Find the maximum tension allowed.

[SAUMYA B]

A light rope fixed at one end to the wooden clamp on the ground passes over a tree branch and hangs on the other side. It makes an angle 30° 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 360N is applied to it. Find the maximum acceleration in the upward direction with which the man can climb safely. Neglect friction at the tree branch. Take g ≈ 10m /s^2.

Let t be tension in rope. Then upward force on the clamp is tsin30°≈t/2≈360N.
T≈720N
Let a be acceleration of man in upward direction then,
T-600N≈60 kg . a
Therefore maximun acceleration of the man is
a≈ (720-600)N/60 kg. ≈ 2m/s^2.


Now my question is if we draw the tree body diagram of the wooden clamp we should include the normal contact force by the rope and the ground.Therefore upward force on the clamp is the resultant of the of the normal contact force by the rope and the tension in it. Why then we only consider the tension in the rope as the only upward force.

 

[SteamKing]

From your description, it appears that the wooden clamp is thrust into the ground, and then the rope is tied to the clamp and thrown over a tree branch. I don't know what you mean by "the normal contact force by the rope and the ground."

Could you include a sketch of the problem?

 

[haruspex]

I think I understand the question. You are supposing that the clamp goes over the rope end, so a short horizontal section of rope is clamped between the ground below it and the clamp above it. Is that right?
But it could equally be that the clamp encompasses the rope and grips it perfectly well without needing any normal force from the ground.
Since there's no way to estimate what such a normal force would be in the first model, I think it's fair to take the second.