Forces and equilibrium - Orthopaedic Biomaterials in Research and Practice - page 10

M L at P = -250 N × 0.05 m = -12.5 Nm
M K at P = 50 N × 0.2 m = 30 Nm

L

P

O

K

1 cm = 0.1 m
1 cm = 50 N
= 1.25 Nm

Q

FIGUre 1.6 moments.

may be obtained graphically by constructing a rectangle with the length
equal to the length of each force vector and the width equal to the length
of the moment arm and then by dividing the area up into squares of known
moment. Thus, the rectangle constructed on L contains 10 squares, each
with an area of 0.25 cm 2 , representing a moment of 1.25 N ∙ m, for a total
moment of 12.5 N ∙ m. It is noted from the example that the moment arm
is always drawn perpendicular in a line normal to the force vector to
the point on which the moment is being taken around. For example, the
moment arm of force L around P is drawn from point O to P, as opposed
to point Q to P, or any other point. The force vector can be graphically
translated in a direction parallel to its vector direction without changing
the problem. In more complicated problems, it may be convenient to uti-
lize the trigonometric relationships described above to construct a set of
equations that will allow solving for the unknown reactions in a structure
at equilibrium. Finally, it is obvious that a force that passes through a
point may exert no moment at that point.
The existence of moments imposes a second condition for equilibrium:

Rotational equilibrium exists if the sum of moments about any point on
an object is equal to zero.

Moment calculations may be greatly simplified since it can be shown
that if the sum of the moments is equal to zero at any point on an object,
it is equal to zero at all points on the object. When an object is in rota-
tional equilibrium, a point may be selected that eliminates the moment
of one force (i.e., is on the line of action of that force), which may reduce
the number of equation unknowns for which to solve.
If the conditions for both translational and rotational equilibrium are
met, then Newton's first law is satisfied and the object is either at rest or
moving with a constant velocity.

PROBLEM 1.4
Is the boomerang in Figure 1.7 in equilibrium?