Just some of the forces are shown here. т, Sketch т, 30° T, T 4.45° (a) (b) Only forces on the system are shown. т. T2y т, 30° 45° 30° 45° System of interest- T, (c) (d) Free-body diagram The net vertical force is zero, so Ty + Tay = -W TIx The net horizontal force is zero, so Tix = -Tx (e) Figure 4.24 A traffic light is suspended from two wires. (b) Some of the forces involved. (c) Only forces acting on the system are shown here. The

University Physics Volume 1
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Author:William Moebs, Samuel J. Ling, Jeff Sanny
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Chapter6: Applications Of Newton's Laws
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Consider the traffic light (mass 15.0 kg) suspended from two wires. Find the tension in each wire, neglecting the masses of the wires

Just some of the forces are
shown here.
т,
Sketch
т,
30°
T, T
4.45°
(a)
(b)
Only forces on the system
are shown.
т.
T2y
т,
30°
45°
30°
45°
System of
interest-
T,
(c)
(d)
Free-body diagram
The net vertical
force is zero, so
Ty + Tay = -W
TIx
The net horizontal
force is zero, so
Tix = -Tx
(e)
Figure 4.24 A traffic light is suspended from two wires. (b) Some of the forces involved. (c) Only forces acting on the system are shown here. The
Transcribed Image Text:Just some of the forces are shown here. т, Sketch т, 30° T, T 4.45° (a) (b) Only forces on the system are shown. т. T2y т, 30° 45° 30° 45° System of interest- T, (c) (d) Free-body diagram The net vertical force is zero, so Ty + Tay = -W TIx The net horizontal force is zero, so Tix = -Tx (e) Figure 4.24 A traffic light is suspended from two wires. (b) Some of the forces involved. (c) Only forces acting on the system are shown here. The
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