Consider muscle fibers that are 8 cm long and that develop a maximum of 20 N/cm2

Question

Consider muscle fibers that are 8 cm long and that develop a maximum of 20 N/cm2 force. Consider a muscle that has a volume of 20 cm3 with the fibers aligned with the direction of the tendon ( = 0) a. What is the maximum force developed by this muscle? b. If partial recruitment results in activation of 10% of the muscle fibers (assuming they are al of equal size), how much force could the muscle generate? C. If the muscle fibers contract 15% of their length in 50ms under no load, what is the maximum muscle velocity? d. Suppose the muscle fibers were oriented at 15 relative to the tendon axis, with the same volume of muscle. What is the maximum force delivered to the tendon?

Explanation / Answer

a) the given values are: lenth =8cm, volume = 20cm^3, force provided maximum is 20 N/cm^2.

So physiological cross section area (PCSA) = muscle volume/fibre length = muscle mass/p.fibre length where p = density of the muscle

=m/(m/v)*l, = 20cm^3/8cm , PCSA=2.5 cm^2.

Therefore Total force is directly proportional to PCSA,

Since specific tension of the muscle fibre is given, So total force = PCSA*specific tension

=> total force = 2.5 * cos theta(0) = 2.5 N/cm^2

b) since Total force is directly proportional to PCSA, if the size is not changing due partial increment of 10% muscle fibres, The amount of force generate will be same.

c) initial length is 8 cm, under 15% contract length is 50 ms i.e 5000 cms, velocity = distance/time

velocity = 5000/t, Therefore velocity is directly proportional to length or distance, so velocity will increase by 15% too.

d) Total force = PCSA * specific tension => Total force = 2.5 * cos theta(15), total force = 2.5 * (-0.75) = -1.875 N/cm^2.

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