1120
Chapter 11
11.57.
I
DENTIFY
:
Apply
0
z
τ
=
∑
to the beam.
S
ET
U
P
:
The freebody diagram for the beam is given in Figure 11.57.
E
XECUTE
:
0, axis at hinge
z
τ
Σ
=
, gives
(6.0 m)(sin40
)
(3.75 m)(cos30
)
0
T
w
° −
° =
and
7600 N
T
=
.
E
VALUATE
:
The tension in the cable is less than the weight of the beam.
sin40
T
°
is the component of
T
that is
perpendicular to the beam.
Figure 11.57
11.58.
I
DENTIFY
:
Apply the first and second conditions of equilibrium to the drawbridge.
S
ET
U
P
:
The freebody diagram for the drawbridge is given in Figure 11.58.
v
H
and
h
H
are the components of
the force the hinge exerts on the bridge.
E
XECUTE
:
(a)
0
z
τ
=
∑
with the axis at the hinge gives
(7.0 m)(cos37
)
(3.5 m)(sin37
)
0
w
T
−
+
=
°
°
and
5
cos37
(45,000 N)
2
1.19
10
N
sin37
tan37
T
w
=
=
=
×
°
°
°
(b)
0
x
F
=
∑
gives
5
h
1.19
10
N
H
T
=
=
×
.
0
y
F
=
∑
gives
4
v
4.50
10
N
H
w
=
=
×
.
2
2
5
h
v
1.27
10
N
H
H
H
=
+
=
×
.
v
h
tan
H
H
θ
=
and
20.7
θ
=
°
. The hinge force has magnitude
5
1.27
10
N
×
and is
directed at 20.7
°
above the horizontal.
E
VALUATE
:
The hinge force is not directed along the bridge. If it were, it would have zero torque for an axis at
the center of gravity of the bridge and for that axis the tension in the cable would produce a single, unbalanced
torque.
Figure 11.58
11.59.
I
DENTIFY
:
Apply the first and second conditions of equilibrium to the beam.
S
ET
U
P
:
The freebody diagram for the beam is given in Figure 11.59.
Figure 11.59
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Equilibrium and Elasticity
1121
E
XECUTE
:
(a)
0,
z
τ
=
∑
axis at lower end of beam
Let the length of the beam be
L
.
(sin 20 )
cos40
0
2
L
T
L
mg
⎛
⎞
°
= −
° =
⎜
⎟
⎝
⎠
1
2
cos40
2700 N
sin20
mg
T
°
=
=
°
(b)
Take
y
+
upward.
0
y
F
=
∑
gives
sin60
0
n
w
T
−
+
° =
so
73.6 N
n
=
0
x
F
=
∑
gives
s
cos60
1372 N
f
T
=
° =
s
s
,
f
n
μ
=
s
s
1372 N
19
73.6 N
f
n
μ
=
=
=
E
VALUATE
:
The floor must be very rough for the beam not to slip. The friction force exerted by the floor is to
the left because
T
has a component that pulls the beam to the right.
11.60.
I
DENTIFY
:
Apply
0
z
τ
=
∑
to the beam.
S
ET
U
P
:
The center of mass of the beam is 1.0 m from the suspension point.
E
XECUTE
:
(a)
Taking torques about the suspension point,
(4.00 m)sin30
(140.0 N)(1.00 m)sin30
(100 N)(2.00 m)sin30
w
+
=
°
°
°
. The common factor of sin30
°
divides out,
from which
15.0 N.
w
=
(b)
In this case, a common factor of sin45
°
would be factored out, and the result would be the same.
E
VALUATE
:
All the forces are vertical, so the moments are all horizontal and all contain the factor sin
θ
, where
θ
is the angle the beam makes with the horizontal.
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 Spring '06
 Buchler
 Physics, Force, Friction, Sin, freebody diagram, tan θ

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