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Mechanical Properties Of Fluids

Question
CBSEENPH11018104

Derive an expression for the radius of rope used in a crane to lift the mass M kg. The elastic limit of the material of rope is σ N/m2, and safety factor of crane is K.

Solution
Given mass = M kg
Elastic limit of the material = σ N/m2
Safety factor for crane = K

The design of the rope is such that it should tolerate a load of KM kg.
Let r be the radius of rope required to be used in the crane.
So, using the formula, 

   Elastic limit =Maximum loadArea of cross sectioni.e.            σ=KMgπr2            r = KMgπσ