33\\ -0.8& 0.267& 0\end{array}\right|\\ =2\text{i}+\text{j}+\text{i}\left(0-0.889\right)-\text{j}\left(0+2.664\right)+\text{k}\left(0\right)\\ =2\text{i}+\text{j}-0.889\text{i}-2.664\text{j}\\ =\left(\text{1}\text{.111i}-1.664\text{j}\right)\text{m/s}\end{array}$

Determine the magnitude of the velocity of point F.

$v_F=\sqrt{{\left(v_F\right)}_x^2+{\left(v_F\right)}_y^2}$

Substitute 1.111 m/s for ${\left(v_F\right)}_x$ and –1.664 m/s for ${\left(v_F\right)}_y$.

$\begin{array}{c}{v}_F=\sqrt{{\left(1.111\right)}^2+{\left(-1.664\right)}^2}\\ =2.0\text{m/s}\end{array}$

Determine the direction of the velocity of point F.

$\theta ={\tan }^{-1}\left(\frac{{\left(v_F\right)}_y}{{\left(v_F\right)}_x}\right)$

Substitute 1.664 m/s for ${\left(v_F\right)}_y$ and 1.111 m/s for ${\left(v_F\right)}_x$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{1.664}{1.111}\right)\\ =56.3°\left(\text{IV}-\text{Quadrant}\right)\end{array}$

Therefore, the velocity of point F is $\underset{\_}{2\text{m/s}}$ with the angle of $\underset{\_}{56.3°\left(\text{IV}-\text{Quadrant}\right)}$.