Question 4 of 4

Question 4 (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. Net external knee load $F_y=650\ \text{N}$ (axial), $M_x=+21\ \text{N}\cdot \text{m}$ (frontal-plane/ab-adduction), $M_z=-23\ \text{N}\cdot\text{m}$ (flexion-extension), with $F_x=F_z=M_y=0$; Figure 4 gives five candidate load-bearing structures, each acting parallel to the tibial long axis (y): the quadriceps/patellar tendon (Quad, 0.04 m anterior of the flexion axis), the lateral ligament L_L (0.05 m lateral), lateral joint contact force J_L (0.02 m lateral), medial joint contact force J_m (0.02 m medial), and medial ligament L_m (0.05 m medial).

[Figure not reproduced: Figure 4 (redrawn) — five candidate axial load paths crossing the knee, viewed from above the tibial plateau: quadriceps/patellar tendon (green, 0.04 m anterior of the flexion axis) and, lateral→medial, the lateral ligament, lateral joint contact, medial joint contact, and medial ligament. See the official exam paper.]

Find. a) Quadriceps tension; b) the M_x and F_y equilibrium equations in the four remaining unknowns; c),d) two solved structure-pairs and a physiological-plausibility check.

Approach. Each structure's force is parallel to y, so its moment about the flexion–extension (z) or ab-adduction (x) axis is simply its own out-of-plane offset times its force magnitude; balancing $M_z$ (using the quad's 0.04 m anterior offset) isolates the quad tension directly, then $M_x$ and $F_y$ give two more equations in the four remaining structural unknowns — statically indeterminate until two of the four are chosen to carry the whole load.

1. a) Quadriceps tension from the M_z balance. Only the quad has a non-zero x-offset (0.04 m, anterior) among the axial structures, so: $$M_{z,ext} + F_{quad}\,(0.04) = 0 \;\Rightarrow\; F_{quad} = \frac{-(-23)}{0.04} = \boxed{575\ \text{N}} \ \text{(tension)}$$
2. b) M_x and F_y equations (four unknowns: L_L, J_L, J_m, L_m). Using the ML offsets (lateral $+$, medial $-$: 0.05, 0.02, $-$0.02, $-$0.05) and the now-known quad tension: $$M_x:\quad 21 - 0.05F_{L_L} - 0.02F_{J_L} + 0.02F_{J_m} + 0.05F_{L_m} = 0$$ $$F_y:\quad 650 + 575 + F_{L_L}+F_{J_L}+F_{J_m}+F_{L_m} = 0 \;\Rightarrow\; F_{L_L}+F_{J_L}+F_{J_m}+F_{L_m} = \boxed{-1225\ \text{N}}$$ Two equations, four unknowns — the system is indeterminate until a specific pair of structures is assumed to carry the entire load (parts c, d).
3. c) Choose the two joint-contact forces (J_L, J_m), ligaments slack (L_L=L_m=0). Solving the pair simultaneously: $$21 - 0.02F_{J_L} + 0.02F_{J_m} = 0, \qquad F_{J_L}+F_{J_m} = -1225$$ $$F_{J_L} = \boxed{-87.5\ \text{N}}, \qquad F_{J_m} = \boxed{-1137.5\ \text{N}}$$ Both are negative (compressive in this convention) — physiologically possible: the axial load is carried entirely as bone-on-bone contact through the tibial plateau, with the medial condyle much more heavily loaded than the lateral (consistent with the frontal-plane moment $M_x$), a typical picture for normal stance-phase gait.
4. d) Choose one joint force and the medial ligament (J_L, L_m), the rest zero. $$21 - 0.02F_{J_L} + 0.05F_{L_m} = 0, \qquad F_{J_L}+F_{L_m} = -1225$$ $$F_{J_L} = \boxed{-575\ \text{N}}, \qquad F_{L_m} = \boxed{-650\ \text{N}}$$ $F_{L_m}$ comes out negative — i.e., the medial ligament would need to be compressive to satisfy equilibrium, which is not physiologically possible (a ligament can only pull, never push). This pairing is therefore an invalid load-sharing assumption: it shows why, for a predominantly axial/compressive stance-phase load, the joint's own bony contact — not the collateral ligaments — must carry the bulk of the load; ligaments only engage in tension when a genuine distraction/opening tendency exists on their own side of the joint.