Equilibrium and Elasticity
11-15
EVALUATE:
We assumed that the wing force was upward and the tail force was downward. When we solved for
these forces we obtained positive values for them, which confirms that they do have these directions. Note that the
rear stabilizer provides a
downward
force. It does not hold up the tail of the aircraft, but serves to counter the
torque produced by the wing. Thus balance, along with weight, is a crucial factor in airplane loading.
Figure 11.43
11.44.
IDENTIFY:
Apply the first and second conditions of equilibrium to the truck.
SET UP:
The weight on the front wheels is
f
n
, the normal force exerted by the ground on the front wheels. The
weight on the rear wheels is
r
n
, the normal force exerted by the ground on the rear wheels. When the front wheels
come off the ground,
f
0
n
→
. The free-body diagram for the truck without the box is given in Figure 11.44a and
with the box in Figure 11.44b. The center of gravity of the truck, without the box, is a distance
x
from the rear
wheels.
EXECUTE:
0
y
F
=
∑
in Fig.11.44a gives
rf
8820 N 10,780 N 19,600 N
wn n
=+=
+
=
0
τ
=
∑
in Fig.11.44a, with the axis at the rear wheels and counterclockwise torques positive, gives
f
(3.00 m)
0
nw
x
−=
and
f
(3.00 m)
10,780 N
(3.00 m) 1.65 m
19,600 N
n
x
w
⎛⎞
==
=
⎜⎟
⎝⎠
.
(a)
0
=
∑
in Fig.11.44b, with the axis at the rear wheels and counterclockwise torques positive, gives
box
f
(1.00 m)
(3.00 m)
(1.65 m)
0
wn
w
+−
=
.
f
(3600 N)(1.00 m) (19,600 N)(1.65 m)
9,580 N
3.00 m
n
−+
0
y
F
=
∑
gives
r
f
box
nnw w
+= +
and
r
3600 N 19,600 N 9580 N 13,620 N
n
=+
. There is 9,580 N on the
front wheels and 13,620 N on the rear wheels.
(b)
f
0
n
→
.
0
=
∑
gives
box
(1.00 m)
(1.65 m)
0
ww
−
=
and
4
box
1.65
3.23 10 N
×
.
EVALUATE:
Placing the box on the tailgate in part (b) reduces the normal force exerted at the front wheels.
Figure 11.44a, b
11.45.
IDENTIFY:
In each case, to achieve balance the center of gravity of the system must be at the fulcrum. Use
Eq.(11.3) to locate
cm
x
, with
i
m
replaced by
i
w
.
SET UP:
Let the origin be at the left-hand end of the rod and take the
x
+
axis to lie along the rod. Let
1
255 N
w
=
(the rod) so
1
1.00 m
x
=
, let
2
225 N
w
=
so
2
2.00 m
x
=
and let
3
wW
=
. In part (a)
3
0.500 m
x
=
and
in part (b)
3
0.750 m
x
=
.
EXECUTE:
(a)
cm
1.25 m
x
=
.
11
2 2
33
cm
123
wx
x
www
++
=
gives
12
c
m1
2
3
3c
m
()
wwx w
xw
x
w
xx
−
=
−
and
(480 N)(1.25 m) (255 N)(1.00 m) (225 N)(2.00 m)
140 N
0.500 m 1.25 m
W
−−
−
.
(b)
Now
3
140 N
and
3
0.750 m
x
=
.
cm
(255 N)(1.00 m)
(225 N)(2.00 m)
(140 N)(0.750 m)
1.31 m
255 N
225 N 140 N
x
.
W
must be moved
1.31 m 1.25 m
6 cm
to the right.
EVALUATE:
Moving
W
to the right means
cm
x
for the system moves to the right.