A mass M is hung with a light inextensible string as shown in Find the tension in the horizontal part of the string .

Explanation

`T_(1)` is tension in horizontal part PA of the string and `T_(2)` is tension in part PB of the string A load M is attached at `T_(2)` has two rectangular components `T_(2) cos 30^(@)` along AP and `T_(2) sin 30^(@)` along the vertical direction
`Mg = T_(2) sin 30^(@) = (T^(2))/(2)`
or `T_(2) = 2 Mg`
For equilibrium along horizontal ,
` T_(1) = T_(2) cos 30^(@) = T_(2) sqrt (3)/(2) `
using (i) `T_(1) = 2 Mg sqrt3/(2) = sqrt(3) mg`