Question

A 2.05-m-long cylindrical steel wire with a cross-sectional diameter of 4.00 mm is placed over a light, frictionless pulley. An object of mass m1 = 5.80 kg is hung from one end of the wire and an object of mass m2 = 3.10 kg from the other end as shown in the figure below. The objects are released and allowed to move freely. Compared with its length before the objects were attached, by how much has the wire stretched while the objects are in motion?

Answer 1

Given

length of the wire is L = 2.05 m ,

diameter d = 4 mm = 0.004 m

m1 = 5.80 kg, m2 = 3.10 kg

writing the forces acting on the mases vertically  

T - m1*g = -m1*a

T - m2*g = m2*a

saubstituting the values

T - 5.80*9.8=-5.80*a ==> T = -5.80*a+5.80*9.8= 5.80(9.8-a)

T -3.10*9.8= 3.10*a ==> 5.80(9.8-a) - 3.10*9.8= 3.10*a ==> a= 2.9730 m/s2

now the tension is

T = 5.80(9.8-a) = 5.80(9.8-2.9730) N = 39.5966 N

the young's modulus of the steel is y = 20*10^10 N/m2

by definition y = F*L/(A*dL)

20*10^10 = 39.5966*2.05/(pi*0.002^2*L)

dL = 3.229*10^-5 m

so the the wire stretched while the objects are in motion is 3.229*10^-5 m