Question 1 (paper Question I) — Equivalent Force-Moment System for Three Cables (Part A · Statics, equal value): 04-BS-3, December 2016, Q1

Question 1 (paper Question I) — Equivalent Force-Moment System for Three Cables (Part A · Statics, equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Check — coordinate reconstruction. The back mounting face is a 100 mm (height) × 100 mm (depth) square in the plane x = 0; the two stacked 50 mm dimensions locate A at the centre of this face, A(0, 50, 50) mm — a natural reduction point. The wedge-shaped fin tapers from the top-front corner of that face, B(0, 100, 0) mm, forward to its tip D(100, 0, 0) mm (using the bottom 100 mm dimension as the fin's reach); C sits 75 mm along the B→D ridge, C = B + 75 Â· (D−B)/|D−B| = (53.0, 47.0, 0) mm. The 45°/45° pair at C is read as elevation above the horizontal (x–z) plane / azimuth from +x within that plane; the 30°/60° pair at D is the same 2-D direction (complementary angles) lying entirely in the x–y plane. These readings are used consistently below; the method (resolve each cable into a Cartesian force, then sum forces and moments about A) is unaffected by the exact millimetre placement of B and C.

Given. A(0, 50, 50) mm, B(0, 100, 0) mm, C(53.0, 47.0, 0) mm, D(100, 0, 0) mm, E(150, −50, 100) mm. Cable B→E: 600 N. Cable at C: 1200 N, 45° above the horizontal plane and 45° azimuth from +x. Cable at D: 1500 N, 30° from vertical (60° from +x), lying in the x–y plane.

Given data
Point	x (mm)	y (mm)	z (mm)
A (reduction pt.)	0	50	50
B (600 N cable)	0	100	0
C (1200 N cable)	53.0	47.0	0
D (1500 N cable)	100	0	0
E (anchor of B–E cable)	150	−50	100

Find. The resultant force $\mathbf{F}_R$ and resultant couple moment $\mathbf{M}_R$ that replace the three cable forces, acting at A.

Figure 1 — bracket with cables at B (600 N), C (1200 N) and D (1500 N); dashed line shows the B–E cable running to the given anchor point E.

Approach. Write each cable force as a Cartesian vector (direct unit vector for B–E; elevation/azimuth decomposition for the other two), sum the three forces for $\mathbf{F}_R$, then sum $\mathbf{r}\times\mathbf{F}$ about A for $\mathbf{M}_R$.

Cable B→E (600 N). $\mathbf{r}_{E/B}=(150,-150,100)\ \text{mm}$, $|\mathbf{r}_{E/B}|=234.5\ \text{mm}$. $$\mathbf{u}_{BE}=(0.6396,\,-0.6396,\,0.4264)\quad\Rightarrow\quad \mathbf{F}_1=600\,\mathbf{u}_{BE}=\boxed{(383.8,\,-383.8,\,255.8)\ \text{N}}$$
Cable at C (1200 N). Elevation 45° above horizontal, azimuth 45° from +x: $$F_{1y}=1200\sin45^{\circ}=848.5\ \text{N},\quad F_{1,\text{horiz}}=1200\cos45^{\circ}=848.5\ \text{N}$$ $$F_{2x}=848.5\cos45^{\circ}=600\ \text{N},\quad F_{2z}=848.5\sin45^{\circ}=600\ \text{N}\ \Rightarrow\ \mathbf{F}_2=\boxed{(600,\,848.5,\,600)\ \text{N}}$$
Cable at D (1500 N). 30° from vertical / 60° from horizontal, in the x–y plane: $$\mathbf{F}_3=1500(\sin30^{\circ},\,\cos30^{\circ},\,0)=\boxed{(750.0,\,1299.0,\,0)\ \text{N}}$$
Resultant force. Substituting connectors, $\mathbf{F}_R=\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3$: $$\mathbf{F}_R=(1733.8,\,1763.8,\,855.8)\ \text{N},\qquad |\mathbf{F}_R|=\boxed{2617\ \text{N}}$$
Position vectors from A. $\mathbf{r}_{B/A}=(0,50,-50)$, $\mathbf{r}_{C/A}=(53.0,-3.0,-50)$, $\mathbf{r}_{D/A}=(100,-50,-50)$ mm.
Resultant moment about A. Summing $\mathbf{r}\times\mathbf{F}$ for each cable (in N·mm, then converting): $$\mathbf{M}_R=\mathbf{r}_{B/A}\times\mathbf{F}_1+\mathbf{r}_{C/A}\times\mathbf{F}_2+\mathbf{r}_{D/A}\times\mathbf{F}_3=(99162,\,-118508,\,195036)\ \text{N}\cdot\text{mm}$$ $$\mathbf{M}_R=\boxed{(99.2\,\mathbf{i}-118.5\,\mathbf{j}+195.0\,\mathbf{k})\ \text{N}\cdot\text{m}},\qquad |\mathbf{M}_R|=248.8\ \text{N}\cdot\text{m}$$

Equivalent force-moment system at A
Quantity	Value
Resultant force $\mathbf{F}_R$	(1733.8, 1763.8, 855.8) N
$\|\mathbf{F}_R\|$	2617 N
Resultant couple $\mathbf{M}_R$ about A	(99.2, −118.5, 195.0) N·m
$\|\mathbf{M}_R\|$	248.8 N·m

Topic: Reduction of a 3-D force system to an equivalent force-couple system (Cartesian vector method).

Key relations: Cartesian force $\mathbf{F}=F\mathbf{u}$ where $\mathbf{u}$ is the line-of-action unit vector; resultant $\mathbf{F}_R=\sum\mathbf{F}_i$; resultant couple about a point A, $\mathbf{M}_{R/A}=\sum(\mathbf{r}_{i/A}\times\mathbf{F}_i)$.

Why this works: any system of forces (here, three cable tensions with distinct lines of action) is statically equivalent, at any chosen point, to a single force equal to the vector sum plus a couple equal to the sum of the moments — this lets a designer check both translational and rotational tendency of the bracket in one calculation.

Common pitfall: mixing up which reference line an elevation/azimuth angle pair is measured from (horizontal plane vs. a sloped edge), and forgetting the position vector must run from the moment point (A) to the force's own application point, not to some other labelled point on the figure.

Source: Hibbeler, Engineering Mechanics: Statics, 14th ed., Ch. 4 (Force System Resultants), §4.6–4.8.

Question 1 of 6: Question 1 (paper Question I) — Equivalent Force-Moment System for Three Cables (Part A · Statics, equal value)

Question 1 (paper Question I) — Equivalent Force-Moment System for Three Cables (Part A · Statics, equal value)