Chapter 15.3, Problem 15.92P

(a)

To determine

Find the angular velocity of the member ABD.

Answer to Problem 15.92P

The angular velocity of the member ABD is $\underset{\_}{6.08\text{rad/s}\left(\text{Counterclockwise}\right)}$.

Explanation of Solution

Given information:

The angular velocity of the arm DE is ${\omega }_{DE}=3\text{rad/s}\left(\text{Clockwise}\right)$.

Calculation:

Show the member ABD as shown in Figure 1.

Refer to Figure 1.

Calculate the velocity at D as follows:

$v_D=\left(DE\right){\omega }_{DE}$

Substitute $160\text{mm}$ for DE and $3\text{rad/s}\left(\text{Clockwise}\right)$ for ${\omega }_{DE}$.

$\begin{array}{c}{v}_D=160\text{mm}\times 3\text{rad/s}\left(\text{Clockwise}\right)\\ =480\text{mm/s}\end{array}$

Consider the velocity of the point B as follows:

${\text{v}}_B=v_B(\downarrow )$

Consider the velocity at point B and D are denoted by ${\text{v}}_B$ and ${\text{v}}_D$.

The perpendiculars to ${\text{v}}_B$ and ${\text{v}}_D$ are denoted by BC and DC.

The perpendicular BC and DC intersect at C. Then,

The point C is the instantaneous centre of ABD.

Refer to Figure 1.

Calculate the distances BD and DK as follows:

$BD=120\text{mm}$

$\begin{array}{c}DK=BD\cos 30°\\ =120\cos \left(30°\right)\\ =103.923\text{mm}\end{array}$

Calculate the value of the angle $\beta$ as follows:

$\cos \beta =\frac{DK}{ED}$

Substitute $103.923\text{mm}$ for DK and 160 mm for BD.

$\begin{array}{l}\cos \beta =\frac{103.923\text{mm}}{160\text{mm}}\\ \beta ={\cos }^{-1}\left(\frac{103.923}{160}\right)\\ \beta =49.495°\end{array}$

Calculate the value of the angle $\varphi$ as follows:

$\begin{array}{c}\varphi =180°-30°-\beta \\ =180°-30°-49.495°\\ =100.505°\end{array}$

Consider the triangle BCD.

Apply law of sine.

$\begin{array}{l}\frac{CD}{\sin 30°}=\frac{BC}{\sin \varphi }=\frac{BD}{\sin \beta }\\ \frac{CD}{\sin 30°}=\frac{BC}{\sin \left(100.505°\right)}=\frac{120\text{mm}}{\sin \left(49.495°\right)}\end{array}$

Calculate the distance BC and CD as follows:

$\begin{array}{c}\frac{CD}{\sin 30°}=\frac{120\text{mm}}{\sin \left(49.495°\right)}\\ CD=120\text{mm}\times \frac{\sin \left(30°\right)}{\sin \left(49.495°\right)}\\ CD=78.911\text{mm}\end{array}$

$\begin{array}{l}BC=120\text{mm}\times \frac{\sin \left(100.505°\right)}{\sin \left(49.495°\right)}\\ BC=155.177\text{mm}\end{array}$

Consider the triangle ABC.

Apply law of cosine.

${\left(AC\right)}^2={\left(BC\right)}^2+{\left(AB\right)}^2-2\left(AB\right)\left(BC\right)\cos \left(150°\right)$

Substitute $155.177\text{mm}$ for $BC$ and $200\text{mm}$ for $AB$.

$\begin{array}{l}{\left(AC\right)}^2={\left(155.177\right)}^2+{\left(200\right)}^2-2\left(200\right)\left(155.177\right)\cos \left(150°\right)\\ A{C}^2=117,834.791\text{mm}\\ AC=\sqrt{117,834.791}\text{mm}\\ AC=343.27\text{mm}\end{array}$

Apply law of sine.

$\begin{array}{l}\frac{\sin \gamma }{AB}=\frac{\sin 150°}{AC}\\ \sin \gamma =\sin \left(150°\right)\times \left(\frac{AB}{AC}\right)\end{array}$

Substitute $343.27\text{mm}$ for $AC$ and $200\text{mm}$ for AB.

$\begin{array}{l}\sin \gamma =\sin \left(150°\right)\times \left(\frac{200\text{mm}}{343.27\text{mm}}\right)\\ \sin \gamma =0.2913158\\ \gamma ={\sin }^{-1}\left(0.2913158\right)\\ \gamma =16.9°\end{array}$

Calculate the angular velocity of the member ABD as follows:

${\omega }_{ABD}=\frac{v_D}{CD}$

Substitute $480\text{mm/s}$ for $v_D$ and $78.911\text{mm}$ for $CD$.

$\begin{array}{c}{\omega }_{ABD}=\frac{480\text{mm/s}}{78.911\text{mm}}\\ =6.08\text{rad/s}\left(\text{Counterclockwise}\right)\end{array}$

Thus, the angular velocity of the member ABD is $\underset{\_}{6.08\text{rad/s}\left(\text{Counterclockwise}\right)}$.

(b)

To determine

Find the velocity of A.

Answer to Problem 15.92P

The velocity at A is $\underset{\_}{\text{2}\text{.09}\text{mm/s}}$.acting at $73.1°$ counter clockwise below the horizontal.

Explanation of Solution

Given information:

Calculation:

Refer to Part (a).

Calculate the velocity of A using the relation:

$v_A=\left(AC\right)\times {\omega }_{ABD}$

Substitute $343.27\text{mm}$ for $AC$ and $6.08\text{rad/s}\left(\text{Counterclockwise}\right)$ for ${\omega }_{ABD}$.

$\begin{array}{c}{v}_A=343.27\text{mm}\times 6.08\text{rad/s}\\ =\text{2087}\text{.08}\text{mm/s}\\ =\text{2087}\text{.08}\text{mm/s}\times \frac{1\text{m}}{1,000\text{mm}}\\ =2.09\text{mm/s}\end{array}$

The velocity $v_A$ acts at angle $\left(180°-\gamma \right)$ counter-clockwise below the horizontal. Then,

$\begin{array}{c}\left(180°-\gamma \right)=\left(180°-16.9°\right)\\ =73.1°\end{array}$

Thus, the velocity at A is $\underset{\_}{\text{2}\text{.09}\text{mm/s}}$.acting at $73.1°$ counter-clockwise below the horizontal.