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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

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

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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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