Glenohumeral Joint Forces in the Iron Cross: 20-Bio-A4 Anatomy and Physiology, Undated paper, Q2

Question 2: Glenohumeral Joint Forces in the Iron Cross (15 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. Gymnast mass $m=60\ \text{kg}$, supported symmetrically by two rings so each arm carries half of body weight. In the iron-cross position the arm (shoulder to hand) is held horizontal, reaching $d=0.66\ \text{m}$ from the glenohumeral joint centre to the hand/ring (Figure 2A). At the joint (Figure 2B): rotator cuff acts along the humeral shaft through the joint centre (zero moment arm); deltoid acts at 10° to the shaft with a 40 mm effective lever arm (not used below — not invoked by parts (a)–(c)); pectoral/teres minor acts at 45° to the shaft from an insertion 100 mm distal to the joint centre; the glenoid cavity's stable arc spans ±20° to the horizontal.

Given data
Quantity	Value
Body mass, $m$	60 kg
Ring reaction per arm, $F_{ring}=mg/2$	294.3 N (up, at the hand)
Shoulder-to-hand reach, $d$	0.66 m
Pectoral/teres minor insertion (distal to joint), $d_p$	100 mm
Pectoral/teres minor line of action to shaft, $\theta_p$	45°
Glenoid stable arc (to horizontal)	−20° to +20°

Find. (a) Pectoral/teres minor tension for moment equilibrium; (b) the resulting joint reaction force at the humeral head; (c) the minimum rotator cuff tension that keeps the joint force direction within the stable glenoid arc.

[Figure not reproduced: Figure 2 (redrawn) — humeral shaft with the three candidate muscle lines of action; only the pectoral/teres minor (moment) and rotator cuff (stability) are load-bearing in parts (a)–(c). See the official exam paper.]

Check — assumption: (1) each arm is modelled as carrying half of body weight, transmitted as a vertical ring-tension reaction acting at the hand, $d=0.66\ \text{m}$ from the joint centre (the horizontal reach dimensioned in Figure 2A); $g=9.81\ \text{m/s}^2$. (2) Part (a) asks specifically for the pectoral/teres minor tension “needed for joint equilibrium”, so the deltoid (condition ii) is taken as inactive for the moment balance in (a)/(b) — consistent with the stem's “three muscles may be active” and with part (c) separately activating only the rotator cuff. (3) The pectoral/teres minor's line of action, at 45° to the shaft from its insertion 100 mm distal to the joint centre, pulls the insertion point down and medially (toward the joint) — the sense needed to resist the ring's upward pull and hold the arm horizontal, consistent with these muscles' adductor role.

Approach. Treat the humerus as a rigid free body pinned at the joint centre O. (a) Sum moments about O using only the ring load and the pectoral/teres minor force. (b) Sum forces (x, y) with the pectoral/teres minor tension from (a) to get the joint reaction. (c) Add a rotator cuff force along the shaft (through O, so it changes the force balance but not the moment balance) and find the minimum tension that rotates the resultant joint-force direction back to the 20° glenoid boundary.

