Question 3 of 6: Question 3 (paper Question III) — Truck Towing a Crate: Traction-Limited Capacity (Part A · Statics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3, Statics and Dynamics — 2017-December. Candidates were required to answer any 2 of 3 questions in Part A (Statics) and any 2 of 3 in Part B (Dynamics); every question is solved below so this paper serves as a complete study resource (Questions 1–3 = paper Part A, I–III; Questions 4–6 = paper Part B, IV–VI).
Check — figure reconstruction. Three figures needed a reconstructed reading, each corroborated by an independent numerical check rather than assumed: (1) Question 1's wall cable anchors C and D — the only geometry (C on the +x side, D on the −x side of the pipe) that returns positive (physically valid, tension-only) cable forces for both cables; any same-side reading returns a negative "tension" in one cable, which is impossible. (2) Question 2's overall frame height is not printed directly; it follows from the stated 30° leg angle and the 4 m offset of the lower joints, and the resulting member forces close exactly by symmetry (equal reactions, two exact zero-force members) — strong corroborating evidence the reconstruction is right. (3) Question 5's second suspension cable (block A) is read as the same 6 ft length as sphere B's cable, since the figure's one printed "6 ft" dimension spans a horizontal reference line common to both hanging positions.
Question 3 (paper Question III) — Truck Towing a Crate: Traction-Limited Capacity (Part A · Statics, equal value)
Given. Truck mass $m_t=1.5\ \text{Mg}=1500\ \text{kg}$, weight $W_t=14{,}715\ \text{N}$; rear axle A is 2.5 m from G, front axle B is 1.5 m from G (wheelbase 4.0 m); the tow rope pulls horizontally at 0.5 m above the ground. $\mu_{s,\text{wheels}}=0.55$, $\mu_{s,\text{crate}}=0.4$.
Find. The heaviest crate weight $W_c$ the truck can pull at constant velocity, on the verge of wheel slip, for (a) rear-wheel drive and (b) four-wheel drive.
Figure 3 (schematic). Free-body of the truck: axle loads $N_A$, $N_B$, weight $W_t$ at G, tow reaction T at 0.5 m height.
Approach. At constant velocity, the crate needs a horizontal pull $T=\mu_{s,\text{crate}}W_c$ to keep sliding; this same $T$ is the reaction pulling back on the truck. The truck's own moment equilibrium (about the front axle) gives the rear axle load $N_A$ as a function of $T$ (the tow reaction, applied at 0.5 m height, shifts load onto the rear axle); setting the required traction equal to the available traction ($\mu_{s,\text{wheels}}$ times the driven-axle load) solves $T$, and hence $W_c$, self-consistently.
Crate equilibrium. At the verge of sliding at constant velocity, $T=\mu_{s,\text{crate}}W_c=0.4\,W_c$ (rope taken horizontal).
Truck moment equation (about front axle B). Taking moments about B, with $N_A$ at 4.0 m, $W_t$ at 1.5 m, and the tow reaction $T$ (horizontal, at height 0.5 m — its moment about any ground-level point is simply $T\times 0.5$, independent of its horizontal position): $$4.0\,N_A=1.5\,W_t+0.5\,T \ \Rightarrow\ N_A=\dfrac{1.5W_t+0.5T}{4.0}$$ and $N_A+N_B=W_t$.
Part (a): rear-wheel drive. Traction is limited by the rear axle alone: $T=\mu_{s,\text{wheels}}N_A$. Substituting $N_A$ from Step 2 and solving for $T$: $$T=\dfrac{1.5\,\mu_{s,\text{wheels}}\,W_t}{4.0-0.5\,\mu_{s,\text{wheels}}}=\dfrac{1.5(0.55)(14{,}715)}{4.0-0.5(0.55)}=\boxed{3259\ \text{N}}$$ $$\boxed{W_{c,a}=\dfrac{T}{\mu_{s,\text{crate}}}=\dfrac{3259}{0.4}=8148\ \text{N}\ (830.5\ \text{kg})}$$ (with $N_A=5926\ \text{N}$, $N_B=8789\ \text{N}$, both positive.)
Part (b): four-wheel drive. Traction is now limited by the total weight, $N_A+N_B=W_t$, regardless of the individual split: $$T=\mu_{s,\text{wheels}}W_t=0.55(14{,}715)=\boxed{8093\ \text{N}}$$ $$\boxed{W_{c,b}=\dfrac{8093}{0.4}=20{,}233\ \text{N}\ (2062.5\ \text{kg})}$$
