What would the greatest force on the vine be during the swing?

  • Thread starter shell4987
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Homework Statement


Tarzan, who weighs 619 N, swings from a cliff at the end of a convenient vine that is 14 m long (Fig. 8-37). From the top of the cliff to the bottom of the swing, he descends by 4.3 m. The vine will break if the force on it exceeds 1550 N. What would the greatest force on the vine be during the swing?
**image attached**

Homework Equations


Fynet=T-mgcos(theta)=(mv^2)/r
T=mgcos(theta)+(mv^2)/r
0+mgh=(1/2)mv^2+0 (initial and final energies)
v^2=2gh


The Attempt at a Solution


I attempted this problem and plugged everything into the tension equation, but I don't know how to get the angle for the cosine, I put in zero and my tension was 424.44 N and that's obviously not right. Can anyone help me out?
 

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Answers and Replies

  • #2
Astronuc
Staff Emeritus
Science Advisor
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The greatest force on the vine will occur at maximum tangential speed (and therefore maximim kinetic energy). The tension is the resultant the component of Tarzan's weight in the direction parallel with the vine and centrifugal force due to the circular motion of Tanzan's mass.

So with mgy = 1/2 mv2 one can find v, where y is the vertical distance traveled.
 

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