anonymous
  • anonymous
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.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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IrishBoy123
  • IrishBoy123
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 \(\lambda\)

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