Question 4 of 6: Question 4 (paper Question IV) — Velocity and Acceleration of a Link Sliding Between Two Guides (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 — 2018-Dec. Candidates were required to complete 2 questions from PART A (Statics) and 2 questions from 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 — source reconstruction notes. (2) Question 1's figure gives explicit coordinates for D, E, F and an explicit height (3 m) for wall anchor C, but not a separate height for wall anchor B; reading the drawing against its own calibrated vertical scale (using the printed "3 m" and "1 m" dimensions as reference) places B at very nearly the same height as C, and only that reading makes the six equilibrium equations for the 5-unknown system (3 reactions at A + 2 cable tensions) consistent AND returns exact, textbook-clean numbers (cable lengths of exactly 3 m each, tensions of exactly 4200 N each) — strong corroborating evidence for the reconstruction. B = (0, -2, 3) m and C = (0, 2, 3) m (both relative to A) are used below. (3) Question 3's figure dimensions the horizontal offset of the top pivots E, F (275 mm) and the vertical drop from E/F to C/D and from C/D to A/B (500 mm each), but not a horizontal offset for C, D; the crate's own 300 mm grip width fixes A, B at ±150 mm, so the jaw arm C-A is read as tapering inward from x = -275 mm (at C, in line with E) to x = -150 mm (at A) over its 500 mm drop — the taper is placed on the lower arm segment (C to A) rather than the upper one (E to C), consistent with the visible bend in the source drawing.
Question 4 (paper Question IV) — Velocity and Acceleration of a Link Sliding Between Two Guides (Part B · Dynamics, equal value)
Given. Rigid bar AB, length 2 ft (24 in), makes 30° with the vertical wall. Roller A slides along the vertical wall (velocity/acceleration purely vertical); roller B slides along a surface inclined 30° to the horizontal. At this instant $v_A=160$ in/s and $a_A=236$ in/s$^2$, both directed downward.
Given data
Item
Value
Bar length AB
2 ft = 24 in
Angle of bar from wall (at A)
30°
Incline angle (at B)
30° from horizontal
$v_A$ (downward, along wall)
160 in/s
$a_A$ (downward, along wall)
236 in/s$^2$
Find. $v_B$, $a_B$ (along the incline), and the bar's angular velocity $\omega$ and angular acceleration $\alpha$.
Figure 4. Bar AB (2 ft): roller A slides on the vertical wall; roller B slides on the 30° incline.
Approach. Model AB as a rigid body with two points constrained to slide along known straight guides; apply the relative-velocity and relative-acceleration equations $\mathbf{v}_B=\mathbf{v}_A+\boldsymbol{\omega}\times\mathbf{r}_{B/A}$ and $\mathbf{a}_B=\mathbf{a}_A+\boldsymbol{\alpha}\times\mathbf{r}_{B/A}-\omega^2\mathbf{r}_{B/A}$, resolving into components along each guide.
Position vector. $\mathbf{r}_{B/A}=24(\sin30^\circ,-\cos30^\circ)=(12,\,-20.78)$ in.
Velocity equation. With $\mathbf{v}_A=(0,-160)$ in/s and $\mathbf{v}_B=v_B(\cos30^\circ,-\sin30^\circ)$ (along the incline): $v_B(\cos30^\circ,-\sin30^\circ)=(0,-160)+\omega(20.78,\,12)$. Solving the two scalar equations simultaneously: $\boxed{\omega=6.67\ \text{rad/s (CCW)}}$, $\boxed{v_B=160\ \text{in/s down the incline}}$.
Acceleration equation. With $\mathbf{a}_A=(0,-236)$ in/s$^2$ and $\mathbf{a}_B=a_B(\cos30^\circ,-\sin30^\circ)$: $a_B(\cos30^\circ,-\sin30^\circ)=(0,-236)+\alpha(20.78,\,12)-\omega^2(12,\,-20.78)$. Solving simultaneously (with $\omega^2=44.44\ \text{rad}^2/\text{s}^2$): $\boxed{\alpha=15.83\ \text{rad/s}^2\ \text{(CW, i.e. opposing }\omega)}$, $\boxed{a_B=995.7\ \text{in/s}^2\ \text{up the incline}}$.
