Question 3 of 6: Question 3 (paper Question III) — Beam, Rope and Rough Peg (Part A · Statics, 20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination, December 2013 — Statics and Dynamics (04-BS-3). Closed book (one self-prepared 8½×11 in. sheet of notes permitted). Candidates were instructed to complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); every question is solved below as a complete study resource.
Check — the source caption is internally inconsistent (it describes the block resting on the beam yet also has the rope "connected to the 100 N block" past the peg, and gives a horizontal run to the peg immediately followed by "extends downward"). The self-consistent, standard reconstruction adopted here: the beam (pinned at A, 10 m to its far end B) carries the 100 N block on its top surface at distance d from A; a cord ties the block to a rough peg fixed at B (μs = 0.4, horizontal run — this is the interface explicitly named "between the beam and the block"), the cord then makes a 90° turn over the peg (μs = 0.4, "between the rope and the peg") and rises to a fixed overhead anchor that is what actually holds the beam level (matching the statement that the beam itself "is supported by the rope"). This uses every given number exactly once and returns a clean d ≈ 5.00 m — strong evidence the reconstruction matches the intended problem.
Given. Beam weight 50 N (uniform, acts at mid-span, 5 m from A); beam length A to peg B = 10 m; block weight 100 N; μs = 0.4 at both the block–beam interface and the rope–peg interface; peg wrap angle 90° (horizontal cord from the block turning to a vertical cord to the overhead anchor).
Find. The maximum distance d (measured from A) at which the block can be placed and the system remain in equilibrium.
Figure 3 — beam pinned at A, block at distance d, cord routed over the peg at B to a fixed overhead anchor.
Approach. Two friction limits act in series: the block can transmit at most μsN of cord tension before it slides on the beam, and the peg can amplify that tension by at most the capstan factor $e^{\mu_s\beta}$ before the rope slips. The beam's own moment equilibrium about A then converts the maximum deliverable tension into a maximum d.
Maximum tension the block can transmit into the cord. The cord runs horizontally, so it does not change the block's normal force: N = 100 N.
$$T_{c,\max} = \mu_s N = 0.4(100) = \boxed{40.0\ \text{N}}$$
Maximum tension the peg can deliver to the overhead (wall) side via the capstan equation, wrap angle $\beta=90^\circ=\pi/2$:
$$T_{w,\max} = T_{c,\max}\,e^{\mu_s\beta} = 40.0\,e^{0.4(\pi/2)} = \boxed{74.98\ \text{N}}$$
Beam moment equilibrium about A (own weight at 5 m, block's normal reaction at d, overhead-anchor tension acting upward at B, 10 m):
$$\sum M_A = 0:\quad T_w(10) - 50(5) - 100\,d = 0 \Rightarrow d = \dfrac{10\,T_w - 250}{100}$$
Substituting the maximum deliverable tension $T_w=T_{w,\max}$:
$$d_{\max} = \dfrac{10(74.98)-250}{100} = \boxed{5.00\ \text{m}}$$
