Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2013 — 04-Bio-A4 Biomechanics, 3 hours, closed book (approved calculator
allowed). Five questions constitute a complete exam paper; each question is of equal value (15 marks).
This solution follows the
paper's true subject and cites biomechanics references accordingly.
Reference texts: Winter, Biomechanics and Motor Control of Human Movement
(4th ed.); Zatsiorsky, Kinematics of Human Motion; Nordin & Frankel, Basic Biomechanics
of the Musculoskeletal System (5th ed.).
Given. An elderly subject (mass 60 kg) exits a bath tub; the right foot's ground
reaction force (GRF) acts at the centre of pressure P with three components, and Figure 1 gives the
limb geometry relative to P.
Given data (Figure 1)
Quantity
Value
Vertical GRF, Fz
400 N (up)
Side-to-side GRF, Fy
45 N (toward the bath)
Backward GRF, Fx
20 N (backward)
Knee height above P
0.5 m
Hip height above knee
0.4 m
Frontal-plane limb tilt from vertical
15° (at the hip)
Sagittal-plane limb tilt from vertical
30° (at the hip)
P ahead of the vertical ankle/shank line
80 mm
Find. The joint resultant moments at the knee and hip in the sagittal, frontal, and
transverse planes.
[Figure not reproduced: Figure 1 (redrawn) — rear view (frontal-plane geometry) and side view (sagittal-plane geometry) of the support limb, with P as the coordinate origin. See the official exam paper.]
Check — assumption: the printed sketch is hand-drawn; the shank is read as
vertical in both projections (ankle directly below the knee), with the thigh carrying the full 15°
(frontal) and 30° (sagittal) tilt at the hip — the only reading consistent with all of Figure 1's
own dimensions. “Ignore inertial effects” is taken (with no segment-mass-fraction data given)
to also exclude segment weight from this free-body analysis: only the GRF at P is carried into each
joint cut. The subject's 60 kg mass is scene-setting and is not needed for the moment-arm method used
below.
Approach. Cut a free body at each joint (foot+shank for the knee; foot+shank+thigh
for the hip); with segment weight and inertia excluded, static equilibrium requires the joint moment to
balance the moment of the GRF about that joint, Mjoint = r ×
F, where r runs from P to the joint.
Set up coordinates at P. Let x = anterior(+), y = toward-the-bath(+), z = up(+).
The ankle sits 80 mm posterior to P (shank vertical): $x_{ankle}=-0.08\ \text{m}$. The knee is directly
above the ankle: $\mathbf{r}_{knee}=(-0.08,\ 0,\ 0.5)\ \text{m}$.
Locate the hip. The thigh adds a further 0.4 m of vertical rise, tilted 30°
(sagittal) and 15° (frontal) from vertical:
$$\Delta x = 0.4\tan(30^\circ) = 0.231\ \text{m}, \qquad \Delta y = 0.4\tan(15^\circ) = 0.107\ \text{m}$$
so $\mathbf{r}_{hip} = (-0.08+0.231,\ 0.107,\ 0.9) = (0.151,\ 0.107,\ 0.9)\ \text{m}$.
Sagittal-plane moments (part a) — rotation about the mediolateral (y) axis,
$M_{sag} = r_z F_x - r_x F_z$:
$$M_{sag,knee} = (0.5)(-20) - (-0.08)(400) = \boxed{+22.0\ \text{N}\cdot\text{m}}$$
$$M_{sag,hip} = (0.9)(-20) - (0.151)(400) = \boxed{-78.4\ \text{N}\cdot\text{m}}$$
The knee moment is net extensor (positive, resisting further forward collapse); the hip moment is the
larger flexor demand, dominated by the vertical GRF acting well anterior to the hip.
Frontal-plane moments (part b) — rotation about the anteroposterior (x) axis,
$M_{fr} = r_y F_z - r_z F_y$:
$$M_{fr,knee} = (0)(400) - (0.5)(45) = \boxed{-22.5\ \text{N}\cdot\text{m}}$$
$$M_{fr,hip} = (0.107)(400) - (0.9)(45) = \boxed{+2.4\ \text{N}\cdot\text{m}}$$
The knee carries a substantial varus/valgus (ab-adductor) demand from the side-to-side GRF acting on a
0.5 m vertical arm; at the hip the same force's moment is largely cancelled by the vertical GRF acting
through the hip's own frontal offset.
Transverse-plane moments (part c) — rotation about the vertical (z) axis,
$M_{tr} = r_x F_y - r_y F_x$:
$$M_{tr,knee} = (-0.08)(45) - (0)(-20) = \boxed{-3.6\ \text{N}\cdot\text{m}}$$
$$M_{tr,hip} = (0.151)(45) - (0.107)(-20) = \boxed{+8.9\ \text{N}\cdot\text{m}}$$
Transverse-plane (rotational) demand is the smallest of the three at both joints, as expected for a
predominantly vertical GRF with modest horizontal components.
