What angle will the vine break and what's the max tension?

  • #1
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Homework Statement


You come to a river and you see a vine you will use to swing across. Your mass is 75.0 kg, the vine is 18.0 m long, and the vine breaks when under 655 N of tension.
a) at what angle from the vertical will the vine break?
b) what should be the maximum tension supported by the vine for you to safely cross the river?

Homework Equations


T=mgcosθ+ m (v^2/r)
θ=cos^-1((T-m(v^2/r))/mg)
Tmax=mgcos(0)+m(v^2bottom/r)
Tmax=mg+m(v^2bottom/r)

The Attempt at a Solution


How i would solve this problem is by using the formulas like, θ=cos^-1(T/mg)
θ=cos^-1(655/((75.0kg)(9.8m/s^2))
and get θ=27.0 degrees

and Tmax=mg
Tmax=(75.0kg)(9.8m/s^2)
=735 N

is this the correct approach?
 
Last edited:

Answers and Replies

  • #2
haruspex
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In your attempt, you have not included the v2/r term. To find the velocity you need to know what the starting angle is, but you do not have enough information. Is this the exact wording of the question? Is it possible you have left something out?
 
  • #3
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In your attempt, you have not included the v2/r term. To find the velocity you need to know what the starting angle is, but you do not have enough information. Is this the exact wording of the question? Is it possible you have left something out?
This is the EXACT wording of the question i have, this is on an old exam my professor gave the class to study from (without solutions). Am i supposed to assume since the most tenstion will be halfway thru the swing that it'll be 90 degrees? But it doesnt say he's starting with the vine exactly straight either so i feel that'd be incorrect to assume. Maybe he just wanted to keep the velocity term as v and solve and keep a v in the solution?
 
  • #4
haruspex
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This is the EXACT wording of the question i have, this is on an old exam my professor gave the class to study from (without solutions). Am i supposed to assume since the most tenstion will be halfway thru the swing that it'll be 90 degrees? But it doesnt say he's starting with the vine exactly straight either so i feel that'd be incorrect to assume. Maybe he just wanted to keep the velocity term as v and solve and keep a v in the solution?
I think you will have to assume the vine starts horizontal. An alternative would be to specify an initial angle, θ0, and express the answer in terms of that.
 
  • #5
CWatters
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Deleted. I miss read the question.
 

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