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Problem Set 3 (Muscle Function)

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Problem Set 3: Muscle Function
1. Energy in the Achilles tendon
The average Achilles tendon in humans is 65 mm long and has a cross sectional area of 50 mm2 at
rest. Researchers have shown that during running the tendon undergoes 3% strain (stretched by
3%) during each stride. Calculate how much energy is stored in the tendon during each stride.
Calculate how much energy is released from the tendon with each stride. Show your work and
state your assumptions.
Answer
The energy released from the tendon is 1.31 J
2. Stronger not bigger
Ben was self conscious about his small calf muscles and decided to try and bulk up his gastrocnemius
muscles. After three months of working out, his gastrocnemius muscle felt stronger but didnt look
any bigger. He was able to estimate that the force generated by his muscle was approximately 30%
higher. However there was no change in the size, dimensions or mass of the muscle. If his gastroc
had a pennation angle of 15 deg before his workout regiment, what would the new pennation angle
have to be to explain the increase in muscle force/strength.
Answer
The new pennation angle is 19.66 degrees
3. Scaling of stiffness
The dimensions of tendons in some small organisms can scale geometrically. If the modulus of elasticity (Em ) remains constant during growth, by what proportion does the spring constant/stiffness
(k) change as the animal’s body mass doubles.
1
Answer
The spring constant increases by a factor of 1.26
4. Muscle loading tendon
A muscle is operating with an in-series tendon. During a contraction the muscle remains on the
plateau of the force length curve while shortening at 20% of Vmax . The contraction results in 1.6J of
stored elastic energy in the tendon. If the maximum isometric force of the muscle is 15N, Calculate
the stiffness of the tendon.
Answer
Tendon stiffness is 13.9 N/m
2
Equations
Variables
= ∆L/Lr
:strain, L:length, Li :resting length
σ = F/A
σ:stress, F :force, A:area
(F + a)(V + b) = (Fo + a)b
F :muscle force, V :shortening velocity, Fo : max isometric force
b, a:shape constants
Vopt = b(1 + 1/g)1/2 − b
Vopt :optimal velocity b, a:shape constants
g = a/Fo = b/Vmax
Vmax :maximum shortening velocity, Fo :maximum isometric force
H = (m − Bz )
−(|Ls − 2a| − Bz )
H: overlap, m:myosin length a:actin length, Ls :sarc. length Bz :bare zone
F = k∆L
F :force, k:spring constant, L:length
E = (1/2)k(∆L)2
E:strain energy, k:spring constant, L:length
k = Em A/Lr
Em :elastic modulus, k:spring constant, Lr :resting length, A:area
y = amb
m:body mass, b:scaling exponent, a:scaling factor
Fa = 1 − a((L/Lo ) − 1)2
Fa :active muscle force, L:muscle length, Lo :optimal muscle length
a:shape constant
Fp = (ec(L−Lr ) )/(e0.5c )
Fp :passive muscle force, L:muscle length, Lr : resting muscle length
c:shape constant,
FL = a2 − 0.8(1 + V )/(1 − b2 V )
FL : muscle force during active lengthening, a2 : constant (peak force),
V : Lengthening velocity, b2 : shape constant
Fm = nFf cos θ
Fm :muscle force, Ff :fiber force, n:number of fibers, θ:pennation angle
PCSA = m/(ρLf )
PCSA:physiological x-sec area, m mass, ρ:density, Lf :fiber length
3
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