Question 3 of 4

Question 3 (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. Right arm pulling a stiff gear lever handle at H; forearm EH horizontal, 350 mm; upper arm SE, 300 mm, at 60° above the leftward horizontal from E. Force applied by the hand on the lever: 220 N horizontal (toward the driver) + 60 N vertical (up); the hand also applies a 5 N·m clockwise moment to the handle at H.

Find. a) the free-body diagram of the forearm+upper arm; b) the elbow joint moment; c) the shoulder joint moment; d) the active muscle group at each joint.

[Figure not reproduced: Figure 3 (redrawn) — free-body diagram of the forearm+upper arm: the reaction force Fₕ and moment Mₕ applied to the hand at H by the lever (Newton's third law reaction to the driver's 220 N/60 N/5 N·m applied loads), and the unknown reaction force Rₛ and. See the official exam paper.]

a) Free-body diagram. Cutting the arm at the shoulder (diagram above) isolates the forearm+upper arm as one rigid free body loaded by: the known reaction force and moment transmitted to the hand at H (equal and opposite to the 220 N/60 N/5 N·m the driver applies to the lever), and the unknown joint reaction force R_S and moment M_S the torso applies at the shoulder. (Segment weight is excluded — no mass data is given.) For part b), the same H load is used on a smaller free body — the forearm+hand alone, cut at the elbow — with the unknown reaction now at E instead of S.

Approach. By Newton's third law, the lever exerts on the hand the exact reverse of what the hand applies to it: $\mathbf{F}_H=(220,-60)\ \text{N}$ (right, down) and $M_H=+5\ \text{N}\cdot\text{m}$ (CCW). Taking moments about each joint in turn (elbow, then shoulder) and adding the couple $M_H$ directly (a couple's moment is the same about any point) gives that joint's resultant moment.

1. Set up coordinates at the elbow E. Forearm horizontal: $H=(0.350,\,0)$ m. Upper arm 60° above the leftward horizontal at E, i.e. at standard angle $180^\circ-60^\circ=120^\circ$: $$S = 0.300(\cos120^\circ,\ \sin120^\circ) = (-0.150,\ 0.260)\ \text{m (relative to E)}$$
2. b) Elbow moment. $M_{elbow}=r_{E\to H,x}F_{H,y}-r_{E\to H,y}F_{H,x}+M_H$: $$M_{elbow} = (0.350)(-60) - (0)(220) + 5.0 = -21.0+5.0 = \boxed{-16.0\ \text{N}\cdot\text{m}\ (16.0\ \text{N}\cdot\text{m, clockwise/flexor)}}$$
3. c) Shoulder moment. $\mathbf{r}_{S\to H} = H-S = (0.500,\ -0.260)$ m: $$M_{shoulder} = (0.500)(-60) - (-0.260)(220) + 5.0 = -30.0+57.2+5.0 = \boxed{32.2\ \text{N}\cdot\text{m}\ (\text{CCW/extensor})}$$
4. d) Active muscles. Pulling the lever toward the body flexes the elbow against the lever's resistance, so the elbow flexors (biceps brachii, brachialis, brachioradialis) must be active to supply the 16.0 N·m flexor moment found in b). At the shoulder, drawing the extended arm back toward the body is a shoulder-extension/adduction action, so the shoulder extensors/adductors (latissimus dorsi, posterior deltoid, teres major) supply the 32.2 N·m found in c).