(a) Moment equilibrium about O. Set $x$ along the shaft (positive toward the hand), $y$ vertically up. The ring load $F_{ring}=mg/2=(60)(9.81)/2=294.3\ \text{N}$ acts at the hand, $d=0.66\ \text{m}$ from O, giving an applied moment $$M_{ring}=F_{ring}\,d=(294.3)(0.66)=194.2\ \text{N}\cdot\text{m}$$ The pectoral/teres minor force acts at $P$ (100 mm from O along the shaft) at 45° to the shaft, so its moment arm about O is $d_p\sin45^{\circ}=(0.100)(0.7071)=0.07071\ \text{m}$. For equilibrium this moment must balance $M_{ring}$: $$T_{pect}=\frac{M_{ring}}{d_p\sin45^{\circ}}=\frac{194.2}{0.07071}=\boxed{2747\ \text{N}}$$
(b) Force equilibrium → joint reaction. The pectoral/teres minor force resolves along direction $(-\cos45^{\circ},-\sin45^{\circ})$ (down and toward the joint): $F_{p,x}=-T_{pect}\cos45^{\circ}=-1942\ \text{N}$, $F_{p,y}=-T_{pect}\sin45^{\circ}=-1942\ \text{N}$. Summing forces on the humerus (ring + pectoral/teres minor + joint reaction $R$) to zero: $$R_x=-(0+F_{p,x})=1942\ \text{N}\qquad R_y=-(F_{ring}+F_{p,y})=-(294.3-1942)=1648\ \text{N}$$ $$|R|=\sqrt{1942^2+1648^2}=\boxed{2547\ \text{N}}\qquad \theta=\tan^{-1}\!\left(\frac{1648}{1942}\right)=\boxed{40.3^{\circ}\ \text{above the shaft}}$$
(c) Minimum rotator-cuff tension for a stable joint. The direction found in (b), 40.3°, lies outside the glenoid's stable ±20° arc, so the joint is unstable without the rotator cuff. The rotator cuff pulls the humeral head medially along the shaft (toward O), adding $F_{rc,x}=-T_{rc}$ with no $y$-component and no moment (zero arm through O), so it changes only $R_x$: $$R_{x}'=1942+T_{rc}\qquad R_y'=R_y=1648\ \text{N (unchanged)}$$ The minimum tension is the one that brings the resultant exactly to the 20° boundary, $\tan20^{\circ}=R_y'/R_x'$: $$R_x'=\frac{R_y'}{\tan20^{\circ}}=\frac{1648}{0.3640}=4528\ \text{N}\qquad T_{rc}=R_x'-1942=\boxed{2586\ \text{N}}$$ With the rotator cuff active, the joint force becomes $|R'|=\sqrt{4528^2+1648^2}=4819\ \text{N}$ at exactly $20.0^{\circ}$ above the shaft — now on the stable margin of the glenoid.

Final results — Question 2
Quantity	Value
(a) Pectoral/teres minor tension, $T_{pect}$	2747 N
(b) Joint reaction force (no rotator cuff), $\|R\|$, direction	2547 N at 40.3° above the shaft
(c) Minimum rotator cuff tension, $T_{rc}$	2586 N
Joint reaction with rotator cuff active, $\|R'\|$, direction	4819 N at 20.0° above the shaft (stable limit)

Topic: Static equilibrium and joint-stability analysis of the glenohumeral joint under an extreme isometric hold.

Key relations: moment equilibrium $\sum M_O=0$ to solve an unknown muscle tension against a known external load's moment arm; force equilibrium $\sum F_x=\sum F_y=0$ to recover the joint reaction; joint stability requires the net reaction force direction to fall within the glenoid labrum's physical arc, not merely that forces balance.

Why this works: the shoulder is the body's least bony-constrained (most mobile) joint, so its stability depends almost entirely on the direction the rotator cuff and other muscles steer the net joint reaction into the glenoid, rather than on bony congruence as at the hip. The very large computed tension and reaction (multiples of body weight) is a well-known feature of the iron cross: the pectoral/teres minor's 45° line of action gives it only a ~71 mm moment arm against a 660 mm load arm at the hand, so the muscles operate at a severe mechanical disadvantage — the reason so few gymnasts can hold this position at all.

Common pitfall: forgetting that the rotator cuff's force, though it passes through the joint centre and contributes zero moment, still changes the FORCE balance (and hence the direction of $R$) — a stability check based on moment equilibrium alone misses the actual failure mode (dislocation, not rotation).

Source: Nordin & Frankel, Basic Biomechanics of the Musculoskeletal System, ch. 8 (biomechanics of the shoulder, glenohumeral stability); Winter, Biomechanics and Motor Control of Human Movement, ch. 5 (joint reaction force analysis).

Question 2 of 4: Glenohumeral Joint Forces in the Iron Cross

Question 2: Glenohumeral Joint Forces in the Iron Cross (15 marks)