Question 1 of 6: Question 1 (paper Question I) — Pipe Assembly: Ball-and-Socket Joint and Two 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 — 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 1 (paper Question I) — Pipe Assembly: Ball-and-Socket Joint and Two Cables (Part A · Statics, equal value)
Given. A rigid pipe run A–B–E–G–H is fixed to a ball-and-socket joint at A and steadied by two cables from B to fixed wall points C and D. Two vertical loads act on the horizontal run: $F_1=3\ \text{kN}$ at G and $F_2=5\ \text{kN}$ at H, both downward.
Given data — coordinates (m)
Point
x
y
z
Role
A
0
0
0
ball-and-socket
B
0
1
1
cable junction / elbow
E
0
0
1
elbow
G
0
1.5
1.5
$F_1=3\ \text{kN}\downarrow$
H
0
3
2
$F_2=5\ \text{kN}\downarrow$
C
2
0
3
cable BC anchor
D
−2
0
2
cable BD anchor
Find. The reaction components $\mathbf{A}=(A_x,A_y,A_z)$ at the ball-and-socket, and the tensions $T_{BC}$, $T_{BD}$.
Figure 1 (reconstructed). Ball-and-socket at A, cables B–C and B–D, loads at G and H.
Approach. The ball-and-socket carries no moment, so summing moments about A involves only the two unknown cable tensions; force equilibrium then gives the reaction at A.
Unit vectors along the cables. $\overrightarrow{BC}=C-B=(2,-1,2)\ \text{m}$, $|\overrightarrow{BC}|=3\ \text{m}$, $\mathbf{u}_{BC}=(0.6667,-0.3333,0.6667)$. $\overrightarrow{BD}=D-B=(-2,-1,1)\ \text{m}$, $|\overrightarrow{BD}|=\sqrt{6}=2.449\ \text{m}$, $\mathbf{u}_{BD}=(-0.8165,-0.4082,0.4082)$.
Moments about A. Only the cable forces at B and the two loads contribute (A itself produces no moment about A): $$\sum\mathbf{M}_A=\mathbf{r}_B\times(T_{BC}\mathbf{u}_{BC}+T_{BD}\mathbf{u}_{BD})+\mathbf{r}_G\times(0,0,-F_1)+\mathbf{r}_H\times(0,0,-F_2)=\mathbf{0}$$ With $\mathbf{r}_B=(0,1,1)$, $\mathbf{r}_G=(0,1.5,1.5)$, $\mathbf{r}_H=(0,3,2)$, this expands to three scalar equations in $T_{BC}$ and $T_{BD}$; the $x$- and $y$-component equations are independent and the $z$-component is then automatically satisfied (a consistency check that the geometry is self-consistent).
Solve for the cable tensions. Solving the two independent moment equations simultaneously: $$\boxed{T_{BC}=11.70\ \text{kN}},\qquad \boxed{T_{BD}=9.553\ \text{kN}}$$ Both are positive, i.e. genuine tensions — the physical check that confirms the C/D geometry reading.
Reaction at A (force equilibrium). $$\mathbf{A}+T_{BC}\mathbf{u}_{BC}+T_{BD}\mathbf{u}_{BD}+(0,0,-F_1)+(0,0,-F_2)=\mathbf{0}$$ $$\boxed{\mathbf{A}=(0,\ 7.80,\ -3.70)\ \text{kN}}$$
