Question 3 of 6: Question 3 (paper Question III) — Friction with Capstan Pulleys (Part A · Statics, 20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination December 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.
Check — Figure 1 is drawn on a grid of 1 division = 1 m; node positions are measured against it, and each figure's geometry is stated at first use. All six printed questions (I–VI) are answered below.
Question 3 (paper Question III) — Friction with Capstan Pulleys (Part A · Statics, 20 marks)
Given. Block A (15.44 lbf) rests on the ground, block B (20 lbf) rests on top of A; a cord from B runs horizontally to a fixed wall; a separate cord from A runs horizontally to fixed pulley C, wraps 90° up to fixed pulley D, wraps 180° back down to where the vertical force P is applied. Coefficients: μA=0.3 (A–ground), μB=0.4 (B–A), μC=0.4, μD=0.1 (rope–pulley).
Given data
Quantity
Value
WA
15.44 lbf
WB
20.00 lbf
μA (A–ground)
0.3
μB (B–A)
0.4
μC, wrap 90° = π/2 rad
0.4
μD, wrap 180° = π rad
0.1
Find. The largest vertical force P applied at the free end of the D-cord without causing any block or the rope to slip.
Figure 3 — blocks A, B, wall cord, and the capstan pulleys C, D carrying force P.
Approach. Work outward from the two blocks: B is held by its own wall cord, so find the friction A exerts on B (and by reaction, B on A) at impending slip; then take A's free-body (ground friction + reaction from B) to get the cord tension needed at A; finally apply the capstan (belt-friction) equation twice, once per pulley, to find P.
Block B — impending friction from A. B's cord to the wall is horizontal, so the normal between A and B is simply $N_{AB}=W_B=20.0\ \text{lb}_f$. At the verge of A sliding rightward under B (B held fixed by its wall cord), the friction on B from A is at its maximum:
$$f_{AB}=\mu_B N_{AB}=0.4(20.0)=\boxed{8.00\ \text{lb}_f}$$
so the wall cord tension is $T_1=f_{AB}=8.00\ \text{lb}_f$ (by B's own horizontal equilibrium).
Block A — ground friction. The ground carries both weights:
$$N_A=W_A+N_{AB}=15.44+20.0=35.44\ \text{lb}_f$$
$$f_{ground}=\mu_A N_A=0.3(35.44)=\boxed{10.632\ \text{lb}_f}$$
Block A — horizontal equilibrium, impending slip. The cord tension pulling A toward C must overcome the ground friction and (by Newton's third law) the 8.00 lbf reaction friction from B:
$$T_{CA}=f_{AB}+f_{ground}=8.00+10.632=\boxed{18.632\ \text{lb}_f}$$
Capstan equation at pulley C. The rope is about to slip toward the A-side, so the tension on the far (D) side of C must be larger by $e^{\mu_C\beta_C}$, $\beta_C=\pi/2$:
$$T_{C\to D}=T_{CA}\,e^{\mu_C\beta_C}=18.632\,e^{0.4(\pi/2)}=18.632(1.8745)=\boxed{34.92\ \text{lb}_f}$$
Capstan equation at pulley D. The rope wraps 180° = π rad around D before reaching P:
$$P=T_{C\to D}\,e^{\mu_D\beta_D}=34.92\,e^{0.1\pi}=34.92(1.3691)=\boxed{47.82\ \text{lb}_f}$$
Each pulley wrap makes it harder, not easier, to slip the rope — the friction between rope and drum acts like a brake that must be overcome from the P side, so P must exceed the 18.632 lbf that would be needed with frictionless pulleys by a factor of $e^{\mu_C\beta_C+\mu_D\beta_D}=2.567$.
