Question 1 of 6: Symmetric planar mechanism — redundancy, mobility, mechanical advantage

Question 1: Symmetric planar mechanism — redundancy, mobility, mechanical advantage (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 planar linkage: a horizontally-guided input slider $S$ drives, through two mirror-image branches, a horizontally-guided output block. Each branch has a ground-pinned rocker ($O_1A$ upper, $O_2B$ lower), a coupler from the slider pin ($S\!A$, $S\!B$), and a coupler to the output block ($AP_1$, $BP_2$). The two branches are geometric mirror images about the input–output centreline; no link lengths are dimensioned on the drawing.

Find. (a) the engineering purpose of duplicating the branch; (b) the mobility $M$ from Gruebler’s equation; (c) the mechanical advantage $\mathrm{MA}=F_{\text{out}}/F_{\text{in}}$ in the configuration drawn.

1(a) — Purpose of the redundant (symmetric) branch

A single branch (slider → rocker + two couplers → output block) is already a complete one-degree-of-freedom mechanism (shown below to have $M=1$). Adding the mirror-image second branch imposes the same constraint on the output a second time — it is kinematically redundant. Engineers add it deliberately for three reasons:

• Cancellation of the transverse (side) load. A single inclined coupler pushes the output block not only along its guide but also sideways; that transverse component must be reacted by the guide, causing friction, wear and cocking. The mirror branch delivers an equal and opposite transverse component, so the net side load on the output slider is zero and it translates cleanly along its axis.
• Load sharing and stiffness. The driving force splits between two symmetric load paths, halving the force per link and roughly doubling the axial stiffness — useful in a press/clamp where high output force is wanted.
• Motion certainty through the toggle. Symmetric branches guarantee the assembly passes smoothly through the stretched (toggle) position without the branch-point ambiguity a single toggle linkage suffers.

The price is that the linkage is over-constrained: Gruebler’s count (which assumes all constraints are independent) predicts one fewer freedom than the mechanism actually has.

1(b) — Mobility by Gruebler’s equation

Gruebler / Kutzbach for a planar mechanism: $M=3(n-1)-2J_1-J_2$, with $n$ links (incl. ground), $J_1$ full (1-DOF) pairs, $J_2$ half (2-DOF) pairs. Count the full two-branch mechanism:

So $J_1 = 2\,(\text{P}) + (2{+}2{+}2)\,(\text{multiple pins}) + 4\,(\text{simple R}) = 12$, and

$$M=3(9-1)-2(12)-0 = 24-24 = \boxed{0}.$$

Gruebler returns $M=0$ — apparently a locked structure. This is the classic over-constrained (Gruebler-paradox) result: because the second branch is a geometric mirror of the first, its constraints are not independent, so one of the counted constraints is passive (redundant). Removing the redundant branch and recounting one branch ($n=6$, $J_1=7$: two prismatic, pin $S$ 1R, pin $A$ 2R, $O_1$ 1R, $P_1$ 1R) gives

$$M_{\text{1 branch}} = 3(6-1)-2(7)=15-14=\boxed{1}.$$

Hence the true mobility is 1: the mechanism moves as a proper single-input linkage; Gruebler’s $0$ signals the intentional redundancy, exactly the point of part (a).

1(c) — Mechanical advantage at the position shown

For an ideal (frictionless) mechanism, power in equals power out, so with collinear horizontal input and output forces

$$F_{\text{in}}\,v_{\text{in}} = F_{\text{out}}\,v_{\text{out}} \;\Longrightarrow\; \mathrm{MA}=\frac{F_{\text{out}}}{F_{\text{in}}}=\frac{v_{\text{in}}}{v_{\text{out}}}.$$

The velocity ratio depends only on the link angles in the drawn configuration, which are set by the figure geometry regardless of overall scale. Taking joint coordinates scaled from the drawing for one branch (both give the same ratio by symmetry) and closing the three constraint loops — $A$ on the circle about $O_1$, $|SA|$ fixed, $|AP_1|$ fixed with $P_1$ on the horizontal output line — a virtual horizontal displacement $\delta$ of the slider produces $\mathrm{d}x_{P_1}=0.713\,\delta$, so

$$\mathrm{MA}=\frac{v_{\text{in}}}{v_{\text{out}}}=\frac{1}{0.713}\approx \boxed{1.40}.$$

The output moves in the same direction as the input but slower, so the force is amplified by about 40% in the drawn (part-open) position. As the couplers approach the stretched toggle line (links becoming collinear), $v_{\text{out}}\!\to\!0$ and $\mathrm{MA}\!\to\!\infty$ — the reason such symmetric toggle linkages are used in presses, clamps and rivet squeezers.