Question 2 (paper Question II): 04-BS-3, December 2014, Q2

Question 2 (paper Question II) (20 marks)

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.

Given. Origin O at the base of the vertical post. Arm end $A=(1.6,\,0,\,2)$ m; cable anchor $B=(1.5,\,0.8,\,0)$ m (both relative to O). Cable tension $T=1.2$ kN, directed from A to B.

Find. The equivalent force–couple system $(\mathbf F,\ \mathbf M_O)$ acting at O.

Figure 2 — post, arm and cable, with the equivalent system to be reported at O.

Approach. Build a unit vector along the cable from the two given position vectors, scale by the tension to get $\mathbf F$, then take $\mathbf M_O=\mathbf r_{OA}\times\mathbf F$ (the moment of a force is independent of where along its line of action it is applied, so using $\mathbf r_{OA}$ — the position vector to A, where the cable leaves the arm — is valid).

Cable unit vector. $$\mathbf r_{AB} = B-A = (1.5-1.6,\ 0.8-0,\ 0-2) = (-0.1,\ 0.8,\ -2)\ \text{m}, \qquad |\mathbf r_{AB}| = \sqrt{0.1^2+0.8^2+2^2}=2.156\ \text{m}$$ $$\hat u_{AB} = \frac{1}{2.156}(-0.1,\ 0.8,\ -2) = (-0.0464,\ 0.3710,\ -0.9275)$$
Equivalent force at O. $$\mathbf F = T\hat u_{AB} = 1.2(-0.0464,\ 0.3710,\ -0.9275)$$ $$\boxed{\mathbf F = (-0.0556,\ 0.4452,\ -1.1130)\ \text{kN}, \qquad |\mathbf F| = 1.2\ \text{kN}}$$
Equivalent moment at O. With $\mathbf r_{OA}=(1.6,\,0,\,2)$ m, $$\mathbf M_O = \mathbf r_{OA}\times\mathbf F = \begin{vmatrix}\mathbf i & \mathbf j & \mathbf k\\ 1.6 & 0 & 2\\ -0.0556 & 0.4452 & -1.1130\end{vmatrix}$$ $$M_x = (0)(-1.1130)-(2)(0.4452) = -0.8904$$ $$M_y = (2)(-0.0556)-(1.6)(-1.1130) = 1.6693$$ $$M_z = (1.6)(0.4452)-(0)(-0.0556) = 0.7123$$ $$\boxed{\mathbf M_O = (-0.890,\ 1.669,\ 0.712)\ \text{kN}\cdot\text{m}, \qquad |\mathbf M_O| = 2.022\ \text{kN}\cdot\text{m}}$$
Draw and label at O (part b). The equivalent system replacing the cable is the single force $\mathbf F=(-0.056,\,0.445,\,-1.113)$ kN acting through O together with the free couple moment $\mathbf M_O=(-0.890,\,1.669,\,0.712)\ \text{kN}\cdot\text{m}$ shown as a double-headed moment arrow at O — the force pulls the post toward $-x,+y,-z$ (down and toward the cable anchor) while the couple represents the net twisting/bending effect the offset cable exerts about the post base.

Equivalent force–moment system at O
Quantity	x	y	z	Magnitude
Force (kN)	-0.056	0.445	-1.113	1.200 kN
Moment (kN·m)	-0.890	1.669	0.712	2.022 kN·m

Topic: Equivalent force–couple systems in 3-D (Cartesian vector method).

Key relations: $\hat u = \dfrac{\mathbf r_{AB}}{|\mathbf r_{AB}|}$; $\mathbf F=T\hat u$; $\mathbf M_O=\mathbf r\times\mathbf F$ (moment of a force about a point).

Why this works: Any force may be replaced by an equal, parallel force acting at a different point plus a couple that restores the original moment about that point — this is exactly what "equivalent system at O" asks for. Expressing the cable force with Cartesian components avoids ever computing an explicit angle, which is why the vector method is preferred for skew 3-D members.

Common pitfall: Using the wrong position vector in the cross product (must run from the moment centre O to a point ON the force's line of action — either A or B works, but mixing $\mathbf r_{OB}$ with a force built from $A\to B$ direction gives a sign/direction error).

Source: Hibbeler, Engineering Mechanics: Statics, Ch. 4 (Force System Resultants — Moment of a Force, Equivalent Systems).

Question 2 of 6: Question 2 (paper Question II)

Question 2 (paper Question II) (20 marks)