Question 6 of 6: Question 6 (paper Question VI) — Conveyor Plate on an Inclined Rail: D'Alembert's Principle (Part B · Dynamics, 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 6 (paper Question VI) — Conveyor Plate on an Inclined Rail: D'Alembert's Principle (Part B · Dynamics, equal value)
Given. A rigid plate hangs from two shoes that slide, without rotating, along a rail inclined at $\beta$ to the horizontal; the shoes are a apart along the rail, and the plate's centre of gravity is $a/2$ from each shoe (horizontally, along the rail) and a perpendicular distance b below the rail. Numerically: $a=6\ \text{ft}$, $b=3\ \text{ft}$, $\beta=28^\circ$, $W=750\ \text{lbf}$, $\mu_k=0.015$.
Find. (a) The acceleration of the plate; (b) expressions for the shoe normal forces $N_1$, $N_2$; (c) the numerical values.
Figure 6 (schematic). Plate hanging from two shoes on the inclined rail; C.G. midway between the shoes, offset b below the rail.
Approach. Because both shoes are rigidly spaced and both slide along the same straight rail, the plate translates parallel to the rail without rotating — exactly like the two-cable pendulum in Question 5. D'Alembert's principle converts the dynamics into a statics problem by adding a fictitious inertial force $-ma$ at the centre of gravity: force balance along and perpendicular to the rail gives the acceleration and the total normal force, and a moment balance about the centre of gravity (required to be zero, since there is no angular acceleration) splits that total between the two shoes.
Perpendicular-to-rail equilibrium. With no acceleration component perpendicular to the rail (pure translation along it): $$\boxed{N_1+N_2=W\cos\beta}$$
Along-the-rail equation of motion (D'Alembert). Gravity's component down the rail is resisted by kinetic friction at both shoes (up the rail, opposing the slide): $$W\sin\beta-\mu_k(N_1+N_2)=\dfrac{W}{g}a\ \Rightarrow\ \boxed{a=g(\sin\beta-\mu_k\cos\beta)}$$
Moment balance about the C.G. (zero, since $\alpha=0$). With shoe 1 at $a/2$ down-rail and perpendicular offset b, and shoe 2 at $a/2$ up-rail, same offset, each carrying normal force $N_i$ and friction $\mu_kN_i$ (up-rail): summing moments about the C.G. and setting the result to zero gives $$N_1\left(\dfrac{a}{2}+b\mu_k\right)=N_2\left(\dfrac{a}{2}-b\mu_k\right)\ \Rightarrow\ \boxed{N_2=N_1\dfrac{a/2+b\mu_k}{a/2-b\mu_k}}$$ Combined with Step 1, this gives $N_1$ and $N_2$ individually.
Numerical results. $$a=32.2(\sin28^\circ-0.015\cos28^\circ)=\boxed{14.69\ \text{ft/s}^2}$$ $$N_1+N_2=750\cos28^\circ=662.2\ \text{lbf},\qquad \dfrac{N_2}{N_1}=\dfrac{3.0+3(0.015)}{3.0-3(0.015)}=1.0305$$ $$\boxed{N_1=326.1\ \text{lbf}},\qquad \boxed{N_2=336.1\ \text{lbf}}$$ (shoe 2, nearer the top of the rail, carries slightly more load than shoe 1 — the small friction-induced moment shifts a little weight onto the trailing shoe.)
