A 1m long uniform rod with mass 1.5 kg is supported at its ends by wires A and B of equal length. The cross sectional area of A is 2mm^(2) and that of B is 4 mm^(2). Young's Modulus for wire A is 1.80 x 10^(11) Pa; that for B is 1.20x10^(11) Pa. At what point along the rod should a weight W = 29.4 N be suspended to produce equal strain in A and B.

Question

irishboy123 · Answer

From Wiki:  $\large E =  \frac{F L_0} {A_0 \Delta L} $

or $\large \Delta L =  \frac{F L_0} {A_0 E } $

and $\large \Delta L_A = \Delta L_A \implies  \frac{F_A} {A_{oA} E_A } = \frac{F_B } {A_{oB} E_B } $  [$\LARGE \star$]

so you need the relationship between $F_A$ and $F_B$

for that, i would take moments separately about each end of the stick

so around  end A, if the weight is at distance $\lambda$ along rod from A, you have $1.5g \times \dfrac{1}{2} + 29.4 \times  \lambda = F_B \times  1 $

repeat for $F_A$ and put those straight into [$\LARGE \star$] and solve for $\lambda$