A light rope fixed at one end of a wooden clamp on the ground passes over a tree branch and hangs on the other side. It makes an angle of `30^0` with the ground. A man weighing (60 kg) wants to climb up the rope. The wooden clamp can come out of the ground if an upward force greater than 360 N is applied it. Find the maximum acceleration in the upward direction with which the man can climb safely. Neglect friction at the tree branch. Take `g=10 m/s^2`

Explanation

Let T be the tension in the rope. The upward force on the clamp is T sin30^0=T/2`. The maximum tension tht will not detach the clamp from the ground is, therefore, given by
`T/2=360 N`
or, `T=720 N`.
If the acceleration of the man in the upward direction is a the equation of motion of the man is
`T-600 N=(60 kg)a`
The maximum acceleration of the man for safe climbing is therefore,
`a=(720N-600N)/(60 Kg) =2m/s^2`