Page 239 - Fisika Terapan for Engineers and Scientists

14.2 Examples of Static Equilibrium 439

2 m mg
s
tan u 2 m s (14.12)
mg
With 0.40, this yields tan 0.80. With a calculator, we find that the angle
s
with this tangent is
u 39
For any angle larger than this, equilibrium is impossible, because the maximum
frictional force is not large enough to prevent slipping of the ladder.
(a)

A uniform rectangular box 2.0 m high, 1.0 m wide, and 1.0 m
EXAMPLE 5
deep stands on a flat floor. You push the upper end of the box
to one side and then release it (see Fig. 14.10a). At what angle of release will the
box topple over on its side?
SOLUTION: The forces on the box when it has been released are as shown in the
“free-body” diagram in Fig. 14.10b. Both the normal force N and the friction force
f act at the bottom corner, which is the only point of contact of the box with the
Box is tilted and
floor. The weight acts at the center of mass, which is at the center of the box. does not slip.
Since the box rotates about the bottom corner, let us consider the torque about
this point. The only force that produces a torque about the bottom corner is the (b)
weight.The weight acts at the center of mass; for a uniform box, this is at the center
of the box. The torque exerted by the weight can be expressed as d Mg, where d
is the perpendicular distance from the bottom corner to the vertical line through the Weight acts at d
center of mass (see Fig. 14.10b).This torque produces counterclockwise rotation if center of mass.
the center of mass is to the left of the bottom corner, and it produces clockwise rota-
N
tion if the center of mass is to the right of the bottom corner.This means that in the
former case, the box returns to its initial position, and in the latter case it topples w
over on its side. Thus, the critical angle beyond which the box will tip over corre-
f
sponds to vertical alignment of the bottom corner and the center of the box (see Fig. Consider net torque
14.10c).This critical angle equals the angle between the side of the box and the diag- about this corner, where
friction and normal forces
onal.The tangent of this angle is the ratio of the width and the height of the box,
exert no torque.
0.50 m
tan u 0.50 (c)
1.0 m
With our calculator we find that the critical angle is then
u
u 27
COMMENT: In this example we found that the box begins to topple over if its
inclination is such that the center of mass is vertically aligned with the bottom N
corner. This is a special instance of the general rule that a rigid body resting on a w
surface (flat or otherwise) becomes unstable when its center of mass is vertically above u
the outermost point of support.
For smaller u, weight At critical angle,
produces a counter- weight exerts
clockwise torque. no torque.

A uniform rectangular box 2.0 m high, 1.0 m wide, and 1.0 m
EXAMPLE 6 FIGURE 14.10 (a) Box standing on
deep stands on the platform of a truck (Fig. 14.11a). What is
edge. (b) “Free-body” diagram for the box.
the maximum forward acceleration of the truck that the box can withstand with-
(c) “Free-body” diagram if the box is tilted
out toppling over? Assume that the coefficient of static friction is large enough at the critical angle.The center of mass is
that the box will topple over before it starts sliding. directly above the edge.