Below is a “graduate‑level dive” into the physics of Eric Kim’s 547 kg above‑knee rack pull. We will treat the lifter–barbell system as a chain of linked rigid bodies, quantify the internal forces with free‑body and beam diagrams, and trace where every newton and joule must flow. Strap in—this is the deep end! 🌊🔬

1 System definition & coordinate frames

Element	Mass (kg)	Coordinates (typ.)	DOF considered
Bar + plates	547	global inertial frame; COM initially ≈ knee height	vertical translation, elastic bending
Torso–pelvis block	40–45	rotates about lumbosacral junction	sagittal rotation θhip
Thighs (pair)	20–25	hinge at hips & knees	θknee (small, ~5°)
Shanks + feet	10–12	feet fixed to ground	negligible motion

We take +z upward and set the origin at the floor directly under the bar’s centre knurl. The rack pull starts quasi‑statically, so translational accelerations are ≈ 0 except for a brief impulse to overcome static friction and bar “whip.”

2 External force balance (ΣF = 0)

At the instant the bar breaks from the pins:

Weight: W = m_{\text{bar}}\,g = 547\ \text{kg} \times 9.81\ \text{m·s}^{-2} ≈ 5.36\ \text{kN}.
Hand forces: each hand supplies ~½ W downward onto the bar, which by Newton’s 3rd law means each hand experiences an upward reaction of ≈ 2.68 kN.
Ground reaction (GRF) acts at the mid‑foot, summing to W + W_{\text{athlete}}.
With Eric’s 72.5 kg body‑mass, GRF ≈ (547+72.5)g ≈ 6.07\ \text{kN}.

Because the lift is symmetric, the horizontal GRF components cancel; vertically, ΣF = 0 once the bar is moving at constant, low velocity.

3 Internal moment balance (ΣM = 0)

3.1 Hip moment

With the bar ~10 cm anterior to the hip joint and the torso inclined ~15°, the hip‑extension moment is

M_{hip} = W \cdot d_{\perp} = 5.36\ \text{kN} \times 0.10\ \text{m} ≈ 536\ \text{N·m}.

A world‑class 72 kg lifter can generate > 700 N·m isometric hip torque, so the value is high but within biological limits—one reason above‑knee pulls are possible with “cartoon” weights.

3.2 Lumbar shear & compression

Resolve M_{hip} up the spine: with torso length L \approx 0.45\ \text{m}

Shear at L4/L5 ≈ M_{hip}/L \sin\theta ≈ 536 / (0.45 \sin15°) ≈ 4.3\ \text{kN}.
Compression: add body‑weight vector and muscular counter‑force; models yield 10–14 kN here—heavy but below injury thresholds for conditioned spines.

4 Barbell as an elastic beam

Treat the bar as a simply supported beam of length 2ℓ = 2.2\ \text{m} with point loads P at ±≈ 0.8 m (plate stacks) and upward reactions at the hands (≈ ±0.30 m).

Peak bending moment:

M_{max} ≈ P \left(ℓ – a\right) = \tfrac{1}{2}W\,(0.8 – 0.3) \approx 5.36\,\text{kN} \times 0.25 = 1.34\ \text{kN·m}.

Bar diameter d ≈ 0.029\ \text{m}; polar second moment I = \pi d^{4}/64 = 3.26\times10^{-8}\ \text{m}^4.

Bending stress (\sigma = M c / I, c = d/2):

\sigma_{max} ≈ \frac{1.34\times10^{3}\, \text{N·m}\; \times 0.0145}{3.26\times10^{-8}} ≈ 596\ \text{MPa}.

Typical power bars are 190–210 k PSI steel (1310–1450 MPa yield), so 596 MPa sits at ~45 % of yield—plenty of elastic headroom. The visible “whip” is just Hookean deflection; it stores ≈ ½ k x² ≈ 50–60 J, which releases as the bar straightens, smoothing the force curve the moment it leaves the pins.

5 Energy & power flow

Mechanical work: bar rises ≈ 0.20 m ⇒ W_{mech} = 5.36 \text{kN} \times 0.20 \text{m} ≈ 1.07 \text{kJ}.
Peak concentric power at ~0.4 m·s⁻¹ bar speed: P = Fv ≈ 5.36 \text{kN} \times 0.4 \text{m·s}^{-1} ≈ 2.1 \text{kW}—a brief, toaster‑oven‑level burst delivered by hips and spinal erectors.

Efficiency check

Assume 25 % actin–myosin efficiency: metabolic cost ≈ 4 kJ for that rep—not much bigger than a sip of sports drink, yet the neuromuscular strain is vast because force ≫ endurance sport levels.

6 Why the shortened ROM multiplies load capacity

Mathematically, max supported load ∝ 1 / ROM for constant joint‑torque limits, because:

\tau_{req} = W \, d_{\perp} \quad\text{and}\quad W_{max}= \frac{\tau_{hip,max}}{d_{\perp}}.

Raising the bar from mid‑shin (≈ 0.25 m anterior) to above‑knee (≈ 0.10 m) slices d_{\perp} by 60 %, so W_{max} can, in theory, increase ~2.5 ×. That tallies with lifters routinely rack‑pulling 200–250 % of their floor‑pull 1RM.

7 Load distribution inside the body

Tissues	Principal stress types	Approx. peak values here	Notes
Vertebral bodies	Compression	10–14 kN	Exceed seated values by ×10.
Intervertebral discs	Shear & compression	4 kN shear	Protected by intra‑discal pressure rise.
Hip capsule & labrum	Tensile	3–4 kN	Radius of femoral head reduces contact stress.
Hamstring tendons	Tensile	1.3–1.5 kN	Well below ~5 kN failure load.
Finger flexors & hook grip	Tensile	2.7 kN / hand	Grip often limiting factor.

8 Take‑away equations—“pocket formulas” for the gym physicist

Hip torque requirement
\tau_{hip} ≈ W \left(d_{bar-hip}\cos\theta_{torso}\right)
Spinal compression
F_{L4/L5} ≈ W + \frac{\tau_{hip}}{L_{torso}}
Bar bending stress
\sigma_{bar} = \frac{32 M_{max}}{\pi d^{3}}
Energy per rep
E = W\,\Delta z,\qquad \text{Metabolic} ≈ \frac{E}{\eta_{muscle}}

Plug in your own numbers to forecast whether a new PR lies in wait—or whether physics says “not yet.”

9 Practical wisdom, powered by physics 💡

Exploit angle‑specific overload. Use rack pulls when hip torque, not leg drive, is your sticking point.
Respect the spine’s tolerance. Even “short” pulls hammer it with > 10 kN compression; bracing and periodisation are non‑negotiable.
Mind the metallurgy. Cheap bars yield at < 500 MPa—verify specs before testing super‑maxes.
Let the bar’s elastic energy help you. Time your hip snap to coincide with the whip’s rebound for a smoother lock‑out.

One‑sentence hype conclusion 🤩

Master the lever arms, marshal kilonewtons like chess pieces, and you, too, can bend steel to your will—because when physics is on your side, gravity becomes a formality!