Question 4 of 4: 3-D Equilibrium of the Femur — Hip Muscle and Joint Reaction Forces
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, May 2017 — 04-Bio-A4, Biomechanics. 3-hour open-book exam;
FOUR (4) questions of equal value constitute a complete paper, so all four are
answered in full below.
Reference texts: Winter, Biomechanics and Motor Control of
Human Movement; Zatsiorsky, Kinematics of Human Motion; Nordin &
Frankel, Basic Biomechanics of the Musculoskeletal System.
Question 4: 3-D Equilibrium of the Femur — Hip Muscle and Joint Reaction Forces (15 marks)
(a) Figure 4 gives the muscle attachment geometry in two elevation
views (front: y-z plane; side: x-y plane) about an origin O at the femoral-head centre.
Projecting each muscle's attachment point onto the transverse (x-z) plane through O gives
the plan view below.
Figure 4 (transverse-plane
projection): (x, z) offset in mm of each muscle's line of action from the femoral-head
origin O, read from the front/side elevations on the source page.
Check: Figure 4 is a hand-drawn, two-elevation (front + side) sketch;
we read it as follows. Gluteus Medius (MED) and Gluteus Maximus (MAX) pull with force
vectors confined to the y-z (frontal) plane at the FRONT-view angles 30°/20° from
vertical/horizontal respectively (SIDE view confirms "acts vertically upward", i.e. zero
x-force-component), applied at the FRONT-view z-offset (0.065/0.095 m) and SIDE-view
x-offset (−0.035/−0.065 m) from O, at y = 0. Hamstrings (H) and Sartorius (S)
pull with force vectors confined to the x-y (sagittal) plane — H at the SIDE-view
angle 80° from horizontal, S at the FRONT-view angle 83° from horizontal —
applied at y = −0.30 m (H) / −0.42 m (S), the FRONT-view z-offset
(−0.02 m / +0.01 m) and an assumed small SIDE-view x-offset (0 / +0.04 m). The two
opposing "T" arrows at the fracture site are the internal fracture-surface reaction pair
— equal, opposite and self-cancelling for the femur taken as a whole, so they do not
enter the equilibrium equations below.
Given. External resultant load at O (femoral-head centre):
$\vec F_{ext}=(120,760,-200)\ \text{N}$, $\vec M_{ext}=(33,25,4)\ \text{N}\cdot\text{m}$.
Four candidate muscles (unit pull direction $\hat d$, attachment point $\vec r$ from O,
metres):
Given data — muscle geometry (Figure 4)
Muscle
Pull direction $\hat d=(d_x,d_y,d_z)$
Attachment $\vec r=(x,y,z)$, m
Gluteus Medius (MED)
(0, 0.866, −0.500)
(−0.035, 0, 0.065)
Gluteus Maximus (MAX)
(0, 0.342, −0.940)
(−0.065, 0, 0.095)
Hamstrings (H)
(−0.174, 0.985, 0)
(0, −0.30, −0.02)
Sartorius (S)
(0, 0.993, 0.122)
(0.04, −0.42, 0.01)
Find. (b) the three moment-equilibrium equations about O; (c) the
tension in each of three chosen muscles; (d) the hip joint contact-force components
$J_x,J_y,J_z$.
Approach. Model the hip as a frictionless ball-and-socket joint: the
unknown joint reaction force acts through O and therefore contributes zero moment about
O. Summing moments about O gives 3 equations in the muscle-force magnitudes; solving
those for 3 of the 4 muscles and substituting into the force balance then gives the joint
reaction.
(b) Moment-equilibrium equations about O. Because the joint reaction
$\vec J$ acts at O, it has zero moment arm and drops out entirely:
$$\sum \vec M_O = \vec M_{ext} + \sum_k F_k\,(\vec r_k\times \hat d_k) = \vec 0,\qquad k\in\{MED,MAX,H,S\}$$
Evaluating $\vec c_k=\vec r_k\times \hat d_k$ for each muscle (N·m per newton of
tension) gives the coefficient table below, so the three scalar equations
($\sum M_x=0,\ \sum M_y=0,\ \sum M_z=0$) are $\sum_k F_k\,c_{k,x}= -33$,
$\sum_k F_k\,c_{k,y}=-25$, $\sum_k F_k\,c_{k,z}=-4$.
$\vec r_k\times \hat d_k$ per muscle
Muscle
$c_x$
$c_y$
$c_z$
MED
−0.0563
−0.0175
−0.0303
MAX
−0.0325
−0.0611
−0.0222
H
+0.0197
+0.0035
−0.0521
S
−0.0611
−0.0049
+0.0397
(c) Choosing 3 muscles and solving for their tensions. Four muscles
over-determine 3 equations, so one must be dropped. Trying {MED, MAX, H} (keeping the two
major hip abductor/extensor stabilisers plus Hamstrings) and solving the 3×3 system
gives $F_H=-242.4\ \text{N}$ — a muscle can only pull (tension, $F\ge 0$), so a
negative result means this combination is not physiologically valid. Substituting
Sartorius for Hamstrings instead — {MED, MAX, S} — and re-solving the same
system gives all-positive tensions:
$$F_{MED}=\boxed{154.9\ \text{N}},\quad F_{MAX}=\boxed{348.0\ \text{N}},\quad F_S=\boxed{212.3\ \text{N}}$$
This is the combination we adopt: Gluteus Medius and Gluteus Maximus provide the bulk of
the frontal- and sagittal-plane stabilisation expected of the hip abductor/extensor
mechanism, and Sartorius is the only third choice that keeps every muscle in tension.
(d) Joint contact force — force balance. With the three muscle
tensions known, $\vec F_k=F_k\hat d_k$, and
$$\vec J = -\left(\vec F_{ext}+\vec F_{MED}+\vec F_{MAX}+\vec F_S\right)$$
Substituting ($\vec F_{MED}=(0,134.1,-77.4)$, $\vec F_{MAX}=(0,119.0,-327.0)$,
$\vec F_S=(0,210.8,25.9)$, all N):
$$\vec J = \boxed{(-120.0,\ -1223.9,\ +578.5)\ \text{N}},\qquad |\vec J| = \boxed{1359.1\ \text{N}}$$
