Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination — Statics and Dynamics (04-BS-3), December 2014. 3 hours, closed book (one 8.5"×11" self-prepared note sheet permitted). Candidates were required to complete 2 of the 3 questions in Part A (Statics) and 2 of the 3 questions in Part B (Dynamics); every question in both parts is solved below.
Check — figure readings (Questions 1 and 3): (1) In Figure 1's truss only three members meet at joint D (C–D, D–B, D–E) — there is no separate C–E member, and the truss coordinates are A(0,0), B(4,0), C(1,2), D(3,2.6), E(7,4) m. (2) Figure 3A's axial dimension stack (2+4+2+4+2 = 14 in) does not sum to the labelled overall height of 16 in unless the hub's "2 in." label is read as a half-thickness measured from the shaft centreline (giving a full hub thickness of 4 in, which closes the stack exactly: 2+4+4+4+2 = 16 in). Both readings are disclosed here and used consistently in the solutions below.
Given. Collar weight $W=20\ \text{lb}_f$, frictionless vertical rod. Spring: free length $L_0=4$ in, stiffness $k=3\ \text{lb}_f/\text{in}$, fixed pivot 8 in horizontally from the rod. Collar released from rest at position 1 (level with the pivot, spring length $=8$ in) and falls 6 in to position 2.
Find. The collar's speed at position 2.
Figure 5 — collar on a vertical rod, pulled by a pivoted spring.
Approach. Frictionless system with only gravity and a linear spring doing work $\Rightarrow$ apply conservation of energy, $T_1+V_1=T_2+V_2$, with $V=V_{\text{grav}}+V_{\text{spring}}$, taking position 2 as the gravity datum.
Spring lengths and stretch at each position. At 1 the spring is horizontal, length $=8$ in. At 2 the collar is 6 in below the pivot's level and still 8 in off the rod horizontally, so
$$L_2 = \sqrt{8^2+6^2} = 10\ \text{in}$$
$$s_1 = L_1-L_0 = 8-4=4\ \text{in}, \qquad s_2 = L_2-L_0 = 10-4=6\ \text{in}$$
Energy at position 1 (at rest, taking $h=6$ in above the datum at 2):
$$T_1 = 0, \qquad V_1 = Wh+\tfrac12 k s_1^2 = 20(6)+\tfrac12(3)(4)^2 = 120+24=144\ \text{lb}_f\cdot\text{in}$$
Energy at position 2 ($h=0$ at the datum):
$$V_2 = 0+\tfrac12(3)(6)^2 = 54\ \text{lb}_f\cdot\text{in}$$
Solve for the kinetic energy and speed at 2.
$$T_2 = V_1-V_2 = 144-54=90\ \text{lb}_f\cdot\text{in} = \tfrac12\left(\frac{W}{g}\right)v_2^2$$
With $g=386.4\ \text{in/s}^2$: $m=W/g=20/386.4=0.05176\ \text{lb}_f\text{s}^2/\text{in}$.
$$v_2 = \sqrt{\frac{2(90)}{0.05176}} = 58.97\ \text{in/s}$$
$$\boxed{v_2 = 58.97\ \text{in/s} = 4.91\ \text{ft/s}}$$
