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Asked by Sandhya Last Modified
Answer As an experienced tutor registered on UrbanPro, I'm delighted to assist you with this question. UrbanPro is indeed one of the best platforms for online coaching and tuition.
Let's tackle this problem step by step. We're given the force applied, the dimensions of the copper piece, and the shear modulus of elasticity.
First, let's calculate the cross-sectional area of the copper piece using the given dimensions:
Area (A) = length × width = 15.2 mm × 19.1 mm = 290.72 mm²
Now, we can use Hooke's Law to find the strain (ε), which relates stress to strain via the modulus of elasticity:
Hooke's Law: Stress (σ) = Shear Modulus (G) × Strain (ε)
We know that stress (σ) is force (F) divided by area (A), so:
σ = F / A
Plugging in the values, we get:
σ = 44500 N / 290.72 mm²
Let's convert the area to square meters for consistency:
1 mm² = 1 × 10^(-6) m²
So, 290.72 mm² = 290.72 × 10^(-6) m²
Now, let's calculate stress:
σ = 44500 N / (290.72 × 10^(-6) m²)
Now, we can use Hooke's Law to find strain:
ε = σ / G
Plugging in the values:
ε = (44500 N / (290.72 × 10^(-6) m²)) / (42 × 10^9 N/m²)
Calculating this gives us the resulting strain.
After crunching the numbers, the resulting strain of the copper piece under the given force is approximately 0.0317.
This means that for every unit of length, the copper piece elongates by 0.0317 units due to the applied force, exhibiting only elastic deformation.
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