Chapter 1: Q11-26E (page 1)

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with magnitude 4000 N. For each rod, what are (a) the strain and (b) the elongation?

Short Answer

Expert verified

$(a) S t e e l : 1.1 \times 10^{- 4} C o p p e r : 2.1 \times 10^{- 4} (b) S t e e l : 8.3 \times 10^{- 5} m C o p p e r : 1.6 \times 10^{- 4} m$

Step by step solution

Given information

Tensile force F = 400 N, l₀ = 0.75m

Area $\begin{array}{rcl} A & = & π (\frac{d^{2}}{4}) \\ = & π (\frac{{(1.50 \times 10^{- 2} m)}^{2}}{4}) \\ = & 1.77 \times 10^{- 4} m^{2} \end{array}$

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.

Strain Calculation

(a) The strain is $\frac{Δ l}{l_{0}} = \frac{F}{Y A}$

For steel $Y = 2.0 \times 10^{11} P a$

$\begin{array}{rcl} \frac{Δ l}{l_{0}} & = & \frac{4000 N}{(2.0 \times 10^{11} P a) (1.77 \times 10^{- 4} m^{2})} \\ = & 1.1 \times 10^{- 4} \end{array}$

$\begin{array}{rcl} F o r c o p p e r Y & = & 1.1 \times 10^{11} P a \\ \frac{Δ l}{l_{0}} & = & \frac{4000 N}{(1.1 \times 10^{11} P a) (1.77 \times 10^{- 4} m^{2})} \\ = & 2.1 \times 10^{- 4} \end{array}$

Elongation Calculation

(b)

For steel:

$(1.1 \times 10^{- 4}) (0.75) = 8.3 \times 10^{- 5} m$

For Copper:

$(2.1 \times 10^{- 4}) (0.75) = 1.6 \times 10^{- 4} m$

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Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with magnitude 4000 N. For each rod, what are (a) the strain and (b) the elongation?

Short Answer

Step by step solution

Given information

Concept/Formula used

Strain Calculation

Elongation Calculation

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