A horizontal aluminium rod of diameter `4.8 cm` projects `5.3 cm` from a wall. A `1200 kg` object is suspended from the end of the rod. The shear modulus of aluminium is `3.0xx10^(10)N//m^(2)`. Nelecting the mass of te rod find a shearing stress on the rod and b the vertical deflection of the end of the rod.
Text Solution
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a. Shearing stress `=F/A=(Mg)/(pir^(2))=(1200xx9.8)/(3.14xx(0.024)^(2))` `=6.5xx10^(6)Nm^(-2)` b. Let `/_\x` be the vertical defelection of the rod. Then shear modulus, `G=(F/A)/(/_\x/l)` Thus `/_\x=((F/A)l)/G=(6.5xx10^(6)xx0.53)/(3.0xx10^(10))=1.1xx10^(-5)M`
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