Question 1 of 4

Question 1 (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. The whole leg (thigh + lower-leg/foot) rotates as one rigid body about the fixed hip; Figure 1 (redrawn from the printed figure) gives the toe 0.20 m forward and level with the ankle, the door contact point 0.05 m above the floor.

[Figure not reproduced: Figure 1 (redrawn) — combined free-body diagram of the kicking leg at the instant of door contact: hip fixed, knee flexed 28°, shank vertical, foot horizontal to the toe/door contact point. See the official exam paper.]

Find. a) The direction of the toe's tangential acceleration; b) the equations of motion for the leg (forces vertically/horizontally, moment about the hip); c) the door–toe contact force; d) which hip muscle(s) produce the observed flexion moment.

Approach. Treat the whole leg as one rigid body pivoting about the fixed hip; locate the toe from the given segment lengths/angles, then use $\sum M_{hip}=I_{hip}\alpha$ (which eliminates the unknown hip reaction force) to solve for the door contact force directly.

1. a) Locate the toe and find the direction of its tangential acceleration. With the hip at the origin (x = anterior, y = superior), the thigh tilts 28° forward, the shank hangs vertically below the knee, and the foot runs 0.20 m forward from the ankle: $$\mathbf{r}_{toe} = \big(0.51\sin28^\circ,\ -0.51\cos28^\circ\big) + (0,\,-0.49) + (0.20,\,0) = (0.439,\ -0.940)\ \text{m}, \qquad r_{toe}=\boxed{1.038\ \text{m}}$$ This position vector sits 25.0° forward of straight down. The toe's velocity (leg swinging forward into the kick) is tangential to this radius, directed 25.0° above horizontal (forward and up); since the toe decelerates, its tangential acceleration points the opposite way: 25.0° below horizontal, directed backward (posteriorly, away from the door) — i.e., straight back along the line the door pushes the foot, consistent with the door braking the kick.
2. b) Free-body diagram and equations of motion. Cutting the leg free at the hip, the external loads are: the hip joint reaction $(R_x,R_y)$ and the 2 N·m hip-flexor moment applied at the hip; each segment's weight at its own COM; and the door's reaction force $F_t$ at the toe, taken tangential to the toe's circular path (opposing the kick). With $a_{cm}$ the (rotating-frame) acceleration of the whole leg's combined COM: $$\sum F_x:\quad R_x - F_t\sin(65.0^\circ) = (m_1+m_2)a_{cm,x}$$ $$\sum F_y:\quad R_y - (m_1+m_2)g + F_t\cos(65.0^\circ)\big|_{\text{sign per geometry}} = (m_1+m_2)a_{cm,y}$$ $$\sum M_{hip}:\quad M_{hip\,flex} + M_{grav} - F_t\,r_{toe} = I_{hip}\,\alpha$$ Because the hip is fixed, $R_x$ and $R_y$ carry zero moment arm about the hip, so the third equation alone isolates $F_t$ without needing the reaction components — that is the equation used in part c).
3. c) Moment of inertia about the hip. Using each segment's own radius of gyration (parallel-axis to its own COM) then shifting to the hip via the straight-line hip–COM distance: thigh COM at 0.20 m along the thigh, $|\mathbf{r}_{1,COM}|=0.20\ \text{m}$; shank COM at the knee position plus 0.22 m straight down, $|\mathbf{r}_{2,COM}|=0.712\ \text{m}$: $$I_{hip} = \big[m_1k_1^2+m_1(0.20)^2\big] + \big[m_2k_2^2+m_2(0.712)^2\big] = 0.459+2.342 = \boxed{2.801\ \text{kg}\cdot\text{m}^2}$$
4. Angular acceleration and gravity moment. From $a_t=\alpha\, r_{toe}$ (magnitude), $\alpha = 50/1.038 = 48.17\ \text{rad/s}^2$, directed opposite the forward swing (decelerating it). Taking moments of each segment's weight about the hip ($M=r_x\,m g$ for a purely vertical force): $$M_{grav} = (0.094)(7)(9.81)\cdot(-1) + (0.239)(4.4)(9.81)\cdot(-1) = -16.8\ \text{N}\cdot\text{m}$$ (negative in the deceleration-positive sense: with both segment COMs still below and forward of the hip, gravity is pulling the leg back toward hanging vertical, i.e., it opposes this particular forward-and-rising kick, the same sense as the door's braking effect).
5. Solve for the contact force. With the hip flexor assisting the swing ($M_{hip\,flex}=+2\ \text{N}\cdot\text{m}$, opposite sense to the deceleration) and the door force purely tangential (moment $=-F_t\,r_{toe}$): $$I_{hip}\alpha_{(decel)} = M_{grav} + M_{hip\,flex} - F_t\,r_{toe}$$ $$F_t = \frac{M_{grav}+M_{hip\,flex}-I_{hip}\alpha_{(decel)}}{r_{toe}} = \frac{(-16.8)+(2.0)-(2.801)(-48.17)}{1.038} = \boxed{115.8\ \text{N}}$$ directed backward and 25.0° below horizontal — the door pushing back on the toe, opposing the kick.
6. d) Muscles producing the observed hip-flexion moment. A net hip-flexion moment (driving the thigh forward, as required to swing the leg into the kick) is produced chiefly by iliopsoas (iliacus + psoas major, the primary hip flexor), assisted by rectus femoris (which flexes the hip while also crossing the knee) and, for the initial part of the swing, sartorius and tensor fasciae latae; at only 2 N·m the demand is modest, consistent with the leg being mostly through its swing and decelerating on contact rather than being actively driven forward at this instant.