Question 1 of 6: Question 1 (paper Question I) — Ball-and-Socket Pipe Assembly with Two Support Cables (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 — 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 1 (paper Question I) — Ball-and-Socket Pipe Assembly with Two Support Cables (Part A · Statics, equal value)
Given. Pipe A–D–E–F is rigidly welded and supported by a ball-and-socket joint at A; a uniformly distributed load of 700 N/m acts downward along segment F–E (length 3 m). Cables DB and DC anchor the pipe to a wall.
Given data (coordinates in metres, relative to A)
Point
x
y
z
Role
A
0
0
0
ball-and-socket joint
D
1
0
1
pipe bend / cable attachment
E
2.5
0
1.5
pipe bend
F
5.5
0
1.5
free end of loaded segment
B
0
-2
3
cable DB wall anchor
C
0
2
3
cable DC wall anchor
Find. The reaction components $A_x, A_y, A_z$ at the ball-and-socket joint and the cable tensions $T_{DB}$, $T_{DC}$.
[Figure not reproduced: Figure 1 (reconstructed from the source dimensions). Ball-and-socket at A; pipe A–D–E–F carries the 700 N/m load along F–E; cables D–B and D–C anchor to a wall behind the pipe. See the official exam paper.]
Approach. Treat the rigid pipe as one free body; write $\sum\mathbf{F}=0$ and $\sum\mathbf{M}_A=0$ in Cartesian-vector form, using the two cable tensions as the only unknowns in the moment equation (the ball-and-socket carries force only, no moment).
Resultant of the distributed load. $W = 700\ \text{N/m}\times 3\ \text{m} = 2100\ \text{N}$, acting downward at the midpoint of F–E: $\mathbf{r}_{load}=\left(\dfrac{5.5+2.5}{2},0,1.5\right)=(4,0,1.5)\ \text{m}$.
Unit vectors along the cables (from D). $\overrightarrow{DB}=B-D=(-1,-2,2)\ \text{m}$, $|\overrightarrow{DB}|=3\ \text{m}\Rightarrow\mathbf{u}_{DB}=(-0.333,-0.667,0.667)$. $\overrightarrow{DC}=C-D=(-1,2,2)\ \text{m}$, $|\overrightarrow{DC}|=3\ \text{m}\Rightarrow\mathbf{u}_{DC}=(-0.333,0.667,0.667)$.
Moment equilibrium about A. $\mathbf{r}_D=(1,0,1)\ \text{m}$. $\sum\mathbf{M}_A=\mathbf{r}_D\times(T_{DB}\mathbf{u}_{DB}+T_{DC}\mathbf{u}_{DC})+\mathbf{r}_{load}\times(0,0,-2100)=\mathbf{0}$. Evaluating the load moment gives $(0,8400,0)\ \text{N}\cdot\text{m}$; the two nonzero component equations this leaves (x and y) solve simultaneously for the cable tensions: $\boxed{T_{DB}=4200\ \text{N}}$, $\boxed{T_{DC}=4200\ \text{N}}$ (the third, z, equation is then automatically satisfied — confirming the reconstructed geometry).
Force equilibrium at A. $\sum\mathbf{F}=\mathbf{A}+T_{DB}\mathbf{u}_{DB}+T_{DC}\mathbf{u}_{DC}+(0,0,-2100)=\mathbf{0}$. The cables' y-components cancel by symmetry ($T_{DB}=T_{DC}$), so $A_y=0$. $A_x=-4200(-0.333-0.333)=2800\ \text{N}$. $A_z=-4200(0.667+0.667)+2100=3500\ \text{N}$ acting in $-z$, i.e. $\boxed{\mathbf{A}=(2800,\ 0,\ -3500)\ \text{N}}$.
