11.38. A 4.0-m-long steel wire has a cross-sectional area of 0.050 cm2. Its proportional (elastic) limit
has a value of 0.0016 times its Young’s modulus (see Table 11.1). Its breaking stress has a value
of 0.0065 times its Young’s modulus. The wire is fastened at its upper end and hangs vertically.
(a) How great a weight can be hung from the wire without exceeding the proportional (elastic)
limit? (b) How much will the wire stretch under this load? (c) What is the maximum weight that
the wire can support?
Identify: The proportional limit and breaking stress are values of the stress, F /A . Use
Eq.(11.10) to calculate l .
10
Set Up: For steel, Y  20 10 Pa . F  w .
Execute:
3
10
6
2
3
(a) w  (1.6 10 )(20 10 Pa)(5 10 m )  1.60 10 N. = fpYA
 F l
l     0  (1.6  103 )(4.0 m)  6.4 mm
 A Y
(b)
3
10
6
2
3
(c) (6.5 10 )(20 10 Pa)(5 10 m )  6.5 10 N. = fbYA
Evaluate:
At the proportional limit, the fractional change in the length of the wire is 0.16%.