A square metal block with mass (m) is suspended by five wires with the same length. One of these wires is made of iron at the center of the block and has a radius (1.2 mm) while the other four wires have the same cross-sectional area and they are made of copper at the block corners. If the tension in the iron wire equals (0.3 mg), find the radius of a copper wire. (Take Yiron = 2×1011, Ycopper =1. 2×1011 N/m2 )

1.18 mm

Explanation:

The mass of the block is m, so the weight of the block is mg.  If the tension in the iron wire is 0.3 mg, then the total tension in the 4 copper wires is 0.7 mg.  So the tension in each copper wire is 0.175 mg.

For each wire:

ΔL/L = F / (EA)

where ΔL is the change in length,

L is the initial length,

F is the force,

E is Young's modulus (modulus of elasticity),

and A is the cross sectional area.

The five wires have the same initial length, and stretch by the same amount.

ΔL/L = ΔL/L

F / (EA) = F / (EA)

(0.3 mg) / (2×10¹¹ Pa × π(1.2 mm)²) = (0.175 mg) / (1.2×10¹¹ Pa × πr²)

0.3 / (2 (1.2 mm)²) = 0.175 / (1.2 r²)

0.36 r² = 0.504 mm²

r = 1.18 mm