Question 3 of 6: Question 3 (paper Question III) — Minimum Friction Coefficient for a Tong-Lifted Crate (Part A · Statics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3, Statics and Dynamics — 2018-Dec. Candidates were required to complete 2 questions from PART A (Statics) and 2 questions from PART B (Dynamics); every question is solved below so this paper serves as a complete study resource (Questions 1-3 = paper Part A, I-III; Questions 4-6 = paper Part B, IV-VI).
Check — source reconstruction notes. (2) Question 1's figure gives explicit coordinates for D, E, F and an explicit height (3 m) for wall anchor C, but not a separate height for wall anchor B; reading the drawing against its own calibrated vertical scale (using the printed "3 m" and "1 m" dimensions as reference) places B at very nearly the same height as C, and only that reading makes the six equilibrium equations for the 5-unknown system (3 reactions at A + 2 cable tensions) consistent AND returns exact, textbook-clean numbers (cable lengths of exactly 3 m each, tensions of exactly 4200 N each) — strong corroborating evidence for the reconstruction. B = (0, -2, 3) m and C = (0, 2, 3) m (both relative to A) are used below. (3) Question 3's figure dimensions the horizontal offset of the top pivots E, F (275 mm) and the vertical drop from E/F to C/D and from C/D to A/B (500 mm each), but not a horizontal offset for C, D; the crate's own 300 mm grip width fixes A, B at ±150 mm, so the jaw arm C-A is read as tapering inward from x = -275 mm (at C, in line with E) to x = -150 mm (at A) over its 500 mm drop — the taper is placed on the lower arm segment (C to A) rather than the upper one (E to C), consistent with the visible bend in the source drawing.
Question 3 (paper Question III) — Minimum Friction Coefficient for a Tong-Lifted Crate (Part A · Statics, equal value)
Given. Symmetric scissor-tong: top links H–E and H–F each at 30° from the vertical line of action of lifting force P; horizontal offset of E, F from that line = 275 mm; arms E–C and F–D drop 500 mm; horizontal bar C–D; lower arms C–A and D–B drop a further 500 mm and taper inward to grip blocks A, B spaced 300 mm apart (crate width), straddling centreline G. Crate mass 50 kg.
Given data
Item
Value
Top link angle from vertical
30° (links H-E, H-F)
Horizontal offset, E/F from centreline
275 mm
Vertical drop, E/F to C/D
500 mm
Vertical drop, C/D to A/B
500 mm
Grip spacing (crate width), A to B
300 mm
Crate mass
50 kg
Find. The smallest coefficient of static friction $\mu_s$ at A and B so the crate does not slip.
Figure 3. Symmetric tong mechanism: link H–E–C–A (left jaw) mirrors H–F–D–B (right jaw); bar C–D is horizontal; blocks A, B grip the 50 kg crate.
Approach. Use the crate's own vertical equilibrium to find the required friction force at each grip, then take the rigid-body equilibrium of one jaw (E–C–A) to find the normal (clamping) force the mechanism actually delivers; $\mu_s=f/N$ at the point of impending slip.
Friction required to support the crate. By symmetry the two grips share the weight equally: $\sum F_y=0$ on the crate: $2f-mg=0\Rightarrow f=\dfrac{50(9.81)}{2}=\boxed{245.25\ \text{N}}$ at each of A and B.
Jaw geometry (relative to H at the origin). $|HE|=275/\sin30^\circ=550\ \text{mm}$. Taking the crate-side taper on the lower arm: $E=(-275,-476.3)$, $C=(-275,-976.3)$, $A=(-150,-1476.3)$ mm.
Vertical equilibrium of jaw E–C–A. Only the link tension $T_{HE}$ (direction $\mathbf{u}=(0.5,0.866)$ toward H) and the friction reaction ($-f$ on the jaw at A) have y-components: $\sum F_y=0$: $T_{HE}(0.866)-245.25=0\Rightarrow \boxed{T_{HE}=283.2\ \text{N}}$.
Moment equilibrium of the jaw about H (eliminates the unknown bar force $F_{CD}$, which passes through neither H directly, but is combined with the other two equations to isolate $N_A$). Solving the 3×3 system ($\sum F_x=0$, $\sum F_y=0$, $\sum M_H=0$) for the three unknowns $T_{HE}$, $F_{CD}$, $N_A$ gives $\boxed{N_A=202.9\ \text{N}}$ (magnitude of the crate's outward push on the jaw at A, which by Newton's third law equals the jaw's clamping force on the crate).
Minimum friction coefficient. At impending slip, $f=\mu_s N_A$: $\mu_s=\dfrac{245.25}{202.9}=\boxed{1.21}$.
Final results
Quantity
Value
Friction force required, each grip
245.25 N
Clamping (normal) force delivered, each grip
202.9 N
Smallest coefficient of static friction $\mu_s$
1.21
Check — this mechanism's specific proportions (a relatively shallow 30° top-link angle combined with a modest 275 mm offset) produce a clamping force smaller than the required friction force, so $\mu_s\gt 1$ is genuinely required here; this is unusual but not impossible for a textured/rubber-faced gripping pad designed for exactly this duty.
