Chapter 1: Q11-29E (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 ^* 10^-3 cm² 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 10.0kg. For each wire, compute (a) the tensile strain and (b) the elongation.

Short Answer

Expert verified

$(a) U p p e r w i r e : 3.1 \times 10^{- 3} L o w e r w i r e : 2.0 \times 10^{- 3} (b) U p p e r w i r e : 1.6 m m L o w e r w i r e : 1.0 m m$

Step by step solution

Given information

$L e n g t h I_{0} = 0.50.0 m, A r e a A = 2.5 \times 10^{- 3} c m^{2}, A l u m i n u m s p h e r e m a s s m_{A} = 6 k g, B r a s s c u b e m a s s m_{B} = 10 k g$

Concept/Formula used

$Y = \frac{l_{0} F}{AΔl}$

Where, Y is Young’s modulus, I₀ 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

$\begin{array}{rcl} T_{2} & = & m_{B} g \\ = & (10 k g) (9.80) \\ = & 98 N \end{array}$

Tension in the above steel wire

$\begin{array}{rcl} T_{1} & = & m_{A} g + T_{2} \\ = & (6 k g) (9.80) + 98 N \\ = & 157 N \end{array}$

Calculation for Tensile strain

(a) Strain in upper wire

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

Strain in lower wire

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

Calculation for elongation in wires

(b) Elongation in upper wire

$\begin{array}{rcl} Δ l_{U} & = & l_{0} \times ε_{U} \\ = & (0.50 \times m) (3.1 \times 10^{- 3}) \\ = & 1.6 \times 10^{- 3} m = 1.6 m m \end{array}$

Elongation in Lower wire

$\begin{array}{rcl} Δ l_{L} & = & l_{0} \times ε_{L} \\ = & (0.50 \times m) (2.0 \times 10^{- 3}) \\ = & 1.0 \times 10^{- 3} m = 1.0 m m \end{array}$

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Short Answer

Step by step solution

Given information

Concept/Formula used

Tension calculation in wires

Calculation for Tensile strain

Calculation for elongation in wires

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