Statics equilibrium?

Calling

`f_s=(sqrt2/2,sqrt2/2)t_s`
`f_h=(-1,0)t_h`
`p=(0,-Mg)`

the equilibrium equation is

`f_s+f_h+p=(0,0)`

or

`{(sqrt2/2 t_s-t_h=0),(sqrt2/2 t_s-Mg=0):}`

solving for `t_s,t_h` we obtain

`t_s = sqrt2 M g` and
`t_h=Mg`

`tan beta=(Mg)/F`

`F=(Mg)/tan beta`

`tan 45^o=1`

`F=(Mg)/1`

`F=Mg`

Let `AB` be initial position of mass `M` and `AB'` be its final position.

We see that the mass has moved a horizontal distance `=CB'`
and consequently it has moved through a vertical distance of `BC`
Due to movement in the vertical direction it has gained
Gravitational potential energy`=Mg|vec(BC)|` ......(1)
Due to Law of conservation of energy this must be equal to work done by an effective Horizontal force `vecF` through a distance `vec(CB')`

I have used effective as position of the force keeps on changing as the mass moves up.
Work done by the force`=vecFcdotvec(CB')`

As shown in the fig above, angle between the two is `0^@`
`=>` Work done by the force`=|vecF||vec(CB')|` .......(2)

Equating (1) and (2) we get
`Mgbar(BC)=|vecF|bar(CB')`
`=>|vecF| =(Mgbar(BC))/ bar(CB')` ......(3)

Let `L` be length of the string. In `Delta ACB'`
`bar(CB')=L sin45^@=L/sqrt2`
and `bar(BC)=L-L/sqrt2` (`Delta ACB'` is an isosceles triangle)

Inserting these values in (3) we get
`|vecF| =(Mg) (L-L/sqrt2)/ (L/sqrt2)`
`=>|vecF| =Mg (sqrt2-1)`