22-Mec-A4 Design and Manufacture of Machine Elements · December 2018

Question 4 of 6: Pin-Joined Tongs — Least Coefficient of Friction (25 marks)

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

Question 4: Pin-Joined Tongs — Least Coefficient of Friction (25 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. The smallest coefficient of static friction $\mu_s$ between the gripping pads and the crate for which the tongs can raise the crate without it slipping out.

Approach. Work outward from the crate: vertical equilibrium of the crate fixes the friction force each pad must develop; joint H fixes the thrust in the inclined links; a moment equation about C on one arm then delivers the pad normal force, and the required coefficient is the ratio of the two.

1. Weigh the crate. The lifting force applied at H must equal the weight, since the tongs themselves are taken as weightless: $$W = mg = (125)(9.81) = 1226.25\ \text{N} = P$$
2. Vertical equilibrium of the crate sets the friction demand. The pads touch the crate on vertical faces, so the normal forces are horizontal and cancel one another; only friction can carry the weight. With two pads sharing it equally, $$2F_f = W \quad\Rightarrow\quad F_f = \frac{W}{2} = 613.1\ \text{N per pad}$$
3. Resolve joint H to get the link thrust. Pin H carries only the vertical force $P$ and the two symmetric links, each inclined at $\beta = 30^{\circ}$ below the horizontal. Summing forces vertically at H, $$2S\sin\beta = P \quad\Rightarrow\quad S = \frac{P}{2\sin 30^{\circ}} = \frac{1226.25}{1.0} = 1226.25\ \text{N}$$ The links are in compression: they push the tops of the two arms apart, which is what closes the jaws below.
4. Set up the arm geometry with C as the origin. Working in millimetres and taking $x$ positive toward the crate centreline and $y$ positive upward, the two loaded points on the left arm are at $$E = (250,\,+450) \qquad A = (300,\,-450)$$ The force the link applies at E acts outward and upward along the link, $\mathbf{S} = S(-\cos\beta,\,+\sin\beta)$. The crate pushes back on the pad at A with $(-N,\,-F_f)$: the normal reaction points away from the crate and the friction reaction points down, because the pad is holding the crate up.
5. Take moments about C. Bar CD is a two-force member pinned at C and D with no load between, so its force is purely horizontal and passes through C — it drops out of a moment equation about C. Using $M_C = xF_y - yF_x$ for each force, $$S\,(250\sin\beta + 450\cos\beta) - 300F_f - 450N = 0$$ $$1226.25\,(125.0 + 389.7) - 300(613.1) - 450N = 0$$
6. Solve for the pad normal force. Evaluating the two known terms, $631{,}157 - 183{,}938 = 447{,}219\ \text{N}\cdot\text{mm}$, so $$N = \frac{447{,}219}{450}$$ $$\boxed{N = 993.8\ \text{N at each pad}}$$
7. Form the friction requirement. Slipping is impending when $F_f = \mu_s N$, and any larger coefficient will also hold, so the least admissible value is $$\mu_s = \frac{F_f}{N} = \frac{613.1}{993.8}$$ $$\boxed{\mu_{s,\min} = 0.617}$$
8. Check the arm vertically as an independent test. Because CD is horizontal it can supply no vertical force to the arm, so the vertical components at E and A must balance on their own: $$S\sin\beta = (1226.25)(0.5) = 613.1\ \text{N} = F_f \quad\checkmark$$ The arm closes, which confirms the assumed directions of the pad reactions and the link thrust.

A coefficient of 0.617 is high — well above clean steel on dry timber (about 0.3) and above steel on steel — so these tongs would need a serrated, rubber-faced or otherwise high-grip pad to lift this crate reliably. The reason is geometric: the pads sit 300 mm out from the pivot on a 450 mm arm, and the mechanism's clamping advantage is set by the ratio of the link and pad lever arms, not by the load.