Composite wire of uniform diameter 3.0 mm consisting of A copper wire ...
Given data:
- Diameter of composite wire (d) = 3.0 mm
- Length of copper wire (Lc) = 2.2 m
- Length of steel wire (Ls) = 7.6 m
- Elongation of composite wire (ΔL) = 0.7 mm
- Young's modulus of copper (Ec) = 1.1 × 10^11 Pa
- Young's modulus of steel (Es) = 2.0 × 10^11 Pa
Calculating cross-sectional area of composite wire:
- The diameter of the composite wire is given as 3.0 mm.
- We know that the diameter is twice the radius (r), so the radius of the composite wire is 1.5 mm or 0.0015 m.
- The cross-sectional area (A) of the composite wire can be calculated using the formula A = πr^2.
- Substituting the value of the radius, we get A = π(0.0015)^2 = 7.065 × 10^-6 m^2.
Calculating total elongation:
- The total elongation (ΔL) of the composite wire is given as 0.7 mm or 0.0007 m.
- The total elongation is the sum of the elongations of the copper and steel wires, ΔL = ΔLc + ΔLs.
Calculating elongation of copper wire:
- The elongation of a wire can be calculated using the formula ΔL = FL/AE, where F is the force applied, A is the cross-sectional area, and E is the Young's modulus.
- Rearranging the formula, we get F = ΔL × A × E.
- Substituting the values, we get Fc = (0.0007) × (7.065 × 10^-6) × (1.1 × 10^11) = 0.005415 N.
Calculating elongation of steel wire:
- Similarly, we can calculate the elongation of the steel wire using the same formula.
- Fs = (0.0007) × (7.065 × 10^-6) × (2.0 × 10^11) = 0.01083 N.
Calculating total load:
- The total load is the sum of the loads on the copper and steel wires, Ft = Fc + Fs.
- Substituting the values, we get Ft = 0.005415 N + 0.01083 N = 0.016245 N.
Answer:
The load on the composite wire is approximately 0.016245 N.
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