Question 3 of 6: Question 3 (paper Question III) — Wedge, Block and Capstan Friction (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 — 2019-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 isometric figure gives explicit 1 m / 1.5 m dimensions along the rod, but plotting them on a single straight line makes the six rigid-body equilibrium equations inconsistent (no solution exists). Reading the printed figure, the horizontal run bends twice: A→C is 2 m along the first horizontal direction (with the vertical 1.5 m branch C→D), C→E is 1 m along the perpendicular horizontal direction (this is where the smooth journal bearing sits), and E→F is a further 1.5 m continuing the first horizontal direction to the point that carries the 250 N cylinder. This reading is the only one of the plausible alternatives that returns a statically-determinate, non-degenerate system, and it lands on the clean values reported below — strong corroborating evidence for the reconstruction. (3) Question 4's dimensions place the pin O exactly 90 mm above the floor, equal to the sphere radius (90 mm R); the sphere therefore contacts the floor exactly when the rod is horizontal, which is used as the impact configuration. (4) Question 6 numbers the four-bar linkage O-A-B-C from the source's own 100 mm / 250 mm / 75 mm / 50 mm dimensions; O and C are the fixed pivots, OA and CB are the driven/driving cranks and AB is the coupler.
Question 3 (paper Question III) — Wedge, Block and Capstan Friction (Part A · Statics, equal value)
Given. Wedge A (50 N) rests on the ground C, $\mu_{AC}=0.4$. Block B (30 N) rests on A's 20° inclined top surface, $\mu_{BA}=0.6$. A rope from B runs horizontally to a fixed cylindrical peg ($\mu_{peg}=0.5$, 90° wrap angle) and then straight down to hanging block D.
Find. The greatest weight $W_D$ that keeps the whole system in equilibrium.
Figure 3. Wedge A on ground C, block B on A's incline, rope over a rough peg to hanging block D.
Approach. As $W_D$ grows the rope tension pulling B (horizontal, toward the peg) grows; resolved onto the 20° incline this pull has a down-slope component, so B is on the verge of sliding down A's incline. Find the tension $T_B$ at that verge from B's equilibrium, confirm the wedge does not slide on the ground first, then relate $T_B$ to $W_D$ through the capstan (belt-friction) equation for the rough peg.
Block B equilibrium at the verge of sliding down the incline. Resolve perpendicular and parallel to the incline (rope tension $T_B$ horizontal):
$$N_{BA}=W_B\cos20^\circ-T_B\sin20^\circ,\qquad \mu_{BA}N_{BA}=W_B\sin20^\circ+T_B\cos20^\circ$$
Solving simultaneously for $T_B$:
$$T_B=W_B\cdot\frac{\mu_{BA}\cos20^\circ-\sin20^\circ}{\cos20^\circ+\mu_{BA}\sin20^\circ}=30\cdot\frac{0.6(0.9397)-0.3420}{0.9397+0.6(0.3420)}=\boxed{5.81\text{ N}}$$
Check the wedge A+B does not slide on the ground C first. The only horizontal force on the combined A+B system is $T_B$ (the rope tension); the maximum available ground friction is $\mu_{AC}(W_A+W_B)=0.4(80)=32$ N, far greater than $T_B=5.81$ N, so the B-on-A slip governs (occurs first) as assumed.
Capstan (belt-friction) equation across the peg. With wrap angle $\beta=90^\circ=\pi/2$ and D about to descend (rope feeding from the B-side to the D-side), the D-side is the higher tension:
$$T_D=T_B\,e^{\mu_{peg}\beta}=5.81\times e^{0.5\times\pi/2}=5.81\times2.193=\boxed{12.75\text{ N}}$$
Weight of D. Since D hangs freely, $W_D=T_D=12.75$ N at the verge of motion — the greatest weight that will not cause motion.
Check — assumption check: block B was assumed to reach its down-slope friction limit before the peg slips independently or the wedge slides on the ground. This is confirmed above (32 N >> 5.81 N for the ground check) and is self-consistent because the capstan relation and B's slip condition were solved as a single coupled limiting state (both interfaces at their friction limit simultaneously).
