Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, May 2014 — 07-Mec-A2, Kinematics and Dynamics of Machines. 3 hours, OPEN BOOK. Six questions of 20 marks each; candidates answer any FIVE. All SIX are solved in full below (Part A — mechanisms and machine dynamics, Q1–Q4; Part B — mechanical vibration, Q5–Q6).
Reference texts: R. L. Norton, Design of Machinery, 5th ed. (instant centres and mechanical advantage, cam design, gear-train kinematics, linkage balancing); S. S. Rao, Mechanical Vibrations, 6th ed. (impulsive loading, single- and two-degree-of-freedom free and forced response); J. E. Shigley & C. R. Mischke, Mechanical Engineering Design (planetary-train cross-check).
Given. A six-bar linkage: slider 2 (input, pin A) → coupler 3 (A–B) → ternary link 4 (rigid triangle B–C–D, grounded at pivot C) → coupler 5 (D–E) → output rocker 6 (E–F, grounded at pivot F). The slider translates horizontally with $v_A=10\ \text{m/s}$ to the right. Joint coordinates scaled from the 1:10 drawing (mm, actual, origin at A, $x$ right, $y$ up):
Point
x (mm)
y (mm)
Role
A
0
0
slider pin (2–3)
B
531
226
pin 3–4
C
712
1
frame pivot of link 4
D
727
190
pin 4–5
E
452
131
pin 5–6
F
497
−137
frame pivot of link 6 (output)
Derived link lengths: AB = 577, B–C = 289, C–D = 190, B–D = 199 (link 4 is the rigid triangle B–C–D), DE = 282, EF = 271 mm.
Find. (i) the 15 instant centres, and (ii) the mechanical advantage from slider input to output link 6.
Figure 1.1 — mechanism at the given position with the six pin/pivot centres and the input–output centre I₂₆ on the vertical through F.
Approach. Count the centres with $N=n(n-1)/2$, locate the obvious ones by inspection and the rest by Kennedy’s three-centres theorem; then obtain the velocities by two relative-velocity loops and the mechanical advantage from the input–output centre I₂₆.
Number of instant centres. With $n=6$ links, $$N=\tfrac{n(n-1)}{2}=\tfrac{6\cdot5}{2}=15.$$ Seven follow by inspection: the prismatic slider 1–2 gives I12 at infinity, perpendicular to the slide (vertical); the pins give I23=A, I34=B, I14=C, I45=D, I56=E, I16=F.
Remaining centres by Kennedy’s theorem. The three centres of any three links are collinear. Thus I24 lies on line I23I34 (= AB) and on line I12I14 (the vertical through C); I26 lies on line I12I16 (the vertical through F) and on line I24I46. Working through the pairs locates all eight secondary centres (table below); in particular I26 = (497, 314) mm — on the vertical through F, 451 mm above it.
Velocity loop 1 (links 3–4). The slider gives every point of link 2 the velocity $\mathbf v_A=(10,0)\ \text{m/s}$. Pin B is shared by coupler 3 (about A) and rocker 4 (about C): $$\mathbf v_A+\boldsymbol\omega_3\times\mathbf r_{B/A}=\boldsymbol\omega_4\times\mathbf r_{B/C}.$$ The two components give $\omega_3=+11.3$ and $\omega_4=-33.1\ \text{rad/s}$; then $\mathbf v_D=\boldsymbol\omega_4\times\mathbf r_{D/C}$ with $|\mathbf v_D|=6.28\ \text{m/s}$.
Velocity loop 2 (links 5–6). Pin E is shared by coupler 5 (about D) and rocker 6 (about F): $$\mathbf v_D+\boldsymbol\omega_5\times\mathbf r_{E/D}=\boldsymbol\omega_6\times\mathbf r_{E/F}.$$ Solving, $\omega_5=-5.47\ \text{rad/s}$ and $\omega_6=-22.2\ \text{rad/s}$ (CW), so the output-pin speed is $|\mathbf v_E|=|\omega_6|\,|EF|=6.02\ \text{m/s}$.
Mechanical advantage. With input force $F_{in}$ on the slider (along $v_A$) and the useful output force $F_{out}$ at pin E of rocker 6, an ideal mechanism conserves power, $F_{in}v_A=F_{out}v_E$, so $$\text{MA}=\frac{F_{out}}{F_{in}}=\frac{v_A}{v_E}=\frac{10}{6.02}.$$ The same value follows straight from I26: $\omega_6=v_A/\overline{FI_{26}}=10/0.451=22.2\ \text{rad/s}$, then $v_E=|\omega_6|\,|EF|$. MA = 1.66 — the linkage multiplies the input force about 1.7× at this configuration.
Instant-centre table (coordinates relative to A, mm):
Centre
Location
Centre
Location
I₁₂
at ∞ (⊥ slider, vertical)
I₂₃
A (pin 2–3)
I₃₄
B (pin 3–4)
I₁₄
C (pin 4–1, frame)
I₄₅
D (pin 4–5)
I₅₆
E (pin 5–6)
I₁₆
F (pin 6–1, frame)
I₁₃
(0, 886) mm
I₂₄
(712, 303) mm
I₁₅
(637, -955) mm
I₄₆
(1149, 281) mm
I₂₆
(497, 314) mm
I₃₅
(208, 285) mm
I₃₆
(330, 208) mm
I₂₅
(637, 875) mm
Quantity
Value
Number of instant centres
15 (7 by inspection, 8 by Kennedy)
Input–output centre I26
(497, 314) mm from A, on the vertical through F
$\omega_3,\ \omega_4$
+11.3, −33.1 rad/s
$\omega_5,\ \omega_6$
−5.47, −22.2 rad/s (link 6 CW)
Output-pin speed $v_E$
6.02 m/s
Mechanical advantage
MA = 1.66
Check. Q1 is a scaled graphical problem (Scale 1:10) with no printed dimensions; joint coordinates were read from the printed figure, so the angular rates and MA carry a graphical tolerance of a few percent. MA is dimensionless and hence independent of the 1:10 scale; only the IC distances and $\omega$ values depend on it.
