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.