Question

During a running stride, a person’s foot is on the ground and the ground reaction force is 447 N in the x direction and 1324 N in the y direction. The foot has a mass of 0.71 kg, a moment of inertia of 0.414 kg m2, and is undergoing angular acceleration of 13.9 rad/s2. Its linear acceleration is 8.63 m/s2 in the x direction and 3.19 m/s2 in the y direction. Relative to the center of mass of the foot, the x, y coordinates of the center of pressure of the ground reaction force are 0.049, -0.056 m and the coordinates of the ankle joint are -0.044, 0.052 m. What is the torque (Nm) produced by muscles that span the ankle joint?

Answer 1

Given that :

mass of the foot, m = 0.71 kg

coordinates of the ankle joint, r₁ = (-0.044 m)² + (0.052 m)² = 0.0681 m

The net ground reaction force is given by :

F_net = (447 N)² + (1324 N)²

F_net = (199809 + 1752976) N²

F_net = 1952785 N²

F_net = 1397.4 N

using an equation, we have

_foot = I_net                                                       

_foot = (I + I_CM)                                                                        { eq.1 }

where, I = moment of inertia by foot = 0.414 kg.m²

= angular acceleration produced by foot = 13.9 rad/s²

I_CM = moment of inertia of cente mass = (1/12) m r²

I_CM = (1/12) (0.71 kg) (0.0681 m)²

I_CM = 0.000274 kg.m²

inserting the values in eq.1,

_foot = [(0.414 kg.m²) + (0.000274 kg.m²)] (13.9 rad/s²)

_foot = (0.414274 kg.m²) (13.9 rad/s²)

_foot = 5.75 N.m

The torque produced by muscles that span the ankle joint which is given as :

_net = _foot + _gf                                                                            { eq.2 }

where, _gf = torque produced by ground reaction force = r₂ F

_net = (5.75 N.m) + r₂ F

where, r₂ = coordinates of the reaction force = (0.049 m)² + (-0.056 m)²

r₂ = 0.0744 m

then, we have

_net = (5.75 N.m) + (0.0744 m) (1397.4 N)

_net = (5.75 N.m) + (103.9 N.m)

_net = 109.6 N.m