# Question #392

The figure below illustrates an experiment in which a frog gastrocnemius muscle contracted against varying loads. The pivot was equally as far from the load as it was from the muscle. The lowest trace in the lower figure was obtained with the weight carrier alone (mass = 20 g) suspended from the lever arm. Each subsequent trace was obtained after an additional 20 g mass was added. What was the active tension in the muscle at a length of 60 mm? Counting up from the bottom trace and noting that an additional 20 g mass is added each time, the muscle that actively shortens to a length of 60 mm is carrying a 160 g mass.

Assuming that acceleration due to gravity = 10 m/s2 and applying F = ma, the total tension in this muscle is 0.16 x 10 = 1.6 N.

Counting from the bottom trace again, the unstimulated muscle at a length of 60 mm is carrying a 100 g mass. The passive tension in the muscle must therefore be 0.1 x 10 = 1 N.

Since total tension = passive tension + active tension, the active tension in the stimulated muscle at a length of 60 mm is 1.6 - 1 = 0.6 N.

Note that, if the pivot were twice as far from the muscle as from the mass, there would be twice the moment and forces would therefore be twice as large.