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A light rod of length 2 m is suspended from the ceiling horizontally by mean.s of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross-section 10^(-3)m^(2) and the other is of brass of cross -section 2xx10^(-3)m^(2). Find out the position along of corss-section 2xx10^(-3)m^(2). Find out the position along the rod at which a weight may be hung to produce,(i) equal stress in bothwires (ii) equal strains in both wires Young.s modulus of brass =1xx10^(11)N//m^(2) Young.s modulus of steel =2xx10^(11)N//m^(2) |
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Answer» Solution :Suppose `a_(1) and a_(2)` are the corss - sectional areas, and `Y_(1) and Y_(2)` are the Young.s MODULI of steel and brass wirerespectively. Let `T_(1) and T_(2)` are tensions in the steel and brass wires respectively. Let x is distance of the position of the hanging weight from the steel wire. (i) First case : For EQUAL stress in both wires, wehave `(T_(1))/(a_(1))=(T_(2))/(a_(2)) (or) (T_(1))/(10^(-3))=(T_(2))/(2xx10^(-3))(or) T_(2)=2T_(1)""......(i)` As the whole system is in equilibrium, so `Sigma bartau=0`. Taking moment of all the forces acting on the rod about C, we have `T_(1)x-T_(2)(2-x)=0""...( ii)` Solving equations (i) and (ii), we get `x=(4)/(3)m` (ii) Second case: For equal STRAINS in both the wires `e_(1)=e_(2)` `(T_(1)l)/(a_(1)Y_(1))=(T_(2)l)/(a_(2)Y_(2))(or) (T_(1))/(10^(-3) xx 2xx10^(11))=(T_(2))/(2xx10^(-3)xx10^(11))` (or) `T_(1)=T_(2)""...(iii)` From equations (ii) and (iii), we get x = 1M |
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