Take the maximum active stress a vertebrate skeletal muscle fibre can sustain as and…

PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Biophysics & Medical Physics · Momentum, energy and conservation laws

Forces, Machines and Signals, Ch Bones, blood and body size

Problem

Take the maximum active stress a vertebrate skeletal muscle fibre can sustain as

\sigma_m = 3.0\times10^{5}\,\mathrm{Pa}

and a typical shortening strain

\varepsilon = 0.30

(both from the muscle stress-strain relations used with Hill's force-velocity relation). Take muscle density

\rho_m = 1060\,\mathrm{kg\,m^{-3}}

. The jumping muscles (the leg extensors) make up a fraction

f_m = 0.10

of body mass. Only a fraction

\eta_{\rm eff} = 0.50

of the mechanical work done by these muscles appears as kinetic energy of the whole body at take-off, the rest being lost to tendon compliance and to co-contraction of opposing muscles. ((a)) Using the argument that jump height is set by the work done per unit mass of muscle, write the work available per unit mass of muscle as

w = \sigma_m\varepsilon\eta_{\rm eff}/\rho_m

and evaluate it. ((b)) The work available per unit body mass is

f_m w

. Convert this into a predicted jump height

h = f_m w/g

and compare with a measured adult human standing vertical jump height of about

0.40\,\mathrm{m}

. ((c)) Explain why this height estimate does not depend on the animal's total body mass, provided the same

f_m

\sigma_m

\varepsilon

\rho_m

and

\eta_{\rm eff}

apply.

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