Question 4 of 5

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. Thigh and lower-leg segments each 0.44 m long, COM at the segment midpoint; thigh mass 7.2 kg (radius of gyration 0.14 m), lower-leg mass 4.3 kg (radius of gyration 0.15 m); knee flexed 28°, ankle neutral, toe tangential deceleration (relative to the hip) 50 m/s² at the instant of door contact; hip flexors relaxed to zero action; hip fixed (not translating or accelerating); no relative motion between thigh and lower leg (they rotate together as one rigid body).

Find. a) & b) the free-body diagram of the leg (dimensions, then forces/moments); c) the contact force between the door and the toe.

[Figure not reproduced: Figure 4 (redrawn) — combined free-body diagram: (a) dimensions & kinematics — hip fixed, knee flexed 28° from vertical (shank vertical, thigh tilted), toe 0.20 m forward / 0.05 m below the ankle, angular acceleration α about the hip; (b) forces/moments — segment weig. See the official exam paper.]

a) Dimensions. With the leg treated as one rigid body pivoting about the fixed hip, the shank is taken vertical (ankle directly below the knee, matching Figure 4's own vertical construction line) and the 28° knee-flexion angle is exactly the thigh's forward tilt from vertical at the knee (0° flexion would leave the whole leg vertical). The foot then places the toe 0.20 m forward and 0.05 m below the ankle, per the figure.

b) Forces and moments. Each segment's weight acts at its own COM; the hip carries an unknown reaction force but — because the hip flexors have relaxed to zero action — zero net muscular moment; the door exerts an unknown contact force on the toe, taken tangential to the toe's circular path about the hip (the only component a single moment equation about the hip can resolve).

c) Approach. Take moments about the fixed hip for the whole rigid leg: this eliminates the unknown hip reaction force entirely, leaving one equation, $\sum M_{hip} = I_{hip}\alpha$, in one unknown (the contact force).

1. Locate the segments (hip at the origin). Shank vertical, thigh tilted 28°: $$\mathbf{r}_{knee} = (-0.44\sin28^\circ,\ -0.44\cos28^\circ) = (-0.207,\ -0.388)\ \text{m}$$ $$\mathbf{r}_{ankle} = \mathbf{r}_{knee} + (0,\,-0.44) = (-0.207,\ -0.828)\ \text{m}, \qquad \mathbf{r}_{toe} = \mathbf{r}_{ankle}+(0.20,\,-0.05) = (-0.007,\ -0.878)\ \text{m}$$ so $r_{toe} = |\mathbf{r}_{toe}| = 0.879\ \text{m}$.
2. Angular acceleration from the toe's tangential deceleration. $$\alpha = \frac{a_t}{r_{toe}} = \frac{50}{0.879} = \boxed{56.9\ \text{rad/s}^2}$$
3. Moment of inertia of the whole leg about the fixed hip (parallel-axis theorem on each segment, using its own radius of gyration about its own COM, then the straight-line distance from hip to that COM): $$I_{thigh,hip} = m_1k_1^2 + m_1\left(\tfrac{L_1}{2}\right)^2 = 7.2(0.14)^2+7.2(0.22)^2 = 0.490\ \text{kg}\cdot\text{m}^2$$ $$I_{shank,hip} = m_2k_2^2 + m_2 d_{2,COM}^2 = 4.3(0.15)^2+4.3(0.643)^2 = 1.872\ \text{kg}\cdot\text{m}^2$$ $$I_{hip} = 0.490+1.872 = \boxed{2.362\ \text{kg}\cdot\text{m}^2}$$
4. Moment of gravity about the hip. COMs at $(-0.104,-0.194)$ m (thigh) and $(-0.207,-0.608)$ m (shank); with weight purely vertical, $M_{grav}=x_{COM}\,mg$ for each segment: $$M_{grav} = (-0.104)(7.2)(9.81) + (-0.207)(4.3)(9.81) = \boxed{-16.0\ \text{N}\cdot\text{m}}$$ (the weight moment opposes the assumed positive/decelerating sense — both segments' COMs sit posterior to the hip's vertical, pulling the leg backward).
5. Solve for the contact force. With the hip-flexor moment zero, $\sum M_{hip} = M_{grav} + F_t\,r_{toe} = I_{hip}\alpha$: $$F_t = \frac{I_{hip}\alpha - M_{grav}}{r_{toe}} = \frac{(2.362)(56.9) - (-16.0)}{0.879} = \boxed{171\ \text{N}}$$ directed opposite the kick (the door pushing back on the toe), consistent with the toe decelerating.