Question 3 (paper Question III) — Column, Capstan Friction and Tipping Check (Part A · Statics, equal value): 04-BS-3, Undated paper, Q3

Question 3 (paper Question III) — Column, Capstan Friction and Tipping Check (Part A · Statics, equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Column A: weight 100 N, base 2 m wide, 4 m tall, on ground with $\mu_s=0.25$. A cord attaches at the top-right corner, runs at $30^\circ$ below horizontal down to a fixed peg C ($\mu_{sC}=0.3$), then straight down to hanging cylinder B. Wrap angle at C (between the two cord segments) $\beta=60^\circ=\pi/3$.

Find. The greatest weight $W_B$ that keeps the system in equilibrium, after stating and verifying the governing failure mode (sliding vs. tipping of column A).

Figure 3. Column A (2 m × 4 m) on ground, cord over peg C to hanging cylinder B.

Approach. Initial assumption: column A fails by sliding first. Find the cord tension $T_A$ at the sliding limit; separately check the tipping limit about A's front-bottom edge (moment balance, independent of friction); the governing (smaller) tension sets $W_B$ via the capstan equation.

Sliding limit. The cord pulls A's top corner at $30^\circ$ below horizontal (down-and-out), so $N_A=W_A+T_A\sin30^\circ$. Impending slide: $T_A\cos30^\circ=\mu_s N_A=0.25(100+T_A\sin30^\circ)$. Solving: $T_A(\cos30^\circ-0.25\sin30^\circ)=25\Rightarrow T_A=\boxed{33.74\text{ N}}$.
Verify the assumption — tipping limit. Taking moments about the base's right-bottom edge (pivot at the corner nearest the cord, height 4 m, width 2 m; friction acts at the pivot's own height so it contributes no moment): $W_A(1)-T_A\cos30^\circ(4)=0\Rightarrow T_A=\dfrac{100(1)}{4\cos30^\circ}=\boxed{28.87\text{ N}}$.
Compare and revise the assumption. $28.87\text{ N}<33.74\text{ N}$, so the column tips before it slides — the initial "sliding governs" assumption is invalid; the governing cord tension is $T_A=28.87$ N.
Capstan (belt friction) equation at peg C. With B tending to descend, the B-side tension is the larger: $W_B=T_A\,e^{\mu_{sC}\beta}=28.87\,e^{0.3(\pi/3)}=28.87(1.3691)=\boxed{39.52\text{ N}}$.

Final Results — Question 3
Quantity	Value
Sliding-limit tension	33.74 N
Tipping-limit tension (governs)	28.87 N
Greatest weight $W_B$	39.52 N

Topic: Combined friction (block-on-surface and belt/capstan) with a tipping-stability check.

Key relations: impending slip $F=\mu N$; capstan equation $T_{\text{large}}=T_{\text{small}}\,e^{\mu\beta}$ ($\beta$ in radians); tipping onset when the resultant normal force reaches the pivot edge, $\Sigma M_{\text{edge}}=0$.

Why this works. A block resting on a surface can fail two independent ways — sliding (a force balance) or tipping (a moment balance about the leading edge). The exam's own instruction to "state and verify assumptions" signals exactly this two-mode check: whichever failure mode requires the smaller applied force governs the real capacity.

Common pitfall: computing only the sliding case and stopping — here it overstates the safe tension by 17%; forgetting that friction, acting at the same height as the tipping pivot, contributes zero moment (so it never appears in the tipping equation) is a second common slip.

Source: Hibbeler, Engineering Mechanics: Statics, Ch. 8 (Friction — frictional forces on flat belts) and Ch. 4 (equivalent systems / tipping).

Question 3 of 6: Question 3 (paper Question III) — Column, Capstan Friction and Tipping Check (Part A · Statics, equal value)

Question 3 (paper Question III) — Column, Capstan Friction and Tipping Check (Part A · Statics, equal value)