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14.2 Examples of Static Equilibrium 437

PROBLEM-SOLVING TECHNIQUES STATIC EQUILIBRIUM

From the preceding examples we see that the steps in the static equilibrium condition for torques: the sum
solution of a problem of statics resemble the steps we employed of torques is zero. Establish and maintain a sign con-
in Chapter 5. vention for torques; for example, for an axis pointing
into the plane of the paper, counterclockwise torques to
1 The first step is the selection of the body that is to obey
the equilibrium conditions.The body may consist of a gen- be positive and clockwise torques to be negative.
uine rigid body (for instance, the bridge in Example 1), or 6 As mentioned in Section 14.1, any line can be thought
it may consist of several pieces that act as a single rigid of as an axis of rotation; and the torque about every such
body for the purposes of the problem (for instance, the axis must be zero. You can make an unknown force dis-
bridge plus the locomotive in Example 1). It is often help- appear from the equation if you place the axis of rotation
ful to mark the boundary of the selected rigid body with a at the point of action or on the line of action of this force,
distinctive color or with a heavy line; this makes it easier so that this force has zero moment arm. Furthermore, as
to recognize which forces are external and which internal. illustrated in Example 1, sometimes it is convenient to
2 Next, list all the external forces that act on this body, and consider two different axes of rotation, and to examine
display these forces on a “free-body” diagram. the separate equilibrium conditions of the torques for
each of these axes.
3 If the forces have different directions, it is usually best to
draw coordinate axes on the diagram and to resolve the 7 As recommended in Chapter 2, it is usually best to solve
forces into x and y components. the equations algebraically for the unknown quantities,
and to substitute numbers for the known quantities as a
4 For each component, apply the static equilibrium condi- last step. But if the equations are messy, with a clutter of
tion for forces: the sum of forces is zero.
algebraic symbols, it may be convenient to substitute some
5 Make a choice of axis of rotation, calculate the torque of the numbers before proceeding with the solution of
of each force about this axis ( RF sin ), and apply the the equations.

and

5
F 9.5 10 N
y
The x and y components of the force exerted by the short arm on the tower are
6 5
therefore 3.0 10 N and 9.5 10 N, respectively.

The bottom of a ladder rests on the floor, and the top rests
EXAMPLE 4
against a wall (see Fig. 14.9a). If the coefficient of static friction
between the ladder and the floor is 0.40 and the wall is frictionless, what is
s
the maximum angle that the ladder can make with the wall without slipping?
SOLUTION: Figure 14.9b shows the “free-body” diagram for the ladder, with all
the forces. The weight of the ladder acts downward at the center of mass. If the
ladder is about to slip, the friction force at the floor has the maximum magnitude
for a static friction force, that is,
f m N (14.6)
s 1
If we reckon the torques about the point of contact with the floor, the normal
force N and the friction force f exert no torques about this point,since their moment
1