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A light rod of length 200 cm is suspended from the ceiling horizontally by means of two vertical wires of equallength tied to its ends. One of the wires is made of steel and is of cross- section0.1 cm^(2) and the other of brass of cross-section 0.2 cm^(2). Along the rod at which distance a weight may be hung to produce equal stresses in both the wires ? |
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Answer» ` 4/3` m from steel wire  As stresses are EQUAL, ` T_(1)/A_(1) = T_(2)/A_(2)` ` i.e, T_(1)/T_(2) = A_(1)/A_(2)= (0.1)/(0.2) or T_(2) = 2T_(1)` Now fortranslatory EQUILIBRIUM of the rod, `T_(1) + T_(2) =W` From (i) and (ii) , we get ` T_(1) = W/3 , T_(2) = (2W)/3` Now if x is the distanceof weight W from steel wire, thenfor rotationalequilibriumof rod, ` T_(1)x = T_(2) (2 -x) or W/3 x = (2 W)/ 3 (2 -x) therefore x = 4/3 m` |
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