Chapter 15.3, Problem 15.93P

Two identical rods ABF and DBE are connected by a pin at B. Knowing that at the instant shown the velocity of point D is 200 mm/s upward, determine the velocity of (a) point E, (b) point F.

Fig. P15.93

To determine

Find the velocity of the point E.

Answer to Problem 15.93P

The velocity of the point E is $\underset{\_}{\text{55}\text{.7}\text{mm/s}\left(\text{Acting at angle 36}\text{.7}°\text{couterclockwise above the horizontal}\right)}$.

Explanation of Solution

Given information:

The velocity at point D is $v_D=200\text{mm/s}(\uparrow )$.

Calculation:

Show the rod ABF and DBE as shown in Figure 1.

Refer Figure 1.

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

Consider the perpendiculars to the velocity at B and D intersect at C. Then,

The instantaneous center is at C.

Consider the triangle BCD.

Apply law of sine.

$\frac{CD}{\sin 150°}=\frac{BC}{\sin 15°}=\frac{BD}{\sin 15°}$

Take $180\text{mm}$ for $BD$.

$\frac{CD}{\sin 150°}=\frac{BC}{\sin 15°}=\frac{180}{\sin 15°}$

Calculate the distances CD, BC and BD.

$\begin{array}{l}\frac{CD}{\sin 150°}=\frac{180}{\sin 15°}\\ CD=\left(\frac{180\sin 150°}{\sin 15°}\right)\\ CD=347.73\text{mm}\end{array}$

$\begin{array}{l}\frac{BC}{\sin 15°}=\frac{180}{\sin 15°}\\ BC=\left(\frac{180\sin 15°}{\sin 15°}\right)\\ BC=180\text{mm}\end{array}$

Calculate the angular velocity $\left({\omega }_{DBE}\right)$ of DBE using the relation:

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

Substitute $347.73\text{mm}$ for $CD$ and $200\text{mm/s}(\uparrow )$ for $v_D$.

$\begin{array}{c}{\omega }_{DBE}=\frac{200\text{mm/s}}{347.73\text{mm}}\\ =0.57515\text{rad/s}\left(\text{Clockwise}\right)\end{array}$

Calculate the velocity at B using the relation:

$v_B=\left(BC\right){\omega }_{DBE}$

Substitute $180\text{mm}$ for $BC$ and $0.57515\text{rad/s}\left(\text{Clockwise}\right)$ for ${\omega }_{DBE}$.

$\begin{array}{c}{v}_B=180\text{mm}\times 0.57515\text{rad/s}\left(\text{Clockwise}\right)\\ =103.527\text{mm/s}\end{array}$

Calculate the angular velocity $\left({\omega }_{ABF}\right)$ of DBE using the relation:

${\omega }_{ABF}=\frac{v_B}{AB}$

Substitute $180\text{mm}$ for $AB$ and $103.527\text{mm/s}$ for $v_B$.

$\begin{array}{c}{\omega }_{ABF}=\frac{103.527\text{mm/s}}{180\text{mm}}\\ =0.57515\text{rad/s}\left(\text{Counterclockwise}\right)\end{array}$

Calculate the velocity at F using the relation:

$v_F=\left(AF\right){\omega }_{ABF}$

Substitute $300\text{mm}$ for $AF$ and $0.57515\text{rad/s}\left(\text{Counterclockwise}\right)$ for ${\omega }_{ABF}$.

$\begin{array}{c}{v}_F=300\text{mm}\times 0.57515\text{rad/s}\\ =172.515\text{mm/s}\end{array}$ (1)

Consider the triangle DCE.

Apply law of cosine.

${\left(CE\right)}^2={\left(CD\right)}^2+{\left(DE\right)}^2-2\left(CD\right)\left(DE\right)\cos 15°$

Substitute $300\text{mm}$ for $DE$ and $347.73\text{mm}$ for $CD$.

$\begin{array}{l}{\left(CE\right)}^2={\left(347.73\right)}^2+{\left(300\right)}^2-2\left(347.73\right)\left(300\right)\cos 15°\\ CE=\sqrt{9387.3203}\\ CE=96.88\text{mm}\end{array}$

Calculate the distance EH using the relation:

$EH=DE\sin 15°$

Substitute $300\text{mm}$ for $DE$.

$\begin{array}{c}EH=\left(300\right)\sin 15°\\ =77.6457\text{mm}\end{array}$

Calculate the value of the angle $\beta$ using the relation:

$\cos \beta =\frac{EH}{CE}$

Substitute $77.6457\text{mm}$ for EH and $96.88\text{mm}$ for $CE$.

$\begin{array}{l}\cos \beta =\left(\frac{77.6457\text{mm}}{96.88\text{mm}}\right)\\ \beta ={\cos }^{-1}\left(\frac{77.6457}{96.88}\right)\\ \beta =36.7°\end{array}$

Calculate the velocity at E using the relation:

$v_E=\left(CE\right){\omega }_{DBE}$

Substitute $96.88\text{mm}$ for $CE$ and $0.57515\text{rad/s}$ for ${\omega }_{DBE}$.

$\begin{array}{c}{v}_E=96.88\text{mm}\times 0.57515\text{rad/s}\\ =\text{55}\text{.7}\text{mm/s}\left(\text{Acting at angle 36}\text{.7}°\text{couterclockwise above the horizontal}\right)\end{array}$

Thus, the velocity at point E is $\underset{\_}{\text{55}\text{.7}\text{mm/s}\left(\text{Acting at angle 36}\text{.7}°\text{couterclockwise above the horizontal}\right)}$.

To determine

Find the velocity at point F.

Answer to Problem 15.93P

The angular velocity at pint F is $\underset{\_}{172.5\text{mm/s}\left(\text{Acting 75}°\text{counterclockwise above horizontal}\right)}$.

Explanation of Solution

Given information:

Calculation:

Refer to Part (a).

Refer to Equation (1).

The velocity at point F is $172.5\text{mm/s}\left(\text{Acting 75}°\text{counterclockwise above horizontal}\right)$.

Thus, the velocity at point F is $\underset{\_}{172.5\text{mm/s}\left(\text{Acting 75}°\text{counterclockwise above horizontal}\right)}$.