A load of `10 kN` is supported from a pulley which in turn is supported by a rope of sectional area `1xx10^(3)mm^(2)` and modulus of elasticity `10^(3)Nmm^(-2)`, as shown in figure. Neglecting the friction at the pulley, determine the deflection of the load.
Text Solution
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Let `T` be the tension in the rope. Then `2T=10kNimpliesT=5N` Hence longitudinal stress in the rope is `sigma=T/A=(5kN)/(10^(3)mm^(2))=5N//mm^(2)` Extension in the rope `=("Stress")/YxxL` `=-(5N//mm^(2))/(10^(3)N//mm)xx1500mm=7.5mm` Therefore, deflection of the load `delta=7.5/2=3.75mm`
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CENGAGE PHYSICS-PROPERTIES OF SOLIDS AND FLUIDS-INTEGER_TYPE
