James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

804 Chapter 12 Vectors and the Geometry of Space
50° 32°
ExamplE 7 A 100-lb weight hangs from two wires as shown in Figure 19. Find the
tensions (forces) T 1 and T 2 in both wires and the magnitudes of the tensions.
T¡
T
50° 32°
100
T¡
T
FIGURE 19
100
50° 32°
T¡ T
50° 32°
w
FIGURE 20
SOLUtion We first express T 1 and T 2 in terms of their horizontal and vertical components.
From Figure 20 we see that
50° 32°
T¡ T sin 50° j
5 T 1 − 2| T 1 | cos 50° i 1 | T 1 |
50°
w
32°
6 T 2 − | T 2 | cos 32° i 1 | T 2 | sin 32° j
The resultant T 1 1 T 2 of the tensions counterbalances the weight w − 2100 j and so
we must have
Thus
T 1 1 T 2 − 2w − 100 j
(2| T 1 | cos 50° 1 | T 2 | cos 32°) i 1 (| T 1 | sin 50° 1 | T 2 | sin 32°) j − 100 j
Equating components, we get
2| T 1 | cos 50° 1 | T 2 | cos 32° − 0
| T 1 | sin 50° 1 | T 2 | sin 32° − 100
Solving the first of these equations for | T 2 | and substituting into the second, we get
| T 1 | sin 50° 1 | T 1| cos 50°
cos 32°
sin 32° − 100
| T 1 | Ssin 50° 1 cos 50°
sin 32°
cos 32°D − 100
So the magnitudes of the tensions are
| T 1 | − 100
sin 50° 1 tan 32° cos 50°
< 85.64 lb
and | T 2 | − | T 1 | cos 50°
cos 32°
< 64.91 lb
Substituting these values in (5) and (6), we obtain the tension vectors
T 1 < 255.05 i 1 65.60 j
T 2 < 55.05 i 1 34.40 j ■
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