Question 5 of 10

Question 3 (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 $A_0A$ (link 2) is pinned to ground at $A_0$ and rotates at $\omega_2=150\,\text{rad/s}$ (ccw, constant — $\alpha_2=0$). Pin $A$ carries the coupler/slider block (link 3), which straddles rocker link 4 (pinned to ground at $B_0$) and slides along it. Point $B$ is rigidly fixed to link 3, offset horizontally from $A$. Digitised at "Scale 1:10": $A_0{=}(241,331)$, $A{=}(350,249)$, $B_0{=}(390,371)$, $B{=}(417,244)$ (source units, $y$ measured downward); converting 1 PDF point at this scale to real length ($1\,\text{pt}=0.353\,\text{mm}$ drawn $\times\,10=3.528\,\text{mm}$ real) gives $A_0A=481\,\text{mm}$, current $B_0A=453\,\text{mm}$, $AB=237\,\text{mm}$.

Find. $\omega_4$, the sliding velocity of link 3 relative to link 4, $\vec V_B$, $\vec a^c_{A3/A4}$, and $\vec a^s_{A3/A4}$.

Approach. Because $A$ is a real pin (shared by link 2 and the sliding block link 3) that also rides in the slot of link 4, resolve $\vec V_A$ into a coincident-point-on-link-4 term plus a slip term along $B_0A$; link 3 does not rotate relative to link 4 (it only slides), so $\omega_3=\omega_4$ and $\vec V_B$ follows by the rigid-body relation from $A$. Differentiate the same loop for accelerations, picking out the Coriolis and slip terms.

1. Crank-tip velocity. $\vec r_{A_0A}=A-A_0=(109,82)$ mm (scaled, std frame), $|\vec r_{A_0A}|=481.2\,\text{mm}$: $$\vec V_A=\vec\omega_2\times\vec r_{A_0A}=150\,(-82,109)\ \text{mm/s} = (-12{,}300,\,16{,}350)\,\text{mm/s},\qquad |\vec V_A| = 20{,}460\ \text{mm/s} = 20.46\ \text{m/s}$$
2. Set up the pin-in-slot velocity equation at $A$. With $\vec r_{B_0A}=A-B_0=(-40,122)$ mm ($|\vec r_{B_0A}|=452.9$ mm) and $\hat s=\vec r_{B_0A}/|\vec r_{B_0A}|=(-0.312,0.950)$ (the current slot axis): $$\vec V_A = \omega_4\,\vec k\times\vec r_{B_0A} \;+\; V_{s}\,\hat s$$ two scalar equations in the two unknowns $\omega_4,V_s$.
3. Solve for $\omega_4$ and the slip speed. $$\boxed{\omega_4 = 51.36\ \text{rad/s (ccw)}}\qquad\qquad \boxed{V_{s}\,(A3/A4,\ \text{along link 4}) = 68.33\ \text{m/s}}$$
4. Velocity of $B$ (rigid with link 3, $\omega_3=\omega_4$). $\vec r_{AB}=B-A=(-8,\,18)$ mm (nearly horizontal, matching the drawing): $$\vec V_B = \vec V_A + \omega_4\,\vec k\times\vec r_{AB} = (-44.30,\,69.82)\ \text{m/s},\qquad \boxed{|\vec V_B| = 82.69\ \text{m/s}}$$
5. Crank-tip acceleration (centripetal only, $\alpha_2=0$): $$\vec a_A = -\omega_2^2\,\vec r_{A_0A} = (-8651.9,\,-6508.8)\ \text{m/s}^2,\qquad |\vec a_A| = 10{,}826.8\ \text{m/s}^2$$
6. Coriolis term (link 4 rotates at $\omega_4$ while link 3 slips at $V_s$ along the slot): $$\vec a^c_{A3/A4} = 2\,\omega_4\,\vec k\times (V_s\hat s) = (-6669.1,\,-2186.6)\ \text{m/s}^2,\qquad \boxed{|\vec a^c_{A3/A4}| = 7018.4\ \text{m/s}^2}$$
7. Close the acceleration loop for $\alpha_4$ and the relative slip acceleration. $$\vec a_A = \big(-\omega_4^2\vec r_{B_0A}+\alpha_4\vec k\times\vec r_{B_0A}\big) + \vec a^c_{A3/A4} + a^s\,\hat s$$ two scalar equations, two unknowns ($\alpha_4, a^s$): $$\alpha_4 = 7132.8\ \text{rad/s}^2\ (\text{ccw}),\qquad \boxed{a^s_{A3/A4} = -2294.6\ \text{m/s}^2\ (\text{i.e. } 2294.6\ \text{m/s}^2 \text{ directed } B_0\to A)}$$