Muscle Force - Biomechanics of the Human Body - page 96

X

i ¼ n

! ! 1 þ ! 2 þ ! 3 þþ ! n ¼ ! i ¼

¼ 0, (6.1)

i¼ 1

which is equivalent to stating that their orthogonal components must be zero:

X ! x ¼ X ! y ¼

! x ¼ and ! y ¼ 0 0 (6.2)

:

This condition, however, does not impede the rotation of an object, as in the
case of a couple, presented in Chap. 2 . Hence, it is necessary, but not sufficient
for an object to be in equilibrium.
2. Sufficient condition for a body to not undergo rotation—the sum of all the
torques exerted on the body must be zero:

X

i¼n

M T ¼ M 1 þ M 2 þ M 3 þþ M n ¼ M i ¼

0 : (6.3)

i¼ 1

Remember that all of the torques must be calculated about the same point
(fulcrum).

The principle of rotational equilibrium requires, furthermore, that the imaginary
line passing through the center of gravity must pass through the support area, as
already stated in Chap. 3 .

6.3 System of Parallel Forces

A system of parallel forces occurs when all of the forces acting on a rigid object are
applied perpendicularly to a given segment of a straight line. See the illustration in
the figure of Example 6.1. In this case, the conditions for static equilibrium for the
forces and their respective torques provide algebraic equations from which
the unknowns can be obtained by calculation. The solution of the problem can
begin with the equation of torques. Let us choose as the rotation axis, for conve-
nience, the point of application of an unknown force, since its torque is eliminated
from the equation because the lever arm of this force will be zero. Then, use the fact
that the resultant of all of the forces applied must be zero or another equation of
torque can be written, choosing a new axis of rotation.

Example 6.1 Alice and Paul carry an object with 50 kg mass which rests on a 3.0 m
board, whose weight of W 2 ¼ 100 N is applied at its center of gravity, as shown in

the figure of Example 6.1. The object is at 1.0 m from Paul's end of the board.
Determine the magnitudes of the forces that Alice ( F A ) and Paul ( F P ) must exert to
support this load.