A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in figure. The cross-sectional areas of wires A and B are 1.0 mm² and 2.0 mm², respectively.  At what point along the rod should a mass m be suspended in order to produced (a) equal stresses and (b) equal strains in both steel and aluminium wires.

From tables of standards we know that for steel γ_A = 2×10¹¹ Nm⁻² and for aluminium 

γ_B = 7×10¹⁰ Nm⁻², A_A = 1.0 mm²      and A_B = 2.0 mm².

(a) Let for equal stress in two wires the load m be suspended at x distance from end A.

As stress =F/A, hence F_A/F_B= A_A/A_B

= 1.00 mm²/2.00 mm²

= 1/2

and from principle of moments, we have F_A × x = F_B × (1.05 − x)     

∴    F_A/F_B= (1.05 − x)/x

Combining two results, we have (1.05 − x)/x=  1/2

 or  2.1 − 2x = x  or  2.1 = 3x ⇒ x = 2.1/3 = 0.7 m

(b) Let for equal strains in two wires the load m be suspended at x′ distance from end A.

As strain = stress/Y = F/A.Y = same for two wires, hence

F_A/F_B= A_A Y_A/A_B Y_B

= 1.00 × 2 × 10¹¹/2.00 × 7 × 10¹⁰

= 10/7

and from principle of moments, we have

F_A × x′ = F_B × (1.05 − x′)     or    

F_A/F_B= (1.05 − x′)/x′

Combining the two results, we have

10/7 = (1.05 − x′)/x′ or  10x′ = 7.35 − 7x   ;     

17x′ = 7.35  or  x′ = 7.35/17 = 0.43m.