(a) A box of weight `10sqrt(3)` N is held n equilibrium with the help of two strings OA and OB as shown in figure-1. the string OA is horizontal. Find the tensions in both the stings.

(b) If you can change location of the point A on the wall and hence the orientation of the string OA without altering the orientation of the string OB as shown in figure-II. what angle should the string OA make with the wall so that a minimum tension is developed in it?

Explanation

(a) Free body diagram of the box
Graphical Method: Use triangle law
`T_(2)sin 60^(@)=10sqrt(3)rArrT_(2)=20N`
`T_(1)tan 60^(@)=10sqrt(3)rArrT_(1)=10N`
Analytical Method: Use Cartesian components
`sumF_(x)=0rArr T_(2)cos 60^(@)=T`
`sumF_(y)=0rArr T_(2) sin 60^(@)=10sqrt(3)`

From equation (i) `&` (ii) we have `T_(1)=10N` and `T_(2)=20N`
(b) Free body diagram of the box

Graphical Method: Use triangle law
For `T_(1)` to be minium, it must be perpendicualr to `T_(2)`
From figure `theta=60^(@)`
Analytical Method: Use cartesian components
`sumF_(x)=0rArrT_(2)cos60^(@)=T_(1)sintheta` ....(i)
`sumF_(y)=0rArr T_(1) cos theta+T_(2) sin 60^(@)=10sqrt(3)` ....(ii)

From equation (i) and (ii), we have `T_(1)(10sqrt(3))/sqrt(3)sintheta+costheta`
If `T_(1)` is minimum, `sqrt(3 ) sin theta+cos theta` must be maximum. Maximum value of `sqrt(3) sin theta+cos theta` is 2.
Solving the above equation we get `theta=60^(@)`