A light rod of length 4.0 m is suspended from the ceiling horizontally by means of two vertical wire 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 the rod at which a weight may be hung to produce equal stresses in both wires.

Answer

`66.67`

Explanation

Let weight W be suspended at distance x from steel end.
Stress in steel wire `= (T _(1))/( a _(1))`
Stress in brass wire `= (T _(2))/( a _(2))`
When stresses are equal,
`(T _(1))/( a _(1)) = (T _(2))/( a _(2))`
` therefore (T _(1))/( T _(2)) = (a _(1))/( a _(2)) = (10 ^(-3))/( 2 xx 10 ^(-3)) = 0.5`
As system of rod and wires is in equilibrium,
`T _(1) xx AC = T _(2) xx BC`
`therefore ( T _(i))/( T _(2)) = (BC)/( AC) = ( 1- x)/(x)`
` therefore 0.5 = (1- x)/( x)`
`therefore 1. 5 x = 1 `
` therefore x = (1)/(1.5) = 0. 6667 m` from stell wire
`therefore x = 66. 67 ` cm