QUICK QUESTION, please help test in an hour!?

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To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch of the tree. He then pulls downward on the free end of the rope with a force of 370 N. Neglect any friction between the rope and the branch, and determine the man's upward acceleration.

I don't need a numerical answer, bc I already have one (0.478 m/s^2)

i need to know why the eq. is F=(9.80x72.0) + 2(370) ?
how come when he pulls the rope down with a force of 370 (so -370) and you add it to all the y forces that doesn't cancel out with one of the tension forces. can someone the process in as much detail as possible?
thankyou soo much!
i meant -9.80 in the eq.
*** why don't u take into account the -370 into the total F (since the applied force is downward)?
 
It's like this


....▓▓▓▓▓│▓▓▓▓▓▓▓▓▓▓
...............││
...............││
...............││
..............//.\
.............\(.}//
..........,....▓
............../'/||
............//..IL

Showing the forces acting. (The -370N pull of the man on the rope, that you are saying, is an internal force, unless you cut the rope, it will not be shown as acting.)


....▓▓▓▓▓│▓▓▓▓▓▓▓▓▓▓ FBD of the branch
...............││
...............↓'↓T = 370N
..T = 370N


.T = 370 N.
...............↑'↑T = 370N
...............││
...............││
..............//.\
.............\(.}//........FBD of the man
..........,....▓
............../'/||
............//..IL
............... ↓W = mg

Now, the 370N pull exerted downwards, which is an "internal" force in the rope is now shown. That is what gives more tension to the rope, otherwise, the T on each rope would just be equal to ½W. And since the branch has no friction, that force is transmitted to the other end. So, when we "cut" the rope to show the forces acting, the "free-body diagram is as shown in the figure.

Unbalanced force = F

F = 2T - W
F = 2(370) - (9.8 x 72.0)

F = ma
2(370) - (9.8 x 72.0) = (72.0)a
740 - 705.6 = (72)a
34.4 = (72)a
a = 0.4777777m-sec‾ ²
 
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