Question

A rigid rod of mass 0.2 kg and length 1.5 m is suspended by wires at each end. The wireshave equal rest length. The wire on the left end of the rod has Young’s modulus4*10^11Pa, while the one on the right end has Young’s modulus2*10^11Pa. A 2 kg mass is hung fromthe beam. How far from the left end of the rod should the mass be hung such that the bar is horizontal?

I was told I need to use tension some how but I can't seem to get it.

Answer 1

I guess the wires have equal area A

delta L = L * F/(EA) same for each
so
F/E must be the same for each
F1/4 = F2/2

F1 = 2 F2
now moments about distance x from left where m is hung
F1 * x = F2 (1.5-x)
2 F2 x = F2(1.5-x)
3 x = 1.5
x = .5 meters