Free-Body Analysis of the Pulling Arm: 20-Bio-A4 Anatomy and Physiology, Undated paper, Q4

Question 4: Free-Body Analysis of the Pulling Arm (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.

The two solutions were derived independently and agree exactly (elbow moment 23.0 N·m, shoulder moment 3.82 N·m), which cross-checks both.

(a) Isolating the forearm (E–H) and the upper arm (S–E) as a two-link chain, the only external load is applied at H. By Newton's third law the handle pushes back on the hand with a force equal and opposite to what the hand applies to it — 120 N to the right and 80 N downward — plus a 5 N·m counter-clockwise (CCW) reaction moment. Each joint (E, S) then carries an internal reaction force and moment supplied by the proximal segment to keep its distal neighbour in equilibrium; the diagram below shows the reaction load at H and the two computed joint moments (Steps 1–2).

[Figure not reproduced: Figure 4 (redrawn): forearm/upper-arm link-segment model with the reaction load at H and the solved elbow (M_E) and shoulder (M_S) joint moments, both counter-clockwise. See the official exam paper.]

Given. Applied hand-to-lever load: 120 N left, 80 N up, 5 N·m clockwise, all at H. Forearm E→H = 0.350 m, horizontal. Upper arm S→E = 0.300 m at 60° above the leftward (backward) horizontal through E.

Find. The net internal moment at the elbow (b) and at the shoulder (c).

Approach. Apply Newton's third law to get the reaction load on the hand, then sum moments about each joint in turn — first for the forearm alone (elbow), then for the whole limb (shoulder), since the elbow reaction becomes internal once the free body is enlarged.

(b) Elbow moment — forearm free body about E. Reaction on the hand at H: $\vec F_H=(+120,-80)\ \text{N}$, $M_H=+5\ \text{N}\cdot\text{m}$ (CCW). With $\vec r_{H/E}=(0.350,0)\ \text{m}$: $$\sum M_E = M_{elbow}+M_H+(\vec r_{H/E}\times \vec F_H)_z = 0$$ $$(\vec r_{H/E}\times \vec F_H)_z = (0.350)(-80)-(0)(120) = -28.0\ \text{N}\cdot\text{m}$$ Solving, $$M_{elbow} = -M_H-(\vec r_{H/E}\times \vec F_H)_z = -5.0-(-28.0) = \boxed{23.0\ \text{N}\cdot\text{m}\ \text{(CCW, flexor sense)}}$$
(c) Shoulder moment — whole-limb free body about S. The elbow reaction is now internal, so only the reaction at H and the shoulder's own moment remain. Placing E at the origin, $S=(-0.150,\,+0.2598)\ \text{m}$, so $\vec r_{H/S}=H-S=(0.500,\,-0.2598)\ \text{m}$. The same reaction load at H now has a longer, tilted moment arm about S: $$(\vec r_{H/S}\times \vec F_H)_z = (0.500)(-80)-(-0.2598)(120) = -40.0+31.18=-8.82\ \text{N}\cdot\text{m}$$ $$M_{shoulder} = -M_H-(\vec r_{H/S}\times \vec F_H)_z = -5.0-(-8.82) = \boxed{3.82\ \text{N}\cdot\text{m}\ \text{(CCW, flexor sense)}}$$

Final results — Question 4
Quantity	Value
Elbow joint moment, $M_{elbow}$	23.0 N·m (CCW, flexor sense)
Shoulder joint moment, $M_{shoulder}$	3.82 N·m (CCW, flexor sense)

(d) Both solved joint moments act in the flexion sense — they resist the reaction load's tendency to extend (straighten) the limb as the driver pulls the lever toward himself. The elbow moment (23.0 N·m) is carried by the elbow flexors: biceps brachii and brachialis, with brachioradialis as a synergist. The smaller shoulder moment (3.82 N·m) is carried by the shoulder flexors/horizontal adductors: anterior deltoid, pectoralis major and coracobrachialis, contracting isometrically to hold the upper arm against the transmitted reaction load.

Topic: Sequential free-body analysis of a linked-segment limb chain.

Key relations: $\sum F=0$, $\sum M=0$ per segment; Newton's third law converts an applied external force/moment into the reaction the limb must resist.

Why this works: treating each segment as an independent rigid body isolates exactly the one unknown joint reaction at that joint. Because the segments are pinned in series, the distal joint's reaction becomes internal (self-cancelling) once the free body is enlarged to include it, leaving only the reaction at the still-external contact point (H) and the moment at the new, more proximal joint.

Common pitfall: forgetting that the given load is what the HAND applies to the lever, not what the lever applies back to the hand — skipping the Newton's-third-law sign flip understates every downstream joint moment; a second common error is a sign slip in the 2-D cross product $(r_x F_y - r_y F_x)$.

Source: Winter, Biomechanics and Motor Control of Human Movement, ch. 3 (link-segment/free-body method); Nordin & Frankel, Basic Biomechanics of the Musculoskeletal System, ch. 8 (joint reaction forces).

Question 4 of 4: Free-Body Analysis of the Pulling Arm

Question 4: Free-Body Analysis of the Pulling Arm (15 marks)