Question 3 of 6: Question 3 (paper Question III) — Belt (Capstan) Friction on an Incline (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-May. 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. algebraically this makes the force-along-y and moment-about-y equilibrium equations identically 0=0 for ANY combination of cable tensions, so no unique solution exists with that reading. Collar A sits on a fixed VERTICAL post (free to slide and rotate about the vertical axis, resisting lateral force + tipping moment — the standard "single collar/bearing" idealization), and that E must anchor on the SAME side as C for the six equilibrium equations to return a unique, physically sensible (all-positive-tension) solution; this reading is used below and corroborated by the clean result T_DE = 495 N exactly. (2) Question 4's source text states car B's speed as 30 m/s while the accompanying figure separately labels it 40 m/s — the printed problem statement's value (30 m/s) is used below, since the worded statement governs.
Question 3 (paper Question III) — Belt (Capstan) Friction on an Incline (Part A · Statics, equal value)
Given. Incline angle $\theta=30^\circ$. Plate B: mass 40 kg, $\mu_B=0.3$ (between B and the incline). Block A: $\mu_A=0.2$ (between A and the incline), sits higher up the same incline than B. The cable runs from B up the incline, makes a 180° hairpin turn around the fixed rod at C (at the top), and returns down the same incline to A; $\mu_C=0.3$ between the cable and the rod.
Given data
Quantity
Value
Incline angle, $\theta$
30°
Mass of plate B
40 kg
$\mu_B$ (plate B / incline)
0.3
$\mu_A$ (block A / incline)
0.2
$\mu_C$ (cable / rod)
0.3
Cable wrap angle at C, $\beta$
180° ($\pi$ rad)
Find. The smallest mass of block A that prevents plate B from sliding down the incline.
Figure 3 (reconstructed). Blocks A and B on the 30° incline, cable hairpinned around the fixed rod at C.
Approach. With B on the verge of sliding down, find the cable tension $T_B$ needed to hold it (using B's own friction, which assists). Apply the capstan (belt-friction) equation across the rod at C to find the tension $T_A$ delivered to block A — friction at the rod reduces the tension A must supply, so the minimum case relies on the full capstan advantage. Finally solve block A's own equilibrium (on the verge of sliding up, pulled by $T_A$) for the minimum weight.
Tension needed to hold plate B. B is on the verge of sliding down, so friction acts at its maximum value, up the incline: $T_B+\mu_B W_B\cos\theta=W_B\sin\theta$. With $W_B=40(9.81)=392.4$ N: $T_B=392.4(\sin30^\circ-0.3\cos30^\circ)=392.4(0.5-0.2598)$, so $\boxed{T_B=94.25\text{ N}}$.
Capstan (belt friction) equation at rod C. Because B tends to slide down (pulling the cable over the rod on B's side), friction at the rod makes the B-side tension the larger one: $T_B=T_A\,e^{\mu_C\beta}$. With $\beta=\pi$: $e^{\mu_C\beta}=e^{0.3\pi}=2.566$, so $T_A=T_B/2.566$, giving $\boxed{T_A=36.73\text{ N}}$.
Equilibrium of block A (minimum weight). At the minimum mass, A is on the verge of being pulled UP the incline by $T_A$, so its own friction acts at maximum value, down the incline, opposing that: $T_A=W_A\sin\theta+\mu_A W_A\cos\theta=W_A(\sin\theta+\mu_A\cos\theta)$.
Solve for the minimum mass. $W_A=\dfrac{36.73}{\sin30^\circ+0.2\cos30^\circ}=\dfrac{36.73}{0.5+0.1732}=54.55$ N, so $m_A=W_A/g=54.55/9.81$, giving $\boxed{m_{A,\min}=5.56\text{ kg}}$.
