EXAMPLE 9-8 Force exerted by biceps muscle. How much force must the biceps muscle exert when a 5.0-kg ball is held in the hand (a) with the arm horizontal as in Fig. 9–13a, and (b) when the arm is at a 45° angle as in Fig. 9–13b? The biceps muscle is connected to the forearm by a tendon attached 5.0 cm from the elbow joint. Assume that the mass of forearm and hand together is 2.0 kg and their cG is as shown. FIGURE 9-13 Example 9–8, forces on forearm. APPROACH The free-body diagram for the forearm is shown in Fig. 9–13; the forces are the weights of the arm and ball, the upward force FM exerted by the muscle, and a force F, exerted at the joint by the bone in the upper arm (all assumed to act vertically). We wish to find the magnitude of FM , which can be done using the torque equation and by choosing our axis through the joint so that F, contributes zero torque. SOLUTION (a) We calculate torques about the point where F, acts in Fig. 9–13a. The Er = 0 equation gives CG (2.0 kg)g (5.0 kg)g -15 cm- (a) 35 cm (0.050 m)FM – (0.15 m)(2.0 kg)g – (0.35 m)(5.0 kg)g 0. We solve for F: (0.15 m)(2.0 kg)g + (0.35 m)(5.0 kg)g FM (41 kg)g = 400N. 0.050 m (b) The lever arm, as calculated about the joint, is reduced by the factor cos 45° for all three forces. Our torque equation will look like the one just above, except that each term will have its lever arm reduced by the same factor, which will cancel out. The same result is obtained, FM = 400 N. NOTE The force required of the muscle (400 N) is quite large compared to the weight of the object lifted (= mg = 49 N). Indeed, the muscles and joints of the body are generally subjected to quite large forces. NOTE Forces exerted on joints can be large and even painful or injurious. Using EF, = 0 we calculate for this case F, = FM - (2.0 kg)g – (5.0 kg)g = 330 N. (b) 5.0 cm
EXAMPLE 9-8 Force exerted by biceps muscle. How much force must the biceps muscle exert when a 5.0-kg ball is held in the hand (a) with the arm horizontal as in Fig. 9–13a, and (b) when the arm is at a 45° angle as in Fig. 9–13b? The biceps muscle is connected to the forearm by a tendon attached 5.0 cm from the elbow joint. Assume that the mass of forearm and hand together is 2.0 kg and their cG is as shown. FIGURE 9-13 Example 9–8, forces on forearm. APPROACH The free-body diagram for the forearm is shown in Fig. 9–13; the forces are the weights of the arm and ball, the upward force FM exerted by the muscle, and a force F, exerted at the joint by the bone in the upper arm (all assumed to act vertically). We wish to find the magnitude of FM , which can be done using the torque equation and by choosing our axis through the joint so that F, contributes zero torque. SOLUTION (a) We calculate torques about the point where F, acts in Fig. 9–13a. The Er = 0 equation gives CG (2.0 kg)g (5.0 kg)g -15 cm- (a) 35 cm (0.050 m)FM – (0.15 m)(2.0 kg)g – (0.35 m)(5.0 kg)g 0. We solve for F: (0.15 m)(2.0 kg)g + (0.35 m)(5.0 kg)g FM (41 kg)g = 400N. 0.050 m (b) The lever arm, as calculated about the joint, is reduced by the factor cos 45° for all three forces. Our torque equation will look like the one just above, except that each term will have its lever arm reduced by the same factor, which will cancel out. The same result is obtained, FM = 400 N. NOTE The force required of the muscle (400 N) is quite large compared to the weight of the object lifted (= mg = 49 N). Indeed, the muscles and joints of the body are generally subjected to quite large forces. NOTE Forces exerted on joints can be large and even painful or injurious. Using EF, = 0 we calculate for this case F, = FM - (2.0 kg)g – (5.0 kg)g = 330 N. (b) 5.0 cm
Related questions
Question
(I) Suppose the point of insertion of the biceps muscle into
the lower arm shown in Fig. 9–13a (Example 9–8) is 6.0 cm
instead of 5.0 cm; how much mass could the person hold
with a muscle exertion of 450 N?
Transcribed Image Text:EXAMPLE 9-8 Force exerted by biceps muscle. How much force must
the biceps muscle exert when a 5.0-kg ball is held in the hand (a) with the arm
horizontal as in Fig. 9–13a, and (b) when the arm is at a 45° angle as in Fig. 9–13b?
The biceps muscle is connected to the forearm by a tendon attached 5.0 cm from
the elbow joint. Assume that the mass of forearm and hand together is 2.0 kg
and their cG is as shown.
FIGURE 9-13 Example 9–8, forces
on forearm.
APPROACH The free-body diagram for the forearm is shown in Fig. 9–13; the
forces are the weights of the arm and ball, the upward force FM exerted by the
muscle, and a force F, exerted at the joint by the bone in the upper arm (all
assumed to act vertically). We wish to find the magnitude of FM , which can be done
using the torque equation and by choosing our axis through the joint so that
F, contributes zero torque.
SOLUTION (a) We calculate torques about the point where F, acts in Fig. 9–13a.
The Er = 0 equation gives
CG
(2.0 kg)g
(5.0 kg)g
-15 cm-
(a)
35 cm
(0.050 m)FM – (0.15 m)(2.0 kg)g – (0.35 m)(5.0 kg)g
0.
We solve for F:
(0.15 m)(2.0 kg)g + (0.35 m)(5.0 kg)g
FM
(41 kg)g = 400N.
0.050 m
(b) The lever arm, as calculated about the joint, is reduced by the factor cos 45°
for all three forces. Our torque equation will look like the one just above, except
that each term will have its lever arm reduced by the same factor, which will
cancel out. The same result is obtained, FM = 400 N.
NOTE The force required of the muscle (400 N) is quite large compared to the
weight of the object lifted (= mg = 49 N). Indeed, the muscles and joints of the
body are generally subjected to quite large forces.
NOTE Forces exerted on joints can be large and even painful or injurious. Using
EF, = 0 we calculate for this case F, = FM - (2.0 kg)g – (5.0 kg)g = 330 N.
(b)
5.0 cm
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