Question 2 of 7: Velocity analysis of a six-bar mechanism

Question 2: Velocity analysis of a six-bar mechanism (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Crank 2 pinned to ground at $B$, rotating CCW at $\omega_2=30\ \text{rad/s}$, driving pin $D$. Ternary link 3 is the shaded triangle $C\,D\,E$; its corner $C$ is a block (slider 4) that slides vertically in the ground guide. Link 5 connects $E$ to pin $F$ on block 6, which slides along an inclined ground guide; $G$ is a point fixed on link 6. Dimensions scaled from the drawing (1:5): $BD\approx214\ \text{mm}$, triangle $CD\approx242,\ DE\approx151,\ CE\approx387\ \text{mm}$, $EF\approx292\ \text{mm}$.

Find. $v_6$ (linear velocity of slider 6), $\omega_5$, and $\mathbf v_G$.

Approach. Work through the loop with the relative-velocity equation $\mathbf v_Q=\mathbf v_P+\boldsymbol\omega\times\mathbf r_{Q/P}$: get $\mathbf v_D$ from the crank; use the vertical-slider constraint at $C$ to find $\omega_3$; carry $\omega_3$ to $E$; then close link 5 with the inclined-slider constraint at $F$ to get $\omega_5$ and $\mathbf v_F$.

1. Velocity of the crank pin $D$. With $\omega_2=30\ \text{rad/s}$ (CCW, $+\hat z$) and $r_{BD}=0.214\ \text{m}$, $$v_D=\omega_2\,r_{BD}=30(0.214)=6.42\ \text{m/s},\quad \perp BD.$$
2. Angular velocity of the ternary link 3. Point $C$ (slider 4) may move only vertically, so the horizontal component of $\mathbf v_C=\mathbf v_D+\boldsymbol\omega_3\times\mathbf r_{C/D}$ must vanish. Solving that scalar condition, $$\boxed{\omega_3=23.5\ \text{rad/s (CW)}},\qquad v_C=7.89\ \text{m/s (vertical, upward)}.$$ Thus the slider 4 velocity is $v_C=7.89\ \text{m/s}$.
3. Velocity of $E$ on link 3. Using the same $\omega_3$, $$\mathbf v_E=\mathbf v_D+\boldsymbol\omega_3\times\mathbf r_{E/D}\;\Rightarrow\; v_E=8.71\ \text{m/s}.$$
4. Close link 5 with the inclined-slider constraint. $\mathbf v_F=\mathbf v_E+\boldsymbol\omega_5\times\mathbf r_{F/E}$, and $\mathbf v_F$ must lie along the inclined guide (direction $F\!\to\!G$). The two scalar equations give $$\boxed{\omega_5=32.6\ \text{rad/s (CW)}},\qquad \boxed{v_6=|\mathbf v_F|=3.87\ \text{m/s}}\ \text{along the guide}.$$
5. Absolute velocity of $G$. Slider 6 forms a prismatic pair with the (straight) ground guide, so $\omega_6=0$ and every point of link 6 has the same velocity: $$\boxed{\mathbf v_G=\mathbf v_F,\quad v_G=3.87\ \text{m/s}}\ \text{(along the inclined guide).}$$