Question

reality the force length equati only valid in a very limited range of extension ratios. The force that a sar- comere (and thus a skeletal muscle) can exert has a maximum, as depicted in the figure below. Sarcomere lengths at several interesting points in the graph are depicted by Li, with i1,....5. L2 100 Maximumm L3 ension 60 (96) L5 40 し4 20 0) 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Sarcomere length Jum) Z disk

Please show me how to do b)

The answer is ≈ 220N

Answer 1

F = c(lambda/lambda_c - 1) L_1 & L_2 can be described by this eq therefore lambda_1 = 1.25 mu m/l_0 & lambda_2 = 1.75/l_0 mu m therefore F_1 = c(1.25/l_c - 1) & F_2 = c(1.75/l_c - 1) F_1 = 0 (from plot) & F_2 = 80 N 0 = c(1.25/l_c - 1) 1.25/l_c = 1 l_c = 1.25 B_0 = c(1.75/1.25 - 1) c = 200 when F = 100 N Then 100 = c(2.05/1.25 - 1) c TildeEqual 156 N