Question 1 of 7: Number synthesis, mobility, mechanical advantage and kinematics

Question 1: Number synthesis, mobility, mechanical advantage and kinematics (40 marks — four 10-mark parts)

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.

1(i) — Number synthesis of a 7-bar with $J_1=8$ R, $J_2=1$ half joint, $M=1$

Given / Find. A planar chain of $n=7$ links using eight revolute pairs and one pin-in-a-slot half joint, with mobility one. Verify feasibility by Gruebler and the link-incidence constraint, enumerate the admissible link-type distributions, and sketch two valid configurations.

Approach. Check Gruebler for the target joint mix, then apply the incidence (constraint) equation that ties the number of binary/ternary/quaternary links to the joints, enumerate integer solutions, and realise two of them as physical skeletons.

1. Gruebler check. For a planar mechanism $M=3(n-1)-2J_1-J_2$. With $n=7,\ J_1=8,\ J_2=1$: $$M=3(7-1)-2(8)-1=18-16-1=\boxed{1}.\ \checkmark$$
2. Incidence (constraint) equation. Every revolute joins two links and the pin-in-slot half joint also joins two links, so the total number of link–joint incidences is $2(J_1+J_2)=2(9)=18$. Writing $n_2,n_3,n_4$ for the numbers of binary, ternary and quaternary links, $$n_2+n_3+n_4=7,\qquad 2n_2+3n_3+4n_4=18.$$
3. Enumerate the valid link sets. The integer solutions are $$(n_2,n_3,n_4)=(3,4,0),\quad(4,2,1),\quad(5,0,2).$$ Each is a legitimate answer to the number synthesis; the first two are the most practical and are sketched below.
4. Realise two configurations. Configuration A uses four ternary links; Configuration B replaces two ternaries by one quaternary link. Both keep eight revolute pins plus one pin-in-slot half joint (drawn as a roller in a slot), so both satisfy $M=1$.

1(ii) — Mobility of the shown mechanism (Gruebler paradox)

Given. A symmetric planar mechanism: crank $A$-$B$ grounded at $A$; coupler $B$-$C$; a vertical slider $C$-$D$ running in a fixed slot; from the junction $D$ two links reach $E$ and $F$; grounded rockers $E$-$G$ and $F$-$H$; and connectors $E$-$M$, $F$-$N$ to a horizontal slider carrying pins $M$ and $N$. Find. the link and joint counts and $M$.

Approach. Number every rigid body, classify each joint (a pin joining $k$ links counts as $k-1$ revolute pairs; each slider is one prismatic pair), apply Gruebler, then interpret the result physically using the mechanism’s mirror symmetry.

1. Count the links ($n=11$). 1 ground; 2 crank $AB$; 3 coupler $BC$; 4 vertical slider $CD$; 5 link $DE$; 6 link $DF$; 7 rocker $EG$; 8 connector $EM$; 9 rocker $FH$; 10 connector $FN$; 11 horizontal slider (pins $M,N$).
2. Count and classify the joints. Revolutes: $A,B,C$ (one each); the triple pins at $D,E,F$ (three links each) give two revolute pairs apiece; and $G,H,M,N$ (one each) — total $3+2+2+2+4=13$ R. Prismatics: the vertical slider 4 in its ground slot and the horizontal slider 11 — $2$ P. So $J_1=15$, $J_2=0$.
3. Apply Gruebler. $$M=3(n-1)-2J_1-J_2=3(11-1)-2(15)-0=30-30=\boxed{0}.$$
4. Decision making — interpret the zero. A raw count of $M=0$ labels the assembly a structure, yet the figure is clearly a moving, symmetric mechanism. The mirror-image right branch ($D$-$F$-$H$, $F$-$N$) imposes exactly the same constraint on the horizontal slider that the left branch ($D$-$E$-$G$, $E$-$M$) already supplies. It is therefore a redundant (idle) constraint. Deleting that one redundant branch (links 6, 9, 10 and their five revolute pairs) gives $$M=3(8-1)-2(10)-0=21-20=\boxed{1}.$$ So the effective mobility is $M=1$ — the crank at $A$ is the single input. The $M=0$ from the plain formula is the classic Gruebler paradox that flags a special (here, mirror-symmetric) geometry.

