Muscle Force - Biomechanics of the Human Body - page 100

to choose the x and y axes, in order to decompose the forces, is that in which one of
the components of the torque becomes zero, not causing rotation, although it can
cause compression or tension on an articulation. The other component will be
perpendicular to the lever arm and will produce rotation.
Example 6.3 Consider a horizontally outstretched arm, as shown in the figure of
Example 6.3. In this posture, the arm is supported by the muscle force F , exerted by
the deltoid muscle attached at 12 cm from the shoulder articulation at an angle of
15 to the humerus. The weight force W of the arm-forearm-hand with a magnitude
of 44 N is exerted at the center of gravity at 30 cm from the shoulder articulation.
There is a third force, of contact N , applied to the humerus at the shoulder
articulation to equilibrate the arm. Determine F and N .

F

N

W = 44 N

F

N

15°

C.G.

12 cm

W = 44 N

30 cm

F

F y

15°

F x

For convenience let us locate the axis of rotation at the shoulder articulation,
because the unknown force N exerts zero torque about this point. We begin by
decomposing the muscle force F in a perpendicular component, F y , to the arm and
the other parallel, F x . The torque of parallel force is zero because its line of action
passes through the axis.
We apply the condition that the net torque must be zero, remembering that the
torque due to N is null, since its lever arm is zero. Hence,

W (0.30 m) ¼ F 0.259 0.12 m,

F ¼ (44 N)(0.30 m)/0.259(0.12 m); F ¼ 425 N.

Now we apply the condition that the resultant of all applied forces must be zero:

W + F + N ¼ 0.

We have to decompose the forces into vertical and horizontal components:

sin15 ¼ F y ¼ (425 N) (425 N) 0.259 ¼ 110.0 N,

cos15 ¼

F x ¼ (425 N) (425 N) 0.966 ¼ 410.5 N,