Please help, I'm stuck on this problem:An aluminum wire and a steel wire, each of length 4m, are hung from the ceiling. A 7 kg mass is suspended from the lower end of each wire. The aluminum wire has a diameter of 1.1 mm. What must be the diameter of the steel wire if it is to stretch the same distance as the aluminum wire, so that the two wires maintain equal lengths after masses are attached? Young's modulus for aluminum is .70 x 10^11 Pa and for steel it is 2.0 x 10^11 Pa.
The cross-sectional area of the steel wire must be (.70 x 10^11 / 2.0 x 10^11) of the aluminum wire. So the area of the Al wire is pi*(0.55mm)^2 = 9.50331778 × 10^-7 m^2 The area of the Steel wire must be 3.32616122 × 10^-7 m^2 now solve for the radius of the steel wire. A = pi*r^2 - r = 0.32538 mm d = 2*r = 0.65077 mm
With the usual symbols E= FL/eA, so F = EeA/L Call the aluminium wire '1' and the steel wire '2'. The tensions are the same (weight of 7kg) so: E?e?A?/L? = E?e?A?/L? Since the extensions and lengths are the same e?=e? and L? = L? so these cancel: E?A? = E?A? A?/A? = E?/E? Let the diameter of the aluminum wire be d? mm. Since the cross-sectional areas are proportional to the diametera squared: d??/d?? = E?/E? d? = 1.1 x √(0.70/2.0) = 0.65mm
