Question 1 of 4: Free-Body Analysis of the Pulling Arm

Question 1: 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.

(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 1 (redrawn): forearm/upper-arm link-segment model with the reaction force 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 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.

1. (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, $$M_{elbow} = -M_H-(\vec r_{H/E}\times \vec F_H)_z = -5.0-(-28.0) = \boxed{23.0\ \text{N}\cdot\text{m}\ \text{(CCW, flexor sense)}}$$
2. (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 $\vec r_{H/S}=H-S=(0.500,\,-0.2598)\ \text{m}$. Substituting connectors: the same reaction load at H now has a longer, tilted moment arm about S: $$(\vec r_{H/S}\times \vec F_H)_z = (0.500)(-80)-(-0.2598)(120) = -40.0+31.18=-8.82\ \text{N}\cdot\text{m}$$ $$M_{shoulder} = -M_H-(\vec r_{H/S}\times \vec F_H)_z = -5.0-(-8.82) = \boxed{3.82\ \text{N}\cdot\text{m}\ \text{(CCW, flexor sense)}}$$

(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.