A 56 kg student runs at 4.5 m/s, grabs a hanging rope (10 m), and swings out...

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...over a lake.? A 56 kg student runs at 4.5 m/s, grabs a hanging rope (10 m), and swings out over a lake. He releases the rope when his velocity is zero .

What is the angle when he releases the rope?
theta = 26 degrees... that much i know

What is the tension in the rope just before he releases it?
Ft = __________ N ?

What is the maximum tension in the rope?
Ft = __________ N ?
 
This is best solved with a diagram:

.........................⁄|\
......................⁄...|.
...'...............⁄..Ø..|
...............⁄..'.......| ..
...........⁄.'............|.(L - h)
........⁄..'..........….| ...
.....⁄...'................|........
☻_--------------------|--------
........‾‾--__..........| ...↕ h
..................‾‾‾---☻←―― V = 4.5m/sec

The student has KE while he was running and this was
converted to PE at the height h as he swings. When his velocity was zero, PE = KE

mgh = ½mVi ² ..<== m cancels
gh = ½Vi ²
2gh = Vi ²
2(9.8)h = (4.5)²
h = 20.25 / 19.6
h = 1.03 m

L-h = 10 - 1.03 = 8.97m

cosØ = (L-h)/10
..........= 8.97 / 10 =.897
Ø = cos‾ ¹(.897) = 26.23°

At the point of zero speed, Fc is zero

TcosØ
.│.........⁄.T........... TcosØ = mg
.│Ø...⁄..'................T(cos26.23°) = 56(9.8)
.│...⁄...'...................T = 611.81 N
☻-------
.│
.│
.↓mg
 
At the extreme position of the string when the velocity becomes zero and the student is momentarily at rest the forces acting on him are
the tension in the string T ,
The component of his weight mgcos θ along the string and opposite to T
The component perpendicular to the above, mgsin θ
For no motion along the string, obviously mgcos θ = T
Maximum tension is mg = Tmax
So if θ is known you can calculate T and Tmax.
 
At the extreme position of the string when the velocity becomes zero and the student is momentarily at rest the forces acting on him are
the tension in the string T ,
The component of his weight mgcos θ along the string and opposite to T
The component perpendicular to the above, mgsin θ
For no motion along the string, obviously mgcos θ = T
Maximum tension is mg = Tmax
So if θ is known you can calculate T and Tmax.
 
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