A steel wire with an initial length of 1 m and a cross-sectional area of 4.00 mm ^2 is secured at both ends such that it remains horizontal and tension-free. When a 2.16 kg load is hung from the wire's midpoint, what will be its vertical depression? Use Y for steel = 2.0 10¹¹ N/m ^2 and g = 10 m/s ^2 .

Step-by-step solution

Let the initial length of the wire be 2l = 1 m , so l = 0.5 m . Let x be the vertical depression at the midpoint. The new length of each half of the wire is l' = l^2 + x^2 . The change in length of each half is l = l^2 + x^2 - l l (1 + x^2 2l^2 ) - l = x^2 2l . The longitudinal strain in the wire is = l l = x^2 2l^2 . The tension T in the wire is given by T = Y A = Y A x^2 2l^2 . For equilibrium at the midpoint, the vertical components of the tension balance the weight of the load: 2 T = m g where = x l' x l . Subs