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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 1: Q29 E (page 1)

In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and $2.5 脳 10^{- 3} c m^{2}$ in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. For each wire, compute (a) the tensile strain and (b) the elongation.

Short Answer

Expert verified

(a) Upper wire: $3.1 脳 10^{- 3}$

Lower wire: $2.0 脳 10^{- 3}$

(b) Upper wire: $1.6 m m$

Lower wire: $1.0 m m$

Step by step solution

01

Given information

Length: $l_{0} = 0.50 m$ ,

Area:localid="1668006176532" $A = 2.5 脳 10^{- 3} c m^{2}$ ,

Aluminum sphere mass: localid="1668006194126" $m_{A} = 6 k g$ ,

Brass cube mass: localid="1668006203126" $m_{B} = 10 k g$ .

02

Concept/Formula used

$Y = \frac{l_{0} F}{鈭� l}$

Where, Y is Young鈥檚 modulus, $l_{0}$ is length of muscle, F is muscle force, A is cross-sectional area and $鈭� l$ is elongation.

03

Tension calculation in wires

Tension in the below steel wire

$T_{2} = m_{B} g = (10 k g) (9.80) = 98 N$

Tension in the above steel wire

$T_{1} = m_{A} g + T_{2} = (6 k g) (9.80) + 98 N = 157 N$

04

(a) Calculation for Tensile strain

Strain in upper wire

$Y = \frac{s t r e s s}{s t r a i n} s t r a i n (蔚_{蠀}) = \frac{s t r e s s}{Y} = \frac{T_{1}}{A Y} 蔚_{蠀} = \frac{157 N}{(2.5 脳 10^{- 7} m^{2}) (2 脳 10^{11} P a)} = 3.1 脳 10^{- 3}$

Strain in lower wire

$s t r a i n (蔚_{L}) = \frac{s t r e s s}{Y} = \frac{T_{2}}{A Y} 蔚_{L} = \frac{98 N}{(2.5 脳 10^{- 7} m^{2}) (2 脳 10^{11} P a)} = 2.0 脳 10^{- 3}$

05

(b) Calculation for elongation in wires

(b) Elongation in upper wire

$鈭� l_{u} = l_{0} 脳蔚_{u} = (0.50 m) (3.1 脳 10^{- 3}) = 1.6 脳 10^{- 3} m = 1.6 m m$

Elongation in Lower wire

$鈭� l_{L} = l_{0} 脳蔚_{L} = (0.50 m) (2.0 脳 10^{- 3}) = 1.0 脳 10^{- 3} m = 1.0 m m$

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