A light rod of length 2.00m is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends….

Q: A light rod of length 2.00m is suspended from the ceiling horizontally by means 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 m² and the other is of brass of cross-section 2 × 10-3 m². Find out the position along the rod at which a weight may be hung to produce, equal stress in both wires

(a) 1.44 m

(b) 1.33 m

(c) 1.55 m

(d) 1.22 m

Ans: (b)

Sol: $ \displaystyle \frac{T_1}{A_1} = \frac{T_2}{A_2} $

$ \displaystyle \frac{T_1}{10^{-3}} = \frac{T_2}{2\times 10^{-3}} $

$\displaystyle T_2 = 2 T_1 $ ….(i)

Let weight is hung at a distance x from one end , Taking moment at this point ,

$ \displaystyle T_1 x = T_2 (2-x) $

$ \displaystyle T_1 x = 2 T_1 (2-x) $

$ \displaystyle x = 4 – 2x $

$ \displaystyle x = \frac{4}{3} = 1.33 m $

Leave a Comment