Mechanical advantage of a mechanical press: 22-Mec-A2 Kinematics and Dynamics of Machines, December 2019, Q2

Question 2 — Mechanical advantage of a mechanical press (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 / Find. A toggle‑type press: input force $F_{in}$ on an input link, output force $F_{out}$ delivered by a slider against the workpiece. Find the mechanical advantage $\mathrm{MA}=F_{out}/F_{in}$ at the drawn position.

Approach. Mechanical advantage is the inverse of the velocity ratio (power in = power out for an ideal, frictionless mechanism), so a velocity analysis at the shown position gives $\mathrm{MA}$ directly — this is why the hint sets $\omega_{in}=1$ rad/s and asks for a velocity polygon. The result is dimensionless, so the 1:1.5 drawing scale cancels.

Virtual‑work / power balance. For an ideal mechanism, the instantaneous input and output powers are equal: $$F_{in}\,v_{in} = F_{out}\,v_{out}\quad\Longrightarrow\quad \mathrm{MA}=\frac{F_{out}}{F_{in}}=\frac{v_{in}}{v_{out}}.$$ where $v_{in}$ is the input‑force point’s velocity along $F_{in}$ and $v_{out}$ the slider velocity along $F_{out}$.
Set the input. Take $\omega_{2}=1$ rad/s on the input crank of length $r_2$. Then the velocity of the input pin is $v_A=\omega_2 r_2$, perpendicular to the crank. Carry it through the coupler(s) with the relative‑velocity equation $\mathbf v_B=\mathbf v_A+\boldsymbol\omega_3\times\mathbf r_{B/A}$, imposing that the output pin travels along the slider guide.
Velocity polygon → velocity ratio. Constructing the polygon (or solving the two loop‑closure velocity equations) yields the slider speed $v_{out}$ for the unit input. Near the press’s toggle position the output link is almost perpendicular to its coupler, so $v_{out}$ is small and the ratio $v_{in}/v_{out}$ is large — the source of the press’s force multiplication.
Representative numbers. (Check; the following uses geometry read approximately from the 1:1.5 drawing to demonstrate the method. Re‑measure on a clean copy for a graded value.) With input crank $r_2\approx60$ mm, an input‑pin velocity $v_A=\omega_2 r_2 = (1)(0.060)=0.060$ m/s, and a measured slider velocity $v_{out}\approx0.011$ m/s at the shown near‑toggle position, $$\mathrm{MA}=\frac{v_{in}}{v_{out}}\approx\frac{0.060}{0.011}\approx \boxed{5.5}.$$ At the exact toggle (coupler and output link collinear) $v_{out}\to0$ and $\mathrm{MA}\to\infty$ — the ideal toggle press.

Quantity	Result
Governing law	$\mathrm{MA}=F_{out}/F_{in}=v_{in}/v_{out}$ (ideal)
Input (assumed)	$\omega_2=1$ rad/s, $v_A=\omega_2 r_2$
Mechanical advantage (representative, flagged)	$\approx 5.5$ at the shown position; $\to\infty$ at toggle

Topic: Mechanical advantage of a linkage via the velocity ratio; toggle mechanisms.

Key relations: ideal power balance $F_{in}v_{in}=F_{out}v_{out}$; $\mathrm{MA}=v_{in}/v_{out}=\omega_{in}/\omega_{out}$ for rotary output; relative velocity $\mathbf v_B=\mathbf v_A+\boldsymbol\omega\times\mathbf r_{B/A}$.

Why this works: With no energy storage or loss, instantaneous input power equals output power, so force ratio is exactly the inverse velocity ratio — a purely kinematic quantity obtainable from one velocity polygon, independent of the actual forces or the drawing scale.

Common pitfall: Forgetting that $\mathrm{MA}$ blows up near a toggle/dead‑point (that is the whole design intent of a press) and that it is scale‑free — the 1:1.5 factor never enters.

Source: Norton, Design of Machinery 6e, §11.11 (mechanical advantage, toggle positions).

Question 2 of 8: Mechanical advantage of a mechanical press

Question 2 — Mechanical advantage of a mechanical press (20 marks)