Answer to Question #121378 in Mechanics | Relativity for neelam

Question

A mass of 1019Kg is supported by two wires A and B each made of a different material. Before loading the wires are of equal length. After loading they are still of equal length. Find the load carried by each wire.
(WA = 3066N) (WB = 6930N)
Wire
Diameter mm
E Nmm-2
A

1

206,000
B
2

116,000

Answer 1

Solution.

$m=1019kg; g=9.81N/kg;$

$d_A=1mm;$

$d_B=2mm;$

$E_A=206000N/mm^2;$

$E_B=116000N/mm^2;$

$F_g=F_A+F_B;$

$F_g=mg;$

$F_g=1019kg\sdot9.81N/kg=9996N;$

$\sigma=\dfrac{F}{S};$ $\sigma=E\epsilon \implies F=ES\epsilon;$

$S=\pi \dfrac{d^2}{4};$

$S_A=3.14\sdot\dfrac{(1mm)^2}{4}=0.785mm^2;$

$S_B=3.14\sdot\dfrac{(2mm)^2}{4}=3.14mm^2;$

$F_g=E_AS_A\epsilon+E_BS_B\epsilon;$

$\epsilon=\dfrac{F_g}{E_AS_A+E_bS_B};$

$\epsilon=\dfrac{9996N}{206000N/mm^2\sdot0.785mm^2+116000N/mm^2\sdot3.14mm^2}=0.019;$

$F_A=206000N/mm^2\sdot0.785mm^2\sdot0.019=3072N;$

$F_B=116000N/mm^2\sdot3.14mm^2\sdot0.019=6924N;$

Answer: $F_A=3072N;$

$F_B=6924N.$