Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in Figure 6.3. Find the tension in each wire, neglecting the masses of the wires.
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Answers
Strategy:
The system of interest is the traffic light, and its free-body diagram is shown in (Figure)(c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in (Figure)(d). There are two unknowns in this problem and
), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero.
Solution
First consider the horizontal or x-axis:
Thus, as you might expect,
This gives us the following relationship:
Thus,
Note that, and
are not equal in this case because the angles on either side are not equal. It is reasonable that
ends up being greater than
because it is exerted more vertically than TThe system of interest is the traffic light, and its free-body diagram is shown in (Figure)(c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in (Figure)(d). There are two unknowns in this problem (
T
1
and
T
2
), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero.
Solution
First consider the horizontal or x-axis:
F
net
x
=
T
2
x
−
T
1
x
=
0.
Thus, as you might expect,
T
1
x
=
T
2
x
.
This gives us the following relationship:
T
1
cos
30
°
=
T
2
cos
45
°
.
Thus,
T
2
=
1.225
T
1
.
Note that
T
1
and
T
2
are not equal in this case because the angles on either side are not equal. It is reasonable that
T
2
ends up being greater than
T
1
because it is exerted more vertically than
T
1
.
Now consider the force components along the vertical or y-axis:
F
net
y
=
T
1
y
+
T
2
y
−
w
=
0.
This implies
T
1
y
+
T
2
y
=
w
.
Substituting the expressions for the vertical components gives
T
1
sin
30
°
+
T
2
sin
45
°
=
w
.
There are two unknowns in this equation, but substituting the expression for
T
2
in terms of
T
1
reduces this to one equation with one unknown:
T
1
(
0.500
)
+
(
1.225
T
1
)
(
0.707
)
=
w
=
m
g
,
which yields
1.366
T
1
=
(
15.0
kg
)
(
9.80
m/s
2
)
.
Solving this last equation gives the magnitude of
T
1
to be
T
1
=
108
N
.
Finally, we find the magnitude of
T
2
by using the relationship between them,
T
2
=
1.225
T
1
, found above. Thus we obtain
T
2
=
132
N
.
Significance
Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker in Newton’s Laws of Motion._1.
Now consider the force components along the vertical or y-axis:
F
net
y
=
T
1
y
+
T
2
y
−
w
=
0.
This implies
T
1
y
+
T
2
y
=
w
.
Substituting the expressions for the vertical components gives
T
1
sin
30
°
+
T
2
sin
45
°
=
w
.
There are two unknowns in this equation, but substituting the expression for
T
2
in terms of
T
1
reduces this to one equation with one unknown:
T
1
(
0.500
)
+
(
1.225
T
1
)
(
0.707
)
=
w
=
m
g
,
which yields
1.366
T
1
=
(
15.0
kg
)
(
9.80
m/s
2
)
.
Solving this last equation gives the magnitude of
T
1
to be
T
1
=
108
N
.
Finally, we find the magnitude of
T
2
by using the relationship between them,
T
2
=
1.225
T
1
, found above. Thus we obtain
T
2
=
132
N
.
Significance
Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker in Newton’s Laws of Motion.