Question 2 of 6: Question 2 (paper Question II) — Boom with Ball-and-Socket and Pulley Cable (Part A · Statics, 20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination May 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.
Coordinates were reconstructed, with A as the origin: boom horizontal along y at A's level, $B=(0,10,0)$ m, load point $H=(0,5,0)$ m (the two 5 m runs); mast anchor $D=(0,0,5)$ m (the 5 m rise); ceiling anchor $C=(0,4,9)$ m (the two 4 m offsets, taken in the same y-z plane as the boom); pulley bracket $E=(0,10,2)$ m (2 m above the boom tip B, matching the vertical "2 m" leader). This coplanar (x = 0) reading makes the moment-about-A equation self-consistent in all three components (the y- and z-moment components both vanish identically, leaving one equation for one unknown T) — strong corroborating evidence for the reconstruction, since an arbitrary geometry would not close this way.
Given. Ball-and-socket joint at A; rigid boom AB, 10 m long, horizontal; downward force F = 1.5 kN applied at H, the midpoint of AB; a single cable runs from a fixed anchor at D, over a pulley bracket E mounted 2 m above B, to a second fixed anchor at C (tension T uniform throughout, frictionless pulley).
Given data (coordinates in m, A at the origin)
Point
x
y
z
A
0
0
0
B
0
10
0
H
0
5
0
D
0
0
5
C
0
4
9
E
0
10
2
Find. The x, y, z components of the reaction at A and the tension T in cable DEC.
Figure 2 — boom AB with ball-and-socket at A, load F at H, and pulley cable D–E–C (reconstructed, coplanar y-z geometry — see check note).
Approach. Take moments about A to eliminate the ball-socket reaction; the cable's net force at E (two segments of equal tension T pulling toward D and toward C) supplies the only other moment about A, giving one equation for T. Then close the force balance for the reaction at A.
Unit vectors along the cable segments.
$$\vec{ED} = D-E = (0,-10,3)\ \text{m},\ |\vec{ED}|=10.440\ \text{m} \Rightarrow \hat{u}_{ED}=(0,-0.9578,0.2874)$$
$$\vec{EC} = C-E = (0,-6,7)\ \text{m},\ |\vec{EC}|=9.220\ \text{m} \Rightarrow \hat{u}_{EC}=(0,-0.6508,0.7593)$$
Moment of the load F about A. $\vec{r}_{H/A}=(0,5,0)$, $\vec{F}=(0,0,-1.5)$ kN:
$$\vec{M}_F = \vec{r}_{H/A}\times\vec{F} = (-7.500,\ 0,\ 0)\ \text{kN}\cdot\text{m}$$
Moment of the cable tension about A, per unit T. $\vec{r}_{E/A}=(0,10,2)$, net cable direction $\hat{u}_{ED}+\hat{u}_{EC}=(0,-1.609,1.047)$:
$$\vec{r}_{E/A}\times(\hat{u}_{ED}+\hat{u}_{EC}) = (13.68,\ 0,\ 0)\ \text{m (per unit T)}$$
Solve for T. $\sum \vec{M}_A = \vec{M}_F + T(13.68,0,0) = 0$:
$$T = \dfrac{7.500}{13.68} = \boxed{0.548\ \text{kN}}$$
The y- and z-components of $\sum \vec{M}_A$ are identically zero (both $\vec{M}_F$ and the cable-moment term have only an x-component here), confirming the moment equation is consistent.
Reaction at A — force balance. $\sum \vec{F}=0$: $\vec{A}+\vec{F}+T(\hat{u}_{ED}+\hat{u}_{EC})=0$:
$$\vec{A} = -\vec{F} - T(\hat{u}_{ED}+\hat{u}_{EC}) = (0,\ 0.882,\ 0.926)\ \text{kN}$$
$$A_x=\boxed{0}, \quad A_y=\boxed{0.882\ \text{kN}}, \quad A_z=\boxed{0.926\ \text{kN}}, \quad |\vec{A}|=1.279\ \text{kN}$$
