A block is pulled slowly as shown in the diagram. Show that force `F` increase as the block moves up. Find the force `F` when the block is at depth `h`.

Explanation

Slowly, i.e. net force`=0`

`2Fcostheta-mg=0`
`F=(mg)/(2costheta)`
As the block moves up, `theta` increases, `costheta` decreases and hence `F` increases.
When the block is at depth `h`

`costheta=(d//2)/(sqrt(d^(2)//4+h^(2)))=(d)/(sqrt(d^(2)+4h^(2)))`
`F=(mg)/(2costheta)=(mgsqrt(d^(2)+4h^(2)))/(2d)`