Question 1 of 6: Question 1 (paper Question I) — Pipe, Cable and Collar Assembly Supporting a Sign (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-May. 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. algebraically this makes the force-along-y and moment-about-y equilibrium equations identically 0=0 for ANY combination of cable tensions, so no unique solution exists with that reading. Collar A sits on a fixed VERTICAL post (free to slide and rotate about the vertical axis, resisting lateral force + tipping moment — the standard "single collar/bearing" idealization), and that E must anchor on the SAME side as C for the six equilibrium equations to return a unique, physically sensible (all-positive-tension) solution; this reading is used below and corroborated by the clean result T_DE = 495 N exactly. (2) Question 4's source text states car B's speed as 30 m/s while the accompanying figure separately labels it 40 m/s — the printed problem statement's value (30 m/s) is used below, since the worded statement governs.
Question 1 (paper Question I) — Pipe, Cable and Collar Assembly Supporting a Sign (Part A · Statics, equal value)
Given. Collar A sits on a fixed vertical post; a rigid rod runs from A along the horizontal (y) direction through anchor points D and B, to which the sign plate (weight 220 N, centre of gravity G) is welded. Cable DE runs from D to a wall anchor E; cable BC runs from B to a wall anchor C.
Given data (coordinates in metres)
Point
x
y
z
Role
A
0
0
4
collar (vertical post)
D
0
-2
4
rod / plate corner
B
0
3
4
rod / plate corner
E
2
-2
4
cable DE anchor
C
3
3
6
cable BC anchor
G
0
0
3
sign centre of gravity, W = 220 N
Find. The cable tensions $T_{DE}$ and $T_{BC}$, and the reaction force and couple-moment components at collar A.
Figure 1 (reconstructed). Vertical-post collar at A, horizontal rod A–D–B carrying the plate, cables D–E and B–C.
Approach. Treat the rod+plate as one rigid body; write $\sum\mathbf{F}=0$ and $\sum\mathbf{M}_A=0$ in Cartesian components. A collar on a vertical post is free to slide and rotate about that (z) axis, so it exerts zero force and moment along z — which turns $\sum F_z=0$ and $\sum M_z=0$ into two independent scalar equations for the two unknown tensions.
Unit vectors along each cable. $\overrightarrow{DE}=E-D=(2,0,0)\text{ m}\Rightarrow\mathbf{u}_{DE}=(1,0,0)$. $\overrightarrow{BC}=C-B=(3,0,2)\text{ m}$, $|\overrightarrow{BC}|=\sqrt{13}\text{ m}\Rightarrow\mathbf{u}_{BC}=(0.832,0,0.555)$.
Position vectors from A. $\mathbf{r}_D=(0,-2,0)$, $\mathbf{r}_B=(0,3,0)$, $\mathbf{r}_G=(0,0,-1)$ m (all relative to A). Since G lies directly on the post's own vertical line through A, the weight contributes zero moment about A.
Force equilibrium in z (collar free along z). $\sum F_z=0$: $T_{BC}(0.555)-220=0$. Solving, $\boxed{T_{BC}=396.6\text{ N}=110\sqrt{13}\text{ N}}$.
Moment equilibrium about z (collar free to rotate about z). $\mathbf{M}_{T_{DE}}=\mathbf{r}_D\times\mathbf{u}_{DE}T_{DE}=(0,0,2T_{DE})$; $\mathbf{M}_{T_{BC}}=\mathbf{r}_B\times\mathbf{u}_{BC}T_{BC}=(1.664T_{BC},0,-2.496T_{BC})$. Setting the z-components to zero: $2T_{DE}-2.496T_{BC}=0$. Substituting $T_{BC}=396.6$ N, $\boxed{T_{DE}=495.0\text{ N}}$.
Remaining reactions at the collar. $\sum F_x=0$: $A_x+T_{DE}+0.832T_{BC}=0\Rightarrow A_x=-495.0-330.0=-825\text{ N}$. $\sum F_y=0$ gives $A_y=0$ (no force in this problem has a y-component). Taking the x- and y-components of the moment sum ($\sum M_x=0$, $\sum M_y=0$) gives the tipping-moment reactions: $M_x=-1.664T_{BC}=-660\text{ N}\cdot\text{m}$, $M_y=0$. $A_z$ and $M_z$ are not applicable — the collar slides and rotates freely along/about the post.
