Question 2 of 6: Question 2 (paper Question II) — 3-D Equilibrium of a Door (Part A · Statics, 20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination, December 2013 — Statics and Dynamics (04-BS-3). Closed book (one self-prepared 8½×11 in. sheet of notes permitted). Candidates were instructed to complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); every question is solved below as a complete study resource.
Check — the AI-vision figure caption placed A on the z-axis with the door lying flat in the horizontal xy-plane, G at 3 ft from A along the direction θ = 45° from the +y axis, B diametrically opposite G through A (also 3 ft from A), and C exactly 3 ft vertically below B. Taken together this makes strut CB a purely vertical two-force member. The exact axis-rotation sense (whether θ is measured toward +x or -x) was not recoverable from the caption; the sign of Ax, Ay is unaffected either way since both terms cancel identically (shown in Step 2), so the reconstruction does not change the answer.
Given. Door weight W = 65 lb (acts at G, vertically down); A the hinge, at the origin; AG = AB = 3 ft in the horizontal plane, with θ = 45° between +y and AG; B diametrically opposite G through A; C is 3 ft vertically below B (strut CB vertical).
Find. The reaction components Ax, Ay, Az at the hinge, and the force in strut CB.
Figure 2 — door geometry: A at the hinge, G the centre of gravity, B/C the strut attachment (oblique 3-D sketch, not to scale).
Approach. Position vectors are read directly from the geometry; take moments about the hinge A to isolate the one unknown strut force, then sum forces to get the hinge reaction.
Position vectors from A.
$$\mathbf{r}_G = 3(\sin45^\circ\,\mathbf{i}+\cos45^\circ\,\mathbf{j}) = (2.121,\ 2.121,\ 0)\ \text{ft}, \qquad \mathbf{r}_B = -\mathbf{r}_G = (-2.121,\ -2.121,\ 0)\ \text{ft}$$
The strut direction (C to B) is $\mathbf{u}_{CB} = (0,0,1)$ (purely vertical, since C is directly below B).
Moment equilibrium about A (weight $\mathbf{W}=(0,0,-65)$ lb, strut force $\mathbf{F}=F_{CB}\,\mathbf{u}_{CB}$ acting at B):
$$\sum \mathbf{M}_A = \mathbf{r}_G\times \mathbf{W} + \mathbf{r}_B\times\mathbf{F} = \mathbf{0}$$
Both cross products reduce to horizontal-axis moments only (no z-component, since both W and F are vertical); expanding the x- and y-components gives the SAME scalar condition (because $\mathbf{r}_B=-\mathbf{r}_G$):
$$-65\,\mathbf{r}_G\times\mathbf{k} - F_{CB}\,\mathbf{r}_G\times\mathbf{k} = 0 \Rightarrow F_{CB} = \boxed{-65\ \text{lb}}$$
The negative sign (force on the door at B acting in $-\mathbf{k}$) means the strut is in tension, magnitude 65 lb — it pulls B down toward C, which is exactly what is needed to hold up the side of the door diametrically opposite the overhanging weight at G.
Force equilibrium for the hinge reaction.
$$\sum \mathbf{F}=0:\quad \mathbf{A} + \mathbf{W} + \mathbf{F} = 0 \Rightarrow \mathbf{A} = -\mathbf{W}-\mathbf{F} = -(0,0,-65) - (0,0,-65)$$
$$A_x = \boxed{0}, \qquad A_y = \boxed{0}, \qquad A_z = \boxed{130\ \text{lb}}$$
Both W and F act purely vertically for this geometry, so the hinge carries no horizontal load at all — it simply supports the remaining 130 lb of vertical weight not carried by the strut.
