Question 1 of 6: Mechanical advantage of an eight-bar mechanism

Question 1: Mechanical advantage of an eight-bar mechanism (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. A symmetric eight-bar force-amplifying mechanism. Input torque $T_{in}$ drives crank AB about frame pivot A; output force $F_{out}$ acts vertically-downward on the guided output block. Coupler BC pushes the vertical slider CD (constrained to the frame slideway); pin D feeds two symmetric branches (D–E–M and the redundant D–F–N), each closed by a grounded rocker (GE, HF) and a link to a pin (M, N) on the vertically-guided output block. Joint coordinates scaled from the 1:15 drawing (drawing units, $x$ right / $y$ down):

Find. The mechanical advantage $\mathrm{MA}=F_{out}/F_{in}$ at the configuration shown.

[Figure not reproduced: Figure 1.1 — Eight-bar mechanism reconstructed to scale from the exam drawing. The right branch ( H, F, N ) mirrors the left and is kinematically redundant — it merely carries the mechanism through the dead-centre (lock-in) position. Blue = moving links, grey = frame/guides. See the official exam paper.]

Approach. Because the two branches are mirror images, the input–output speed ratio is set by a single branch; the redundant branch changes the force sharing but not the kinematic ratio. Mechanical advantage follows from the principle of virtual work — for a lossless machine $T_{in}\,\omega_2 = F_{out}\,v_{out}$, so $\mathrm{MA}=F_{out}/F_{in}=v_B/v_{out}$, a pure velocity ratio that is independent of the 1:15 scale.

1. Reduce the drawing to link lengths. From the coordinates, $r_{AB}=30.9$, $L_{BC}=74.7$, $L_{CD}=44.0$, $L_{DE}=34.6$, $L_{GE}=52.8$, $L_{EM}=37.4$ (drawing units). Symmetry about $x=309.5$ is confirmed by the mirrored $G/H$, $E/F$, $M/N$ pairs, so the left branch is representative.
2. Close the upper loop (crank → coupler → vertical slider). With crank angle $\theta_2=\angle(AB)=29.1^\circ$, pin $B=A+r_{AB}(\cos\theta_2,\sin\theta_2)$. Point $C$ lies on the frame slideway $x=309.5$ with $|C-B|=L_{BC}$, giving $$C_x=309.5,\qquad C_y=B_y+\sqrt{L_{BC}^{2}-(309.5-B_x)^{2}}.$$ The slider is rigid, so $D=(309.5,\;C_y+L_{CD})$ moves purely vertically with $C$.
3. Close the lower loop (branch link + grounded rocker). Pin $E$ is the intersection of the circle of radius $L_{DE}$ about $D$ and the circle of radius $L_{GE}$ about the frame pivot $G$; the branch nearest the drawn point is taken. The output pin $M$ then satisfies $x_M=270$ (it rides on the vertically-guided block) and $|M-E|=L_{EM}$, fixing the block elevation $y_M$. Evaluated at the shown position this reproduces $E=(280,394)$ and $M_y=430$ to within $0.1$ unit, matching the drawn figure.
4. Differentiate the closed chain for the velocity ratio. Holding $\omega_2$ constant and perturbing $\theta_2$, the block velocity per unit crank rotation is $\mathrm{d}y_M/\mathrm{d}\theta_2 = 12.66$ units/rad, while the crank pin speed per unit rotation is $r_{AB}=30.9$ units/rad. Hence $$\boxed{\;\mathrm{MA}=\dfrac{v_B}{v_{out}}=\dfrac{r_{AB}}{\mathrm{d}y_M/\mathrm{d}\theta_2}=\dfrac{30.9}{12.66}\approx 2.44\;}$$
5. Interpret with the scale (optional dimensional form). Restoring the 1:15 factor to the output stroke ($1\ \text{unit}=\tfrac{1}{72}\ \text{in}\times15$), $\mathrm{d}y_M/\mathrm{d}\theta_2 = 0.0670\ \text{m/rad}$, so the torque-to-force ratio is $F_{out}/T_{in}=\omega_2/v_{out}=1/0.0670\approx 14.9\ \text{m}^{-1}$ — i.e. $F_{out}=14.9\,T_{in}$ (N per N·m).