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