A force of 400N is applied to a material of original length of 2m and causes an extension of 4mm. Given that…

Question

A force of 400N is applied to a material of original length of 2m and causes an extension of 4mm. Given that cross sectional area of a material is 80cm². Determine the; (i) Tensile strain (ii) Tensile stress.

ScanSolve AI Tutor · Accepted Answer

Thank you for providing a clearer image. I will re-solve the problem with the correct values.

Identify the given values and convert units to SI.
- Original length ( $L_0$ ) = $2 \text{ m}$
- Extension ( $\Delta L$ ) = $4 \text{ mm} = 4 \times 10^{-3} \text{ m}$
- Cross-sectional area ( $A$ ) = $80 \text{ cm}^2 = 80 \times (10^{-2} \text{ m})^2 = 80 \times 10^{-4} \text{ m}^2 = 8 \times 10^{-3} \text{ m}^2$
- Applied Force ( $F$ ) = $400 \text{ N}$
Calculate the tensile strain.
Tensile strain ( $\epsilon$ ) is defined as the ratio of the extension to the original length.
$\epsilon = \frac{\Delta L}{L_0}$
Substitute the values:
$\epsilon = \frac{4 \times 10^{-3} \text{ m}}{2 \text{ m}}$ $\epsilon = 0.002$
The tensile strain is dimensionless.
$\boxed{\text{Tensile strain} = \textbf{0.002}}$
Calculate the tensile stress.
Tensile stress ( $\sigma$ ) is defined as the ratio of the applied force to the cross-sectional area.
$\sigma = \frac{F}{A}$
Substitute the values:
$\sigma = \frac{400 \text{ N}}{8 \times 10^{-3} \text{ m}^2}$ $\sigma = 50,000 \text{ N/m}^2$ $\sigma = 5 \times 10^4 \text{ Pa}$ $\boxed{\text{Tensile stress} = \textbf{5} \times \textbf{10}^{\textbf{4}} \textbf{ Pa}}$
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A force of 400N is applied to a material of original length of 2m and causes an extension of 4mm. Given that cross sectional area of a material is 80cm². Determine the; (i) Tensile strain (ii) Tensile stress.

Step-by-step solution

Identify the given values and convert units to SI.

Calculate the tensile strain.

Calculate the tensile stress.

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