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 ratio 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 = "Stress"/"Strain" = (F/A)/"Strain" = ((4F)/(pid^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 prop 1/d^2`

Young’s modulus for iron, Y₁ = 190 × 10⁹ Pa

Diameter of the iron wire = d₁

Young’s modulus for copper, Y₂ = 120 × 10⁹ Pa

Diameter of the copper wire = d₂

Therefore, the ratio of their diameters is given as:

`d_2/d_1 = sqrt(Y_1/Y_2) = sqrt((190xx10^9)/(120xx10^9)) = sqrt(19/12) = 1:25:1`

Answer 2

Since each wire is to have same tension, therefore, each wire has the same extension. Moreover, each wire has the same initial length.So, the strain is same for each wire.

Now, `Y = "Stress"/"Strain" = (F/((piD^2)/4))/"Strain"`

or `Y prop 1/D^2 => D prop 1/sqrtY`

`D_"copper"/(D_"iron") = sqrt((Y_""iron)/(Y_"copper")) = sqrt((190 xx 10^9)/(110 xx 10^9)) = sqrt(19/11) =1.314`