Download Solutions to Selected Introductory Problems week4

Solutions to Selected Introductory and Additional Problems: Week 3-4
SOLUTIONS TO SELECTED INTRODUCTORY PROBLEMS (Ch 6)
9. Muscle can generate approximately 90 N of force per square centimeter of cross-sectional
area. If a biceps brachii has a cross-sectional area of 10 cm2, how much force can it exert?
90 N (10 cm2) = 900 N
SOLUTIONS TO SELECTED ADDITIONAL PROBLEMS (Ch 6)
5. If the fibers of a pennate muscle are oriented at a 45° angle to a central tendon, how much
tension is produced in the tendon when the muscle fibers contract with 150 N of force? (106 N)
T = (150 N) cos 45°
T = 106 N
6. How much force must be produced by the fibers of a pennate muscle aligned at a 60° angle to
a central tendon to create a tensile force of 200 N in the tendon? (400 N)
200 N = F cos 60°
F = 400 N
7. What must be the effective minimal cross-sectional areas of the muscles in problems 5 and 6
above, given an estimated 90 N of force producing capacity per square cm of muscle
cross-sectional area? (1.2 cm2, 4.4 cm2)
106 N / 90 N/cm2 = 1.2 cm2
400 N / 90 N/cm2 = 4.4 cm2
8. If the biceps brachii, attaching to the radius 2.5 cm from the elbow joint, produces
250 N of tension perpendicular to the bone, and the triceps brachii, attaching 3 cm away from the
elbow joint, exerts 200 N of tension perpendicular to the bone, how much net torque is present at
the joint? Will there be flexion, extension, or no movement at the joint? (0.25 N-m; flexion)
Torquenet = Torqueflexion - Torqueextension
Tnet = (250 N)(0.025 m) - (200 N)(0.03 m) = 0.25 N-m (flexion)
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9. Calculate the amount of torque generated at a joint when a muscle attaching to a bone 3 cm
from the joint center exerts 100 N of tension at the following angles of attachment: a) 30°, b)
60°, c) 90°, d) 120°, e) 150° (a. 1.5 N-m, b. 2.6 N-m, c. 3 N-m, d. 2.6 N-m,
e. 1.5 N-m)
Torque = F • sin θ • 0.03
a) T = 100 N sin 30° (0.03) = 1.5 N-m
b) T = 100 N sin 60° (0.03) = 2.6 N-m
c) T = 100 N sin 90° (0.03) = 3 N-m
d) T = 100 N sin 120° (0.03) = 2.6 N-m
e) T = 100 N sin 150° (0.03) = 1.5 N-m
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