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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 (θ=0).

a.) What is the maximum force developed by this muscle? (5 points)
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? (5 points)
c.) If the muscle fibers contract 15% of their length in 50 ms under no load, what is the maximum muscle velocity? (5 points)
d.) Suppose the muscle fibers are oriented at 15° relative to the tendon axis, with the same volume of muscle. What is the maximum force delivered to the tendon? What is the maximum velocity? (5 points)

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Answer:

a) [tex]F=50\ N[/tex]

b) [tex]F'=5\ N[/tex]

c) [tex]v=24\ cm.s^{-1}[/tex]

d) [tex]F"=48.2963\ N[/tex]

Explanation:

Given:

  • length of muscle fibers, [tex]l=8\ cm^2[/tex]
  • maximum stress on each fiber, [tex]\sigma=20\ N.cm^{-2}[/tex]
  • Volume of muscles, [tex]V=20\ cm^3[/tex]

∴Area of the muscle:

  • [tex]A=\frac{20}{8} =\frac{5}{2}=2.5 \ cm^2[/tex]

a)

Maximum force developed by this muscle:

[tex]F=\sigma\times A[/tex]

[tex]F=20\times 2.5[/tex]

[tex]F=50\ N[/tex]

b)

Force when the muscles are 10% activated:

[tex]F'=10\%\ of\ F[/tex]

[tex]F'=\frac{50}{10}[/tex]

[tex]F'=5\ N[/tex]

c)

  • contraction in length of muscle, [tex]\Delta l=15\%\ of\ l[/tex]
  • time taken for the contraction, [tex]t=50\times 10^{-3}\ s[/tex]

Now, the speed of the muscle:

[tex]v=\frac{\Delta l}{t}[/tex]

[tex]v=\frac{0.15\times 8}{50\times 10^{-3}}[/tex]

[tex]v=24\ cm.s^{-1}[/tex]

d)

Maximum force delivered when the muscle fibers are oriented at 15° from the tendons:

[tex]F"=50\ cos\ 15^{\circ}[/tex]

[tex]F"=48.2963\ N[/tex]

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