Question 2 of 6: Question 2 (paper Question II) — Equivalent Force-Moment System (Part A · Statics, 20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination December 2015 — 04-BS-3, Statics and Dynamics, 3 hours, closed book (one 8½″×11″ self-prepared note sheet permitted; Casio or Sharp approved calculator only). Candidates complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); all 6 are solved below.
Check — Figure 1 is drawn on a grid of 1 division = 1 m; node positions are measured against it, and each figure's geometry is stated at first use. All six printed questions (I–VI) are answered below.
Question 2 (paper Question II) — Equivalent Force-Moment System (Part A · Statics, 20 marks)
Check — the figure gives C directly (100 mm up the post centreline from the base, so C(0,100,0) mm with the drawn y-axis vertical). D sits at the end of the short horizontal bend at the top of the post, height 300 mm (the printed vertical bracket spans from D's level down to the base), so D(0,300,0) mm; the 19.5 kN force's line of action is drawn from D through a ground-plane point E, located by the two horizontal dimensions 125 mm (x) and 150 mm (z), giving E(125,0,150) mm. This is the standard Hibbeler-style construction for a force defined by two points on its line of action.
Given. Post centreline on the y-axis; C(0,100,0) mm; D(0,300,0) mm; force of magnitude 19.5 kN applied at D, directed from D toward E(125,0,150) mm.
Given data
Point
x (mm)
y (mm)
z (mm)
C (moment centre)
0
100
0
D (force applied)
0
300
0
E (line-of-action point)
125
0
150
Find. The equivalent force F and couple moment MC acting at C.
Figure 2 — post centreline (y-axis), axes relocated to C, and the 19.5 kN force line of action D→E.
Approach. Build the unit vector along DE, scale by 19.5 kN to get F (translating a force along its own line of action changes nothing, so this F is also the equivalent force at C); then take the moment of that force about C using rC→D × F.
Unit vector along the line of action. $\mathbf{r}_{DE}=E-D=(125,-300,150)\ \text{mm}$, $|\mathbf{r}_{DE}|=\sqrt{125^2+300^2+150^2}=357.94\ \text{mm}$.
$$\hat{\mathbf{u}}=\dfrac{(125,-300,150)}{357.94}=(0.3492,-0.8381,0.4191)$$
Equivalent force at C. Scale the unit vector by the 19.5 kN magnitude:
$$\mathbf{F}=19.5\,\hat{\mathbf{u}}=\boxed{(6.81,\,-16.34,\,8.17)\ \text{kN}}, \qquad |\mathbf{F}|=19.5\ \text{kN}$$
Position vector from C to D (the point of application).
$$\mathbf{r}_{C\to D}=D-C=(0,\,300-100,\,0)=(0,\,0.200,\,0)\ \text{m}$$
Couple moment at C.
$$\mathbf{M}_C=\mathbf{r}_{C\to D}\times\mathbf{F}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&0.200&0\\6.81&-16.34&8.17\end{vmatrix}\ \text{kN}\cdot\text{m}$$
$$\mathbf{M}_C=\big(0.200(8.17)-0,\ -(0-0),\ 0-0.200(6.81)\big)=\boxed{(1.634,\,0,\,-1.362)\ \text{kN}\cdot\text{m}}$$
with magnitude $|\mathbf{M}_C|=\sqrt{1.634^2+1.362^2}=\boxed{2.127\ \text{kN}\cdot\text{m}}$.
Part (b): the equivalent system at C is the single force F = (6.81, −16.34, 8.17) kN acting through C, together with a free couple moment MC = (1.634, 0, −1.362) kN·m. Both vectors are drawn with their tails at the relocated origin C: F points generally downward and out of the post (large −y, smaller +x and +z components), while MC has no y-component (it cannot twist the post about its own axis, since the moment arm C→D is itself parallel to y) and instead bends the post about the x- and z-axes simultaneously.
