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. A cyclist applies 450 N to the pedal, directed orthogonal to the crank arm; the ankle joint sits at the pedal axle, the foot is horizontal, and Figure 1 gives the limb/crank geometry.

Find. The moments at the ankle, knee, and hip joints, and the active muscle group at each.

[Figure not reproduced: Figure 1 (redrawn, sagittal plane) — hip–knee–ankle link-segment chain, the pedal crank (BB–ankle), and the pedal's reaction force on the foot at the centre of pressure (COP). See the official exam paper.]

Approach. Set the ankle as the coordinate origin (x = anterior, y = up), locate the knee and hip from the given vertical rises and angles from vertical, then cut a free body at each joint in turn and take moments of the pedal's reaction force about that joint, $M_{joint}=\mathbf{r}_{joint\to COP}\times\mathbf{F}$.

1. Locate the joints and crank centre from the ankle. With the shank 15° from vertical (knee anterior of ankle) and the thigh 75° from vertical (hip posterior of knee): $$\mathbf{r}_{knee}=(0.320\tan15^\circ,\ 0.320)=(0.0857,\ 0.320)\ \text{m}$$ $$\mathbf{r}_{hip}=\mathbf{r}_{knee}+(-0.230\tan75^\circ,\ 0.230)=(-0.7726,\ 0.550)\ \text{m}$$ and the crank centre (BB), 100 mm below the ankle at 20°: $\mathbf{r}_{BB}=(-0.100/\tan20^\circ,\ -0.100)=(-0.2747,\ -0.100)\ \text{m}$, confirming the BB→ankle crank-arm direction is 20° above horizontal as given.
2. Resolve the pedal reaction force. The tangential drive direction (crank rotated 90° clockwise) is $(\sin20^\circ,\,-\cos20^\circ)$, so the foot-on-pedal force is $450(\sin20^\circ,-\cos20^\circ)=(153.9,-422.9)\ \text{N}$; the pedal-on-foot reaction is the negative of this: $$\mathbf{F} = (-153.9,\ 422.9)\ \text{N}, \qquad \mathbf{r}_{ankle\to COP}=(0.120,\ 0)\ \text{m}$$
3. a) Ankle moment. $M_{ankle}=r_xF_y-r_yF_x$: $$M_{ankle} = (0.120)(422.9) - (0)(-153.9) = \boxed{50.7\ \text{N}\cdot\text{m}}$$
4. b) Knee moment. $\mathbf{r}_{knee\to COP} = (0.120-0.0857,\ 0-0.320) = (0.0343,\ -0.320)\ \text{m}$: $$M_{knee} = (0.0343)(422.9) - (-0.320)(-153.9) = 14.5-49.2 = \boxed{-34.8\ \text{N}\cdot\text{m}}$$
5. c) Hip moment. $\mathbf{r}_{hip\to COP} = (0.120-(-0.7726),\ 0-0.550) = (0.8926,\ -0.550)\ \text{m}$: $$M_{hip} = (0.8926)(422.9) - (-0.550)(-153.9) = 377.5-84.6 = \boxed{292.8\ \text{N}\cdot\text{m}}$$ The hip carries by far the largest moment, because the very inclined (75°-from-vertical) thigh puts the hip almost 0.9 m posterior of the COP — a long moment arm for the vertical component of the pedal reaction.
6. d) Active muscles. The reaction force pushes the forefoot up and posteriorly at all three joints, opposing the direction the leg must drive the pedal, so each joint's extensor/plantarflexor group is active to resist it and complete the power stroke: ankle — plantarflexors (gastrocnemius, soleus), resisting the pedal-reaction's dorsiflexing tendency and pressing the forefoot into the pedal; knee — extensors (quadriceps femoris, primarily vastus group), the dominant muscle of the early power stroke; hip — extensors (gluteus maximus, hamstrings), driving the large moment found in c). This hip–knee–ankle extensor synergy is the well-documented EMG pattern for the cycling power (down-stroke) phase.