Chapter 1: Q29E (page 360)

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 $2.5 \times 10^{- 3} c m^{2}$ 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: $3.1 \times 10^{- 3}$

Lower wire: $2.0 \times 10^{- 3}$

(b) Upper wire: 1.6 mm

Lower wire: 1.0 mm

Step by step solution

Given information

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

Area: $A = 2.5 \times 10^{- 3} c m^{2}$ ,

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

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

Concept/Formula used

$Y = \frac{l_{0} F}{∆ l}$

Where, Y is Young’s modulus, l₀ is length of muscle, F is muscle force, A is cross-sectional area and $∆ l$ is elongation.

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

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

(a) Calculation for Tensile strain

Strain in upper wire

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

Strain in lower wire

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