1(iii) — Inverted crank-slider: mechanical advantage, output range, transmission angle

Given. Ground pivots $A$ (crank centre) and $C$ (output-link centre); slider 3 pinned to crank 2 at $B$ and sliding on output link 4. Scaled from the 1:10 drawing (coordinates in drawing units): $A(265,388),\,B(200,138),\,C(885,438)$, giving crank $r_2\approx258$, ground $AC\approx622$, slider distance $CB\approx748$ (drawing units).

Find. (a) $\mathrm{MA}=T_{out}/T_{in}$ at the shown position; (b) the range of motion of output link 4; (c) the range of the transmission angle.

Approach. Since angles and length ratios are scale-invariant, the 1:10 scale cancels. Use the loop-closure velocity ratio for the inverted slider-crank to get $\mathrm{MA}=\omega_2/\omega_4$; the output swing and the transmission-angle extremes follow from the crank/ground geometry.

1. Velocity ratio and mechanical advantage. With $\phi=\theta_2-\theta_4$ the angle between crank 2 and link 4, loop closure gives $\omega_4=\dfrac{r_2\cos\phi}{|CB|}\,\omega_2$, so $$\mathrm{MA}=\frac{T_{out}}{T_{in}}=\frac{\omega_2}{\omega_4}=\frac{|CB|}{r_2\cos\phi}.$$ Substituting $|CB|=748,\ r_2=258,\ \phi=52^\circ$: $$\mathrm{MA}=\frac{748}{258\cos 52^\circ}=\frac{748}{258(0.619)}=\boxed{4.68}.$$
2. Range of motion of the output link (b). As the crank makes a full turn, $B$ traces a circle of radius $r_2$ about $A$; link 4 (through fixed $C$) reaches its angular limits when $CB$ is tangent to that circle, i.e. $AB\perp CB$. The half-swing is $\arcsin(r_2/d)$, so $$\Delta\theta_4=2\arcsin\!\frac{r_2}{d}=2\arcsin\!\frac{258}{622}=\boxed{49^\circ}.$$
3. Range of the transmission angle (c). The transmission angle $\mu$ (between crank 2 and output link 4) is best at the two output extremes, where $AB\perp CB$ gives $\mu=90^\circ$, and degenerates at the two positions where the crank aligns with the ground line $AC$ (dead/toggle points), where $\mu=0^\circ$. Hence $$\boxed{\mu\in[\,0^\circ,\ 90^\circ\,]}\qquad(\mu\approx52^\circ\text{ at the position shown}).$$ The crank fully rotates because $r_2<d$; the two $\mu=0^\circ$ toggles are the singular positions of this inversion.

1(iv) — Translating input driving an oscillating follower

Given. Input block 2 slides horizontally on ground 1 at constant $\dot s=10$ m/s; pin 3 links it to follower 4, pivoted a vertical distance $H=200$ mm below the guide line. Find the follower state when it has rotated $\theta=30^\circ$ CW from vertical.

Find. the angular velocity $\dot\theta$ and angular acceleration $\ddot\theta$ of follower 4.

Approach. The driving pin $P$ (on block 2 and slider 3) has a fixed height $H$ above the pivot and a horizontal position $x=H\tan\theta$; differentiate the constraint $x=H\tan\theta$ twice, using $\dot x=\dot s$ (constant) and $\ddot x=0$.

1. Position constraint. With the pivot below the guide by $H$ and the follower at angle $\theta$ from vertical, the pin lies at horizontal offset $$x=H\tan\theta.$$
2. Angular velocity. Differentiate: $\dot x=H\sec^2\theta\,\dot\theta=\dot s$, so $$\dot\theta=\frac{\dot s\cos^2\theta}{H}=\frac{10\,(\cos30^\circ)^2}{0.200}=\frac{10(0.75)}{0.200}=\boxed{37.5\ \text{rad/s (CW)}}.$$
3. Angular acceleration. Since $\ddot x=0$, differentiating $\dot x=H\sec^2\theta\,\dot\theta$ gives $\ddot\theta=-2\tan\theta\,\dot\theta^2$: $$\ddot\theta=-2\tan30^\circ\,(37.5)^2=-2(0.5774)(1406)=\boxed{-1624\ \text{rad/s}^2}.$$ The negative sign means the follower is decelerating (angular acceleration is CCW, opposing the CW rotation), as expected when a constant-speed input drives an oscillating link toward its extreme.