Page 173 - Schaum's Outline of Theory and Problems of Applied Physics

158 ELASTICITY [CHAP. 13

SOLVED PROBLEM 13.1
A nylon rope 24 mm in diameter has a breaking strength of 120 kN. Find the breaking strengths of similar
ropes (a)12mmand (b) 48 mm in diameter.
Since the breaking stress F/A is the same for all the ropes, their breaking strengths F are in proportion to their
2
2
cross-sectional areas A. The cross-sectional area of a cylinder of diameter d is A = πr = πd /4, and so in each
2
case F varies directly with d .
1 2 1
(a) A rope 12 mm in diameter has an area ( ) = that of a rope 24 mm in diameter. Hence its breaking strength
2 4
is one-fourth as much, or 30 kN.
2
(b) A rope 48 mm in diameter has an area 2 = 4 times that of a rope 24 mm in diameter. Hence its breaking
strength is four times as much, or 480 kN.
YOUNG’S MODULUS
When a tension or compression force F acts on an object of length L 0 and cross-sectional area A (Fig. 13-2),
the result is a change in length L. Below the elastic limit, the ratio between stress and strain in this situation
is called Young’s modulus:
F/A
Y =
L/L 0
longitudinal stress
Young’s modulus =
longitudinal strain
The value of Young’s modulus depends only on the composition of the object, not on its size or shape. The usual
units of Y are newtons per square meter and pounds per square inch.

L 0
DL

A

F

Fig. 13-2

SOLVED PROBLEM 13.2
An aluminum wire 3 mm in diameter and 4 m long is used to support a mass of 50 kg. What is the
elongation of the wire? Young’s modulus for aluminum is 7 × 10 10 Pa.

The cross-sectional area of a wire of radius r = 1.5mm = 1.5 × 10 −3 mis
2
2
A = πr = (π)(1.5 × 10 −3 m) = 7.07 × 10 −6 m 2
The applied force is
2
F = mg = (50 kg)(9.8 m/s ) = 490 N
and so the elongation of the wire is
L 0 F (4m)(490 N) −3
L = = 2 = 3.96 × 10 m = 3.96 mm
Y A (7 × 10 N/m )(7.07 × 10 −6 m )
2
10