Question 6 of 6: Question 6 (paper Question VI) — Four-Bar Linkage Velocity and Acceleration Analysis (Part B · Dynamics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination December 2016 — 04-BS-3, Statics and Dynamics, 3 hours, closed book (one 8½″×11″ self-prepared note sheet permitted; Casio or Sharp approved calculator only). Candidates complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); all 6 are solved below.
Separately, Question 1's bracket figure (Figure 1) gives only one anchor point's coordinates (E) plus two angle pairs for the other two cables; every printed dimension (50 mm×2, 100 mm×2, 75 mm, the two 45° arcs at C, the 30°/60° pair at D) was used, but the exact 3-D position of points B and C required one reasonable reconstruction, disclosed at first use below. All other figures (2–6) reconstruct cleanly and unambiguously from the printed dimensions, confirmed by internal checks that return clean values (member lengths BD = 5 ft and DE = 10 ft in Question 6) — strong evidence the readings are correct.
Question 6 (paper Question VI) — Four-Bar Linkage Velocity and Acceleration Analysis (Part B · Dynamics, equal value)
Given. Fixed pivots A and E; crank AB ($a=3\ \text{ft}$) hangs from A, coupler BD ($b$-arm continues the same vertical member to $B$, $3\ \text{ft}$ below A), rocker DE pivoted at E; horizontal offsets $c=4\ \text{ft}$ (below B to D) and $d=8\ \text{ft}$ (D to below E); A and E are level (dashed reference line). Using this dimension chain: A(0,6), B(0,3), D(4,0), E(12,6) ft, giving $BD=5\ \text{ft}$, $DE=10\ \text{ft}$.
Figure 6 — crank AB, coupler BD and rocker DE; A and E are fixed pivots on the dashed reference line.
Approach. Compute $\mathbf{v}_B$ from the rigid rotation of AB about fixed A, then write the relative-velocity (and, for Part B, relative-acceleration) equation for D two ways — via the BD branch and via the DE branch pivoted at fixed E — and solve the resulting 2×2 linear system for the two unknown angular rates each time.
Part A — velocity of B. $\mathbf{r}_{B/A}=(0,-3)\ \text{ft}$, $\boldsymbol{\omega}_{AB}=10\,\mathbf{k}\ \text{rad/s}$:
$$\mathbf{v}_B=\boldsymbol{\omega}_{AB}\times\mathbf{r}_{B/A}=\boxed{(30,\,0)\ \text{ft/s}}$$
Vector-loop equation for D. $\mathbf{r}_{D/B}=(4,-3)$, $\mathbf{r}_{D/E}=(-8,-6)$ ft.
$$\mathbf{v}_D=\mathbf{v}_B+\boldsymbol{\omega}_{BD}\times\mathbf{r}_{D/B}=\boldsymbol{\omega}_{DE}\times\mathbf{r}_{D/E}$$
Expanding both sides into $x,y$ components gives two equations in $\omega_{BD},\omega_{DE}$; solving:
$$\omega_{DE}=\boxed{2.50\ \text{rad/s (CCW)}},\qquad \omega_{BD}=\boxed{-5.00\ \text{rad/s (i.e. 5.00 rad/s CW)}}$$
Velocity of D. $\mathbf{v}_D=\boldsymbol{\omega}_{DE}\times\mathbf{r}_{D/E}=(6\omega_{DE},-8\omega_{DE})=(15,-20)\ \text{ft/s}$:
$$|\mathbf{v}_D|=\sqrt{15^2+20^2}=\boxed{25.0\ \text{ft/s}},\quad\text{at } 53.1^{\circ}\text{ below the +x axis}$$
Part B — acceleration of B. $\boldsymbol{\alpha}_{AB}=12\,\mathbf{k}\ \text{rad/s}^2$:
$$\mathbf{a}_B=\boldsymbol{\alpha}_{AB}\times\mathbf{r}_{B/A}-\omega_{AB}^2\mathbf{r}_{B/A}=(36,0)-100(0,-3)=\boxed{(36,\,300)\ \text{ft/s}^2}$$
Vector-loop equation for the acceleration of D.
$$\mathbf{a}_D=\mathbf{a}_B+\boldsymbol{\alpha}_{BD}\times\mathbf{r}_{D/B}-\omega_{BD}^2\mathbf{r}_{D/B}=\boldsymbol{\alpha}_{DE}\times\mathbf{r}_{D/E}-\omega_{DE}^2\mathbf{r}_{D/E}$$
With $\omega_{BD}=-5.00$, $\omega_{DE}=2.50\ \text{rad/s}$ already known, expanding into $x,y$ components and solving the 2×2 system for the two unknown angular accelerations:
$$\alpha_{DE}=\boxed{-30.59\ \text{rad/s}^2},\qquad \alpha_{BD}=\boxed{-23.19\ \text{rad/s}^2}$$
Acceleration of D. $\mathbf{a}_D=\boldsymbol{\alpha}_{DE}\times\mathbf{r}_{D/E}-\omega_{DE}^2\mathbf{r}_{D/E}=(-133.6,\,282.3)\ \text{ft/s}^2$:
$$|\mathbf{a}_D|=\sqrt{133.6^2+282.3^2}=\boxed{312.3\ \text{ft/s}^2}$$
