Question 2 of 7: Inverted crank-slider — velocity analysis

Question 2: Inverted crank-slider — velocity analysis (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 ($A_oA$) carries the coupler-block 3 pinned at $A$; block 3 slides in the straight slot of link 4, and link 4 is grounded by a pin at $B_o$. The printed figure draws link 4 as a cranked (Z-jogged) bar: the slot runs parallel to the short arm that reaches $B_o$, but is offset from it, so the slot axis does not pass through $B_o$. Point $B_3$ is a rigid point of coupler 3. Scaling the 1:10 drawing: crank $r_2=A_oA\approx 0.238$ m; $B_oA\approx 0.568$ m, of which $0.553$ m lies along the slot direction and $e\approx 0.129$ m perpendicular to it (so $B_oA$ makes $13.1^\circ$ with the slot); and $A\!\to\!B_3\approx 0.289$ m. Input $\omega_2=100$ rad/s ccw.

Find. (i) $\omega_4$ and the sliding (relative) velocity $v_{3/4}$; (ii) $v_{B_3}$.

[Figure not reproduced: Inverted crank-slider at the drawn position, redrawn from measured joint coordinates. Link 4 (gold) is the cranked bar: straight slot, Z-jog, short arm to $B_o$. Green = $v_A$ ($\perp$ crank), orange = $v_{B_3}$. The red dashed segment $e\approx0.129$ m is the perpendicular offset of the p. See the official exam paper.]

Approach. Get $v_A$ from the crank, then split it into components along and perpendicular to bar 4: the perpendicular part is carried by link 4 (gives $\omega_4$), the parallel part is the slide $v_{3/4}$. Finally add $\omega_3(=\omega_4)$ about $A$ to reach $B_3$.

1. Velocity of the crank pin. $A$ is a pin on crank 2, so $$v_A=\omega_2\,r_2=(100)(0.2376)=\boxed{23.8\ \text{m/s}}\quad(\perp\ A_oA)$$ With $A_o$ at the origin and $A$ up and to the right of it, the ccw crank gives $\mathbf v_A=(-22.10,\ 8.72)$ m/s.
2. The coincident-point equation. Write $A_3$ (the pin, a point of block 3) and $A_4$ (the material point of link 4 instantaneously beneath it): $$\mathbf v_{A_3}=\mathbf v_{A_4}+\mathbf v_{3/4},\qquad \mathbf v_{A_4}=\boldsymbol\omega_4\times\mathbf r_{B_o\to A}\ (\perp B_oA),\qquad \mathbf v_{3/4}\parallel\hat u\ (\text{the slot})$$ Let $\hat u$ be the unit vector along the slot and $\hat p=\hat z\times\hat u$ its normal. Because link 4 is cranked, $\hat u$ is not parallel to $B_oA$ — the two differ by $13.1^\circ$, and that is the whole difficulty of this question.
3. Angular velocity of link 4. Dot the vector equation with $\hat p$: the sliding term drops out, and since $(\hat z\times\mathbf r)\cdot\hat p=\mathbf r\cdot\hat u$, $$\omega_4=\frac{\mathbf v_A\cdot\hat p}{\mathbf r_{B_o\to A}\cdot\hat u}=\frac{12.34\ \text{m/s}}{0.5533\ \text{m}}=\boxed{22.3\ \text{rad/s (ccw)}}$$ The denominator is the component of $B_oA$ measured along the slot, $0.568\cos 13.1^\circ=0.553$ m — it reduces to $\overline{B_oA}$ only when the slot axis passes through the pivot.
4. Relative sliding velocity. Link 4 carries $A_4$ at $\omega_4\,\overline{B_oA}=(22.30)(0.568)=12.67$ m/s perpendicular to $B_oA$, so $$\mathbf v_{3/4}=\mathbf v_A-\mathbf v_{A_4}=(-14.01,\ 18.47)\ \text{m/s}\ \Rightarrow\ \boxed{v_{3/4}=23.2\ \text{m/s}}$$ directed along the slot, outward — block 3 runs away from the jogged end of bar 4. Its direction comes out as exactly $-\hat u$, which is the check that the two-direction resolution was done correctly.
5. Velocity of $B_3$ (velocity image). A block cannot turn relative to its slot, so $\omega_3=\omega_4=22.30$ rad/s. With $\mathbf r_{A\to B_3}=(0.2477,\ 0.1496)$ m, $$\mathbf v_{B_3}=\mathbf v_A+\boldsymbol\omega_3\times \mathbf r_{A\to B_3}=(-25.44,\,14.24)\ \text{m/s}$$ $$|v_{B_3}|=\sqrt{25.44^2+14.24^2}=\boxed{29.2\ \text{m/s}}\quad(\approx 151^\circ\ \text{from}\ +x)$$