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438 CHAPTER 14 Statics and Elasticity

(a) (b)
y
Frictionless wall can exert
only a normal force.

N 2
a
Weight acts at
center of mass.
u
u Floor exerts both
l a normal force and
a friction force.
N 1
w

0 x
f
Ladder is about to slip
FIGURE 14.9 (a) Ladder leaning against a when f = m s N .
1
wall. (b) “Free-body” diagram for the ladder.

arms are zero. The weight w mg acting at the center of mass exerts a counter-
clockwise torque of magnitude (l 2) mg sin , and the normal force N of the
2
wall exerts a clockwise torque of magnitude l N sin , where is the angle
2
between the ladder and the normal force (see Fig. 14.9b); since 90 ,
the sine of equals the cosine of , and the torque equals l N cos .For
2
equilibrium, the sum of these torques must be zero,
l
mg sin u lN cos u 0 (14.7)
2
2
or, equivalently,
1
mg sin u N cos u (14.8)
2
2
We collect the factors that depend on by dividing both sides of this equation by
1
2 mg cos , so
sin u 2N 2

cos u mg
or, since sin cos tan ,
2N 2
tan u (14.9)
mg

To evaluate the angle we still need to determine the unknown N . For this,
2
we use the condition for translational equilibrium: the net vertical and the net hor-
izontal forces must be zero, or

N mg 0 (14.10)
1
N m N 0 (14.11)
2 s 1
From the first of these equations, N mg; therefore, from the second equation,
1
N mg. Inserting this into our expression (14.9) for the tangent of the angle ,
2 s
we obtain the final result