What Calculations Reveal Tarzan's Vine Tension at Point 3?

In summary, Tarzan (mass of 80kg) uses a 18m vine to swing across a gorge to visit Jane's hut. One day, Jane interfered by attaching a branch to the vine below the pivot point. To calculate the tension in the vine at point 3, we use the equations Ug = mgh, ΣE = 0, Ek = 0.5mv^2, and ΣF = ma. By setting Ug = Ek and solving for v, we get a velocity of 18.795 m/s at point 2. Using Fnet = ma and solving for T, we get a tension of 784.8 N, which is 15.34 times greater than Tar
  • #1

Homework Statement

Tarzan (mass of 80kg) has an 18m vine stretched horizontally from his hut to a pivot point P, as shown. He uses it to swing across the gorge to visit Jane's hut, starting from rest to conserve energy. One day Jane spoiled his plans by fastening a strong, thin, smooth, horizontal branch to intercept the vine at B, 15m directly below P. Tarzan hung on and followed the dotted path 1 to 2 to 3. Calculate the tension in the vine as he passed through point 3. How many times greater than his weight is his tension?


The answer is 15.34 (I'm unsure which question this answers)

Homework Equations

Ug =mgh
ΣE = 0
Ek = 0.5mv2
ΣF = ma

The Attempt at a Solution

1: [/B]Ug = 80(9.81)(18)= 14,126.4 J
Ek = 0.5mv2 = 0

2: Ug = 0
Ek = 0.5mv2
14 126.4= 0.5mv2
v = 18.795 m/s

Ft = mg = 784.8 N
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  • #2
it answers the tension in the vine as he passed through point 3. because if it would be the weight then the units would have been mentioned.
  • #3
In your step 2 you calculate a velocity. At what point is that his velocity? At which point do you need to determine his velocity?
As I'm sure you know, Fnet=ma. If the tension at point 3 is T, what is Fnet? What is his acceleration at that point?

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