An upper wire hangs from a ceiling and supports a mass m₂ = 36 kg . A lower wire hangs from m₂ and supports a mass m₁ = 10 kg . A light hanger is attached below m₁ . Both wires are made of the same material having a breaking stress of 8 imes 10^8 N/m ^2 . The area of cross section for the upper wire is 0.006 cm ^2 while that for the lower wire is 0.003 cm ^2 . Calculate the maximum load that can be added to the hange

Step-by-step solution

Let m be the mass of the load added to the hanger. The tension in the lower wire is given by the total weight it supports: T₁ = (m₁ + m)g = (10 + m) 10 The tension in the upper wire is given by the total weight it supports: T₂ = (m₂ + m₁ + m)g = (36 + 10 + m) 10 = (46 + m) 10 The maximum tension the lower wire can withstand before breaking is: T_ 1, max = Breaking stress A₁ T_ 1, max = (8 10^8) (0.003 10⁻⁴) = 240 N The maximum tension the upper wire can withstand before breaking is: T_ 2, max = Breaking stress A₂ T