Bucket Supported by Three Cables (Part A — Statics): 04-BS-3, May 2013, Q2

Question II: Bucket Supported by Three Cables (Part A — Statics) (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. Bucket weight $W=50$ lb hangs from ring D, held by three cables to fixed points A, B, C. From the figure's dimensioned axes (feet, D taken as local origin): A(3, 0, 4.5), B(−1.5, −1.5, 0), C(0, 2.5, 3).

Given data (ft, relative to D)
Point	x	y	z
A	3	0	4.5
B	−1.5	−1.5	0
C	0	2.5	3
D	0	0	0

Figure 2 — three cables converge at ring D; bucket weight acts vertically down at D.

Check: the signs of the dimensions at the A/B corner are not explicit in the figure; the (x,y,z) reading above is the unique sign choice (of those consistent with the printed magnitudes 4.5, 3, 2.5, 1.5, 1.5 ft) that makes all three cable tensions come out positive, as physically required for flexible cables.

Find. The tension in each of cables AD, BD, CD.

Approach. Write a Cartesian unit vector along each cable from D toward its anchor, then solve the vector equilibrium $T_{AD}\hat u_{AD}+T_{BD}\hat u_{BD}+T_{CD}\hat u_{CD}+(0,0,-W)=0$ at ring D as three scalar equations in the three unknown tensions.

Unit vectors. $|AD|=\sqrt{3^2+4.5^2}=5.408$ ft, $\hat u_{AD}=(0.5547,0,0.8321)$. $|BD|=\sqrt{1.5^2+1.5^2}=2.121$ ft, $\hat u_{BD}=(-0.7071,-0.7071,0)$. $|CD|=\sqrt{2.5^2+3^2}=3.905$ ft, $\hat u_{CD}=(0,0.6402,0.7682)$.
Assemble and solve the 3×3 system. Requiring $\Sigma F_x=\Sigma F_y=\Sigma F_z=0$ at D, $$\begin{pmatrix}0.5547 & -0.7071 & 0\\ 0 & -0.7071 & 0.6402\\ 0.8321 & 0 & 0.7682\end{pmatrix}\begin{pmatrix}T_{AD}\\T_{BD}\\T_{CD}\end{pmatrix}=\begin{pmatrix}0\\0\\50\end{pmatrix}$$ Solving gives $$T_{AD}=\boxed{33.4\text{ lb}}, \qquad T_{BD}=\boxed{26.2\text{ lb}}, \qquad T_{CD}=\boxed{28.9\text{ lb}}$$
Check. Substituting back, $T_{AD}\hat u_{AD}+T_{BD}\hat u_{BD}+T_{CD}\hat u_{CD}=(0,0,50)$ to within rounding, and all three values are positive (tension, as a cable must be) — both confirm the solution.

Cable	Tension (lb)
AD	33.4
BD	26.2
CD	28.9

Concept & Theory

Topic. Three-dimensional particle equilibrium using Cartesian vector notation.

Key relations. Position vector $\mathbf r=(x,y,z)$; unit vector $\hat u=\mathbf r/|\mathbf r|$; force vector $\mathbf F=T\hat u$; equilibrium $\Sigma\mathbf F=0$ resolved into $\Sigma F_x=\Sigma F_y=\Sigma F_z=0$.

Why this works. A ring/particle in equilibrium under concurrent forces has no moment equation to satisfy — only force balance — so three scalar equations exactly determine three unknown cable tensions, provided the three cable directions are not coplanar (they span 3-D here).

Common pitfall. Using position vectors measured from the wrong end (anchor to D instead of D to anchor) silently flips every sign; and skipping the positive-tension sanity check hides an inconsistent axis reading, since a cable cannot push.

Source. Hibbeler, Engineering Mechanics: Statics (14th ed.), Ch. 3 — Force Vectors, Cartesian Vector Formulation.

Question 2 of 6: Bucket Supported by Three Cables (Part A — Statics)

Question II: Bucket Supported by Three Cables (Part A — Statics) (20 marks)