A 2m long light metal rod AB is suspended from the ceiling horizontally by means of two vertical wires of - Physics - Mechanical Properties Of Solids

Question

A 2m long light metal rod AB is suspended from the ceiling
horizontally by means of two vertical wires of equal length tied to
its ends One wire is of brass and has cross-sectional area of 0 2
× 10–4 m2 and the other is of steel with 0 1 ×
10–4 m2 cross-sectional area in order to have equal stresses
in the two wires, a weight W is hung from the rod The position of
the weight along the rod from end A should be

Jyoti Pant · Accepted Answer

Given,
The length of rod AB=L=2 mSuppose a weight W is hung at C at a distance x from A
and Ts=Tension in steel wireTb=Tension in brass wireStress in steel wire=Ts×AsStress in brass wire=Tb×AbAlso given,As=0
1×10-4m2Ab=0
2×10-4m2So, for equal stresses,Ts×As=Tb×AbTs×0
1×10-4=Tb×0
2×10-4Ts/Tb=2or Ts=2 TbSince the system is in equilibrium, the moments of force Ts and Tb about C will be equal
Thus,Ts×x=Tb(2-x)or 2 Tb×x=Tb(2-x)or 2x=2-x3x=2or x=2/3=0
67 m Hence to produce equal stress in two wires, the weight should be hung at a distance 0
67 m from the end A