NivaarExam PrepOfficial exam papers ↗

22-Mec-A2 Kinematics and Dynamics of Machines · May 2013

Question 1 of 6: Toggle Press — Mechanical Advantage

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper: National Examinations — 07-Mec-A2 Kinematics and Dynamics of Machines, May 2013. Open-book, 3-hour paper; answer FIVE of six questions, all of equal value (20 marks). Part A (mechanisms & machine dynamics, Q1–Q4) and Part B (mechanical vibration, Q5–Q6). All six questions are solved here.

Reference texts: R. L. Norton, Design of Machinery (6th ed., McGraw-Hill) — velocity analysis, cams, gear trains, engine dynamics; C. E. Wilson & J. P. Sadler, Kinematics and Dynamics of Machinery (3rd ed.); S. S. Rao, Mechanical Vibrations (6th ed., Pearson) — Part B; J. E. Shigley, Mechanical Engineering Design — gear geometry.

Note on the figure-based questions. Q1 (press) is a scaled graphical problem (“Scale 1:15”); its link geometry is read from the printed diagram, so the mechanical advantage carries a graphical tolerance of roughly ±5 %. Q3 (planetary box) is solved from the kinematic topology inferred from the sectioned schematic. Both interpretations are stated explicitly in a check callout so the reader can re-scale from the original drawing.

Question 1: Toggle Press — 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 six-bar linkage press: input handle (link 2) is a bell-crank pinned to frame at O₂, with the 500 N handle force Fin applied vertically at P. Link 3 (coupler A–B) drives pin B, which is also carried by rocker 4 (grounded at ceiling pivot D). Link 5 (B–C) drives the press head (slider 6) vertically. Joint coordinates scaled from the drawing (points, y downward):

PointxyRole
O₂431303frame pivot of link 2
A416274joint 2–3
P230273handle tip (Fin)
B482.5322.5joint 3–4–5
D467.5268frame pivot of rocker 4
C465.6395joint 5–6 (slider)

Find. The output force Fout at the press head for Fin = 500 N.

F_in = 500 NF_outO2 (frame 1)23456ABD
Fig. 1 — six-bar toggle press. Frame pivots O₂ and D (hatched); slider 6 constrained to the vertical guide. The 1:15 scale cancels in the velocity ratio, so only the link geometry matters.

Approach. Give the input link a unit angular velocity, propagate velocities through the loop by the relative-velocity equation to get the slider speed, then apply the principle of virtual work (a lossless machine conserves power) so that the force ratio equals the inverse velocity ratio.

  1. Velocity of A on the input link. With $\omega_2 = 1\ \text{rad/s}$ about O₂, $\mathbf v_A = \boldsymbol\omega_2 \times \mathbf r_{O_2 A}$. Using $\mathbf r_{O_2A}=(-15,\,29)$ (y up), $\mathbf v_A = (-29,\,-15)$ (drawing units/s).
  2. Velocity of B (coupler 3 and rocker 4). B is rigid on link 3 and also swings on rocker 4 about D: $$\mathbf v_B = \mathbf v_A + \boldsymbol\omega_3\times\mathbf r_{AB} = \boldsymbol\omega_4\times\mathbf r_{DB}.$$ Solving the two scalar equations gives $\omega_3 = 0.132$ and $\omega_4 = -0.415\ \text{rad/s}$, hence $\mathbf v_B = (-22.6,\,-6.22)$.
  3. Velocity of the slider C. C is rigid on link 5 and constrained to move vertically, so its horizontal component vanishes: $$\mathbf v_C = \mathbf v_B + \boldsymbol\omega_5\times\mathbf r_{BC},\qquad v_{Cx}=0 \Rightarrow \omega_5 = 0.312\ \text{rad/s}.$$ This yields $v_C = 11.49$ (downward) per rad/s of input.
  4. Input velocity ratio. $F_\text{in}$ is vertical, so its effective moment arm about O₂ is the horizontal offset of P, $d_h=|x_P-x_{O_2}| = 201$. The work-input rate is $F_\text{in}\,d_h\,\omega_2$.
  5. Mechanical advantage by virtual work. For a lossless linkage the input and output powers are equal, $F_\text{in}\,d_h\,\omega_2 = F_\text{out}\,v_C$: $$\text{MA}=\frac{F_\text{out}}{F_\text{in}}=\frac{d_h\,\omega_2}{v_C}=\frac{201}{11.49}=\boxed{17.5}.$$
  6. Output force. $F_\text{out}=\text{MA}\times F_\text{in}=17.5\times 500 = \boxed{8.75\ \text{kN}}$ (directed downward at the press head).
QuantityValue
Slider speed for $\omega_2=1$ rad/s$v_C = 11.5$ units/s (down)
Mechanical advantageMA ≈ 17.5
Output (press-head) forceFout ≈ 8.75 kN
Check: link geometry is read from the 1:15 drawing; MA scales with the measured coordinates, so re-measuring gives MA in the range 16–19 (Fout ≈ 8–9.5 kN). The 1:15 scale itself cancels because MA is a pure velocity ratio.
← Paper overview