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Stage 0 Solutions:
5.1.
Model:
We can assume that the ring is a single massless particle in static equilibrium.
Visualize:
Solve:
Written in component form, Newton’s first law is
(
)
net
1
2
3
0 N
x
x
x
x
x
F
F
T
T
T
= Σ
=
+
+
=
(
)
net
1
2
3
0 N
y
y
y
y
y
F
F
T
T
T
= Σ
=
+
+
=
Evaluating the components of the force vectors from the free-body diagram:
1
x
T
= −
1
T
T
2
x
=
0 N
3
3
cos30
x
T
T
=
°
T
1
y
=
0 N
2
y
T
T
2
=
3
3
sin30
y
T
T
= −
°
Using Newton’s first law:
1
3
cos30
0 N
T
T
−
+
° =
2
3
sin30
0 N
T
T
−
° =
Rearranging:
(
)(
)
1
3
cos30
100 N
0.8666
86.7 N
T
T
=
° =
=
(
)(
)
2
3
sin30
100 N
0.5
50.0 N
T
T
=
° =
=
Assess:
Since
acts closer to the
x
-axis than to the
y
-axis, it makes sense that
.
3
T
r
1
2
T
T
>
5.5.
Visualize:
Please refer to the Figure Ex5.5.
Solve:
Applying Newton’s second law to the diagram on the left,
(
)
net
2
4 N
2 N
1.0 m/s
2 kg
x
x
F
a
m
−
=
=
=
(
)
net
2
3 N
3 N
0 m/s
2 kg
y
y
F
a
m
−
=
=
=
For the diagram on the right:
(
)
net
2
4 N
2 N
1.0 m/s
2 kg
x
x
F
a
m
−
=
=
=
(
)
net
2
3 N
1 N
2 N
0 m/s
2 kg
y
y
F
a
m
−
−
=
=
=