Question III: Belt Friction on a Fixed Rod (Part A — Statics): 04-BS-3, May 2013, Q3

Question III: Belt Friction on a Fixed Rod (Part A — Statics) (20 marks)

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. Block A: $m_A=60$ kg on surface B, $\mu_s=0.25$. Cord from A rises to a fixed rod at C along a 3-4-5 slope (36.87° above horizontal), wraps the rod, then hangs straight down to cylinder D. Cord-rod friction $\mu_s'=0.30$.

Given data
Quantity	Value
$m_A$	60 kg
$\mu_s$ (A–B)	0.25
$\mu_s'$ (cord–C)	0.30
Cord AC slope	3 (vert) : 4 (horiz) : 5 (hyp)

Figure 3 — cord from block A rises at 36.87° to fixed rod C, then drops vertically to cylinder D.

Find. The largest mass $m_D$ such that block A remains in equilibrium (does not slide).

Approach. At the critical (largest-$m_D$) condition, two frictional limits are reached simultaneously: block A is on the verge of sliding toward the rod (maximum $\mu_s$ friction under A), and the cord is on the verge of slipping over the rod toward the D side (capstan/belt-friction equation with $\mu_s'$ and the wrap angle $\beta$). Solve A's equilibrium for the cord tension $T_1$ on the block side, then apply the capstan equation to get $T_2$ on the cylinder side.

Block A equilibrium (verge of sliding). With $\theta=\arctan(3/4)=36.87^\circ$, resolving normal and friction forces on A: $$N=W_A-T_1\sin\theta, \qquad T_1\cos\theta=\mu_s N$$ Substituting and solving for $T_1$ (with $W_A=60(9.81)=588.6$ N), $$T_1(\cos\theta+\mu_s\sin\theta)=\mu_s W_A \;\Rightarrow\; T_1=\boxed{154.9\text{ N}}$$
Wrap angle at the rod. The cord arrives from A along $(4,-3)/5$ (measuring from C) and leaves toward D along $(0,-1)$. The angle between these directions is $$\beta=\cos^{-1}\!\left(\frac{4(0)+(-3)(-1)}{5}\right)=\cos^{-1}(0.6)=53.13^\circ=0.9273\text{ rad}$$
Capstan (belt friction) equation. Because cylinder D tends to pull the cord over the rod (the "heavier", about-to-slip side), $T_2 = T_1 e^{\mu_s'\beta}$: $$T_2 = 154.9\,e^{0.30(0.9273)} = 154.9(1.3208) = \boxed{204.6\text{ N}}$$
Largest mass of D. $$m_D=\frac{T_2}{g}=\frac{204.6}{9.81}=\boxed{20.9\text{ kg}}$$

Quantity	Value
Cord tension $T_1$ (A side)	154.9 N
Wrap angle $\beta$	53.13° (0.9273 rad)
Cord tension $T_2$ (D side)	204.6 N
Largest mass $m_D$	20.9 kg

Concept & Theory

Topic. Combined Coulomb friction (block on a surface) and belt/capstan friction (cord over a fixed rod).

Key relations. Coulomb: $F\le \mu_s N$. Capstan: $T_2=T_1e^{\mu\beta}$, where $\beta$ is the total wrap angle in radians and $T_2$ is the tension on the side about to slip away.

Why this works. At the threshold of motion, both frictional interfaces (block/surface and cord/rod) are simultaneously at their maximum capacity; treating them as two independent, cascading limits (solve the block first for $T_1$, then propagate through the capstan relation to $T_2$) correctly captures the largest hanging mass the system can support without slipping anywhere.

Common pitfall. Using the wrap angle between the cord and the horizontal instead of the true angle between the cord's two departing directions at the rod; and forgetting that pulling up on A (the $\sin\theta$ term) reduces the normal force, which lowers the friction available under A.

Source. Hibbeler, Engineering Mechanics: Statics (14th ed.), Ch. 8 — Friction, §8.5 Frictional Forces on Flat Belts.

Question 3 of 6: Question III: Belt Friction on a Fixed Rod (Part A — Statics)

Question III: Belt Friction on a Fixed Rod (Part A — Statics) (20 marks)