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

Answer

`sqrt(3)Mg`

Explanation

`T_(2) cos 60^(@)=Mg`
`T_(2)=2Mg` ...(i)
and for horizontal equilibium of P
`T_(1)=T_(2) sin 60^(@)=T_(2)(sqrt(3)//2)` ...(ii)
Substituting the value of `T_(2)` from eqn (i) in (ii)
`T_(1)=(2Mg)xx(sqrt(3)//2)`
`=(sqrt(3))Mg`