Solving Giant Swing Horizontal Cable Tension

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Homework Help Overview

The problem involves analyzing the forces acting on a seat in a "Giant Swing" apparatus, where the seat is connected to two cables and swings in a horizontal circle. The objective is to find the tension in the horizontal cable given the weights of the seat and the person sitting in it, as well as the swing's rotational speed.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use free body diagrams (FBD) to analyze the forces on both the person and the chair. They express uncertainty about their calculations, particularly in converting revolutions per minute to meters per second for velocity.
  • Some participants suggest treating the person and chair as a single object to simplify the analysis of vertical and horizontal components of tension.
  • Questions arise regarding the correctness of the conversion from rev/min to m/s and the resulting calculations for velocity.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem. Some guidance has been offered regarding the treatment of the system as a single object and checking unit conversions, but no consensus has been reached on the correctness of the original poster's calculations.

Contextual Notes

There is a noted uncertainty regarding the conversion of rotational speed from revolutions per minute to linear velocity, which is critical for solving the problem. The original poster expresses doubt about their results and seeks clarification on their methodology.

jasonchiang97
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1. Homework Statement
In one of the versions of the "Giant Swing", the seat is connected to two cables, one of which is horizontal (Figure 1) . The seat swings in a horizontal circle at a rate of 33.0 rev/min .

If the seat weighs 295 N and a 869-N person is sitting in it, find the tension in the horizontal cable.
YF-05-58.jpg

Homework Equations


Fc=mv2/r
Period=circumference/velocity so 2πR/v

The Attempt at a Solution


I drew two FBD(free body diagrams). One for the person and one for the chair
For the person : TsinΘ-W-Fn=0
T=(W+Fn)/sinΘ
Fn = Weight of the chair

For the chair: Th+TcosΘ=mv2/r
Th = mv2/r - TcosΘ
mv2/r - [(Wc + Wp)/sinΘ