Question

A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.

Answer 1

The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same.

The relation for Young’s modulus is given as :

$Y = \frac{Stress }{ Strain }= \frac{\frac{F }{ A} }{ Strain }= \frac{\frac{4F}{πd^2} }{ Strain} ....(i)$
Where,
F = Tension force
A = Area of cross-section
d = Diameter of the wire
It can be inferred from equation (i) that $Y ∝ \frac{1 }{ d^2}$
Young’s modulus for iron, Y1 = 190 × 10 9 Pa
Diameter of the iron wire = d1
Young’s modulus for copper, Y2 = 110 × 10 9 Pa
Diameter of the copper wire = d2
Therefore, the ratio of their diameters is given as :

$\frac{d_2 }{ d_1} = \frac{√Y_1 }{ Y_2} = \sqrt\frac{{190 × 10 ^9 }}{ 110 × 10^ 9 }= \sqrt\frac{19 }{11}$ = 1.31:1