Question 1 of 6: Question 1 (paper Question I) — Boom, Cable and Suspended Weight (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-May. 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. Question 5's impact angle (18°) is on the opposite side of vertical from the 20° release angle, not the same side. Both readings are corroborated below by clean, self-consistent numerical results (e.g. member D–F resolves to an exact 2.5 m, and the opposite-side impact reading alone reproduces the rising-then-falling trajectory the figure draws for sphere B).
Question 1 (paper Question I) — Boom, Cable and Suspended Weight (Part A · Statics, equal value)
Given. Boom O–a is rigid and weightless, hinged at the origin O so it swings only within the x-y plane (the pin exerts no moment about the vertical z-axis, its free-rotation axis). Point a lies at (8, 6, 0) m (10 m from O); the cable runs from a to the wall anchor b at (0, 18, 7) m; a vertical rope hangs an 800 N weight straight down from a.
Given data
Point
x (m)
y (m)
z (m)
O (hinge)
0
0
0
a (boom tip)
8
6
0
b (wall anchor)
0
18
7
Find. The cable tension $T_{ab}$, and the reaction force $\mathbf{R}_O$ and reaction couple moment $\mathbf{M}_O$ at the hinge, all as Cartesian vectors.
Figure 1 (reconstructed). Boom O–a in the x-y plane, weight hanging at a, cable a–b to the wall anchor.
Approach. Because the hinge at O is a true pin about the z-axis it carries no moment about z, so summing moments about O and setting the z-component to zero solves directly for the cable tension; force and moment equilibrium then give the reaction at O.
Unit vector along the cable. $\overrightarrow{ab}=b-a=(-8,\,12,\,7)\ \text{m}$, $|\overrightarrow{ab}|=\sqrt{8^2+12^2+7^2}=\sqrt{257}=16.03\ \text{m}$, so $\mathbf{u}_{ab}=(-0.4990,\,0.7485,\,0.4366)$.
Moment of the weight about O. With $\mathbf{W}=(0,-800,0)\ \text{N}$ and $\mathbf{r}_a=(8,6,0)\ \text{m}$: $\mathbf{r}_a\times\mathbf{W}=(0,\,0,\,-6400)\ \text{N}\cdot\text{m}$ (purely about z — a vertical force through a point in the x-y plane produces no moment about a vertical axis).
Moment of the cable force about O, and the pin condition. $\mathbf{r}_a\times\mathbf{u}_{ab}=(2.620,\,-3.493,\,8.982)$, so the z-component of the total applied moment is $-6400+8.982\,T_{ab}$. Since the hinge resists no moment about z, this must vanish: $$T_{ab}=\dfrac{6400}{8.982}=712.5\ \text{N}$$ so $\boxed{T_{ab}=712.5\ \text{N}}$, and $\mathbf{F}_{cable}=T_{ab}\mathbf{u}_{ab}=(-355.6,\,533.3,\,311.1)\ \text{N}$.
Reaction force at O (force equilibrium). $\mathbf{R}_O=-(\mathbf{F}_{cable}+\mathbf{W})=-\big[(-355.6,533.3,311.1)+(0,-800,0)\big]$ $$\boxed{\mathbf{R}_O=(355.6\,\mathbf{i}+266.7\,\mathbf{j}-311.1\,\mathbf{k})\ \text{N}},\quad |\mathbf{R}_O|=542.5\ \text{N}$$
Reaction moment at O (moment equilibrium). The total applied moment is $\mathbf{r}_a\times\mathbf{W}+T_{ab}(\mathbf{r}_a\times\mathbf{u}_{ab})=(1866.7,\,-2488.9,\,0)\ \text{N}\cdot\text{m}$ (the z-component is exactly zero, confirming the pin condition was applied consistently). The reaction is the negative of this: $$\boxed{\mathbf{M}_O=(-1866.7\,\mathbf{i}+2488.9\,\mathbf{j})\ \text{N}\cdot\text{m}},\quad |\mathbf{M}_O|=3111.1\ \text{N}\cdot\text{m}$$
