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

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SUMMARY

The discussion focuses on calculating the tension in Tarzan's vine as he passes through point 3 during his swing across a gorge. Using the mass of Tarzan (80 kg) and gravitational acceleration (9.81 m/s²), the potential energy (Ug) is calculated to be 14,126.4 J. The kinetic energy (Ek) at point 3 is derived from the equation Ek = 0.5mv², leading to a velocity of 18.795 m/s. The tension in the vine at point 3 is determined to be 15.34 times Tarzan's weight, which is 784.8 N.

PREREQUISITES
  • Understanding of gravitational potential energy (Ug = mgh)
  • Familiarity with kinetic energy equations (Ek = 0.5mv²)
  • Knowledge of Newton's second law (ΣF = ma)
  • Basic principles of energy conservation in physics
NEXT STEPS
  • Study the principles of energy conservation in mechanical systems
  • Learn about tension forces in pendulum motion
  • Explore advanced topics in dynamics, specifically centripetal acceleration
  • Investigate the effects of varying mass and height on tension calculations
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding dynamics and tension in swinging motions, particularly in real-world applications like amusement park rides or pendulum systems.

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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?

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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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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.
 
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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