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

Question

4. 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 (00). 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 all 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 15o relative to the tendon axis, with the same volume of muscle. What is the maximum force delivered to the tendon?

Explanation / Answer

A.The cross-sectional area of the muscle can be calculated from V = A x d; A = V /d A = 20 cm 3 / 8 cm = 2.5 cm 2 . The maximum force is 2.5 cm 2 x 20 N cm -2 = 50 N B.

B. 0.1× A = 0.1×2.5 cm2 = 0.25 cm2

activated F = 20N/ cm2 ×0.25 cm2=5 N

Or F = 0.1×Fmax = 0.1×50N = 5N

C. The distance moved in 50 ms is 0.15 x 8 cm = 1.2 cm. Thus the average velocity during this period is 1.2 cm / 0.05 s = 24 cm s -1

D.In this case the muscle fibers are 8 cm long and oriented at 15 o to the tendon. Because the volume is still 20 cm 3 , the cross sectional area is still 2.50 cm 2 , but this cross-sectional area is oriented at an angle. The force is 50 N directed along the axis of the fibers, and this is 50 N x cos 15 o = 48.3 N delivered to the tendon. The shortening velocity is also reduced to 24 cm s -1 x cos 15 o = 23.2 cm s -1.

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