Chapter 15.3, Problem 15.84P

Rod BDE is partially guided by a roller at D that moves in a vertical track. Knowing that at the instant shown the angular velocity of crank AB is 5 rad/s clockwise and that $\beta =25°$ , determine (a) the angular velocity of the rod, (b) the velocity of point E.

To determine

(a)

The angular velocity of rod $BDE$.

Answer to Problem 15.84P

The angular velocity of rod $BDE$ is $2.8395\text{rad}/\text{s}$.

Explanation of Solution

Given information:

The angular velocity of crank $AB$ is $5\text{rad}/\text{s}$ angle $\beta$ is $25°$

Figure-(1) represents the geometry of mechanism.

Figure-(1)

Figure-(2) represents the rod $BDE$.

Figure-(2)

Figure-(2) instantaneous centre of rotation of rod is $C$, it is the intersection point of line drawn perpendicular to velocity at point $B$ and velocity at point $D$.

Write the expression for velocity at point $B$.

$v_B={\omega }_{AB}{l}_{AB}$ ...... (I)

Here, angular velocity of crank $AB$ is ${\omega }_{AB}$ and length of rod $AB$ is $l_{AB}$.

Write the relation for triangle $BDC$.

$\sin \beta =\frac{l_{BC}}{l_{BD}}$ ...... (II)

Here, distance of point $B$ from $C$ is $l_{BC}$ distance of point $D$ from $B$ is $l_{BD}$.

Write the expression for angular velocity of rod $BDE$.

${\omega }_{BDE}=\frac{v_B}{l_{BC}}$ ...... (III)

Calculation:

Substitute $5\text{rad}/\text{s}$ for ${\omega }_{AB}$ and $120\text{mm}$ for $l_{AB}$ in Equation (I).

$\begin{array}{c}{v}_B=\left(5\text{rad}/\text{s}\right)\left(120\text{mm}\right)\\ =600\text{mm}/\text{s}\end{array}$

Substitute $500\text{mm}$ for $l_{BD}$, $25°$ for $\beta$ in Equation (II).

$\begin{array}{l}\sin 25°=\frac{l_{BC}}{500\text{mm}}\\ l_{BC}=500\text{mm}\times \sin 25°\\ l_{BC}=211.3\text{mm}\end{array}$

Substitute $211.3\text{mm}$ for $l_{BC}$ and $600\text{mm}/\text{s}$ for $v_B$ in Equation (III).

$\begin{array}{c}{\omega }_{BDE}=\frac{600\text{mm}/\text{s}}{211.3\text{mm}}\\ =2.8395\text{rad}/\text{s}\end{array}$

Conclusion:

The angular velocity of rod $BDE$ is $2.8395\text{rad}/\text{s}$.

To determine

(b)

The velocity of point $D$.

Answer to Problem 15.84P

The velocity of point $D$ is $1.1817\text{m}/\text{s}$ at an angle of $82.42°$

Explanation of Solution

Given information:

Write the expression for velocity at point $E$.

$v_E={\omega }_{BDE}{l}_{CE}$ ...... (IV)

Here, the distance between point $E$ and instantaneous center of rod is $l_{CE}$.

Write the expression for $l_{CE}$.

$l_{CE}=\sqrt{{\left(CF\right)}^2+{\left(FE\right)}^2}$ ...... (V)

Here, distance of point $F$ from $C$ is $CF$ and distance between point $E$ and $F$ is $FE$.

Write the relation for triangle $DFE$

$\sin \beta =\frac{FE}{DE}$ ...... (VI)

Here, distance between $D$ and $E$ is $DE$.

Write the relation for triangle $DFE$

$\tan \beta =\frac{FE}{DF}$ ...... (VII)

Here, distance between $D$ and $F$ is $DF$.

Write the relation for triangle $BDC$

$\cos \beta =\frac{CD}{BD}$ ...... (VIII)

Here, distance between $D$ and $C$ is $CD$ and distance between $D$ and $B$ is $BD$.

Write the expression for $CF$.

$CF=CD+DF$ ...... (IX)

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

$\gamma =90°-{\tan }^{-1}\left(\frac{FE}{CF}\right)$ ...... (X)

Calculation:

Substitute $25°$ for $\beta$ and $200\text{mm}$ for $DE$ in Equation (VI).

$\begin{array}{l}\sin 25°=\frac{FE}{200\text{mm}}\\ FE=200\text{mm}\times \sin 25°\\ FE=84.5236\text{mm}\end{array}$

Substitute $25°$ for $\beta$ and $84.523\text{mm}$ for $FE$ in Equation (VII).

$\begin{array}{l}\tan 25°=\frac{84.523\text{mm}}{DF}\\ DF=\frac{84.523\text{mm}}{\tan 25°}\\ DF=181.261\text{mm}\end{array}$

Substitute $25°$ for $\beta$ and $500\text{mm}$ for $BD$ in Equation (VIII).

$\begin{array}{l}\cos 25°=\frac{CD}{500\text{mm}}\\ CD=500\text{mm}\times \cos 25°\\ CD=453.15\text{mm}\end{array}$

Substitute $453.15\text{mm}$ for $CD$ and $181.261\text{mm}$ for $DF$ in Equation (IX).

$\begin{array}{l}CF=453.15\text{mm}+181.261\text{mm}\\ CF=634.411\text{mm}\end{array}$

Substitute $84.5236\text{mm}$ for $FE$ and $634.411\text{mm}$ for $CF$ in Equation (V).

$\begin{array}{c}{l}_{CE}=\sqrt{{\left(634.411\text{mm}\right)}^2+{\left(84.5236\text{mm}\right)}^2}\\ =\sqrt{402477.3169{\text{mm}}^2+7144.2389{\text{mm}}^2}\\ =640.016\text{mm}\end{array}$

Substitute $640.016\text{mm}$ for $l_{CE}$ and $2.8395\text{rad}/\text{s}$ for ${\omega }_{BDE}$ in Equation (IV).

$\begin{array}{c}{v}_E=\left(2.8395\text{rad}/\text{s}\right)\times \left(640.016\text{mm}\right)\\ =1817.3254\text{mm}/\text{s}\times \frac{1\text{m}/\text{s}}{1000\text{mm}/\text{s}}\\ =1.817\text{m}/\text{s}\end{array}$

Substitute $84.5236\text{mm}$ for $FE$ and $634.411\text{mm}$ for $CF$ in Equation (X).

$\begin{array}{c}\gamma =90°-{\tan }^{-1}\left(\frac{84.5236\text{mm}}{634.411\text{mm}}\right)\\ =90°-{\tan }^{-1}\left(0.1332316\right)\\ =90°-7.5889°\\ =82.42°\end{array}$

Conclusion:

The velocity of point $D$ is $1.1817\text{m}/\text{s}$ at an angle of $82.42°$