Question 3 of 6: Question 3 (paper Question III) — Friction: Wedge and Belt-Friction Pulley (Part A · Statics, 20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination May 2015 — 04-BS-3, Statics and Dynamics, 3 hours, closed book (one 8½″×11″ self-prepared note sheet permitted; Casio or Sharp approved calculator only). Candidates complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); all 6 are solved below.
Given. Block A (11.25 N) rests on the ground, coefficient of friction μAC = 0.4. Wedge B (6.75 N) rests on a 20° interface on top of A, coefficient of friction μBA = 0.6. A cord from B runs horizontally to a fixed pulley E (belt-friction coefficient μ = 0.5, wrap angle 90°) and down to hanging block D.
Given data
Quantity
Value
Weight of block A
11.25 N
Weight of block B
6.75 N
Interface angle (A–B)
20°
μBA (B on A)
0.6
μAC (A on ground)
0.4
μ at pulley E
0.5
Contact angle at E
90° ($\pi/2$ rad)
Find. The greatest weight of block D that does not cause motion of any part of the system.
Figure 3 — block A on the ground, wedge B on A's 20° interface, cord over pulley E to hanging block D.
Approach. Two independent failure modes are possible as D gets heavier: (a) B slips down the 20° interface relative to A, or (b) A and B slide together on the ground. Solve each mode's critical cord tension T (on the horizontal, B-side segment) independently; the SMALLER one governs, since it is reached first. Then use the capstan (belt-friction) equation to convert that critical T into the corresponding weight of D, since friction at the pulley lets D be heavier than T alone.
Mode (a) — B impends to slide down the interface. Resolving B's weight, the normal force N from A, friction $\mu_{BA}N$ (acting up-slope, opposing the impending down-slope slip) and the horizontal cord tension T along and perpendicular to the 20° interface:
$$N(\cos20^\circ+\mu_{BA}\sin20^\circ) = W_B \;\Rightarrow\; N = \dfrac{6.75}{0.9397+0.6(0.3420)} = 5.896\ \text{N}$$
$$T_a = N(\mu_{BA}\cos20^\circ - \sin20^\circ) = 5.896(0.5638-0.3420) = \boxed{1.308\ \text{N}}$$
Mode (b) — A and B slide together on the ground. The combined weight bears on the ground normal, and T alone must overcome the full ground friction:
$$T_b = \mu_{AC}(W_A+W_B) = 0.4(11.25+6.75) = \boxed{7.2\ \text{N}}$$
Governing mode. $T_a = 1.308\ \text{N} \ll T_b = 7.2\ \text{N}$, so mode (a) — the wedge B slipping on A — governs; the critical cord tension is $T = \boxed{1.308\ \text{N}}$.
Capstan (belt-friction) equation at pulley E. D's weight is the side about to descend (drive the rope over the pulley); friction at the pulley helps hold it, so the tension reaching B is smaller than $W_D$ by the factor $e^{\mu\beta}$:
$$T = W_D\, e^{-\mu\beta} \;\Rightarrow\; W_D = T\, e^{\mu\beta} = 1.308\, e^{0.5(\pi/2)} = 1.308(2.193)$$
$$W_D = \boxed{2.87\ \text{N}}$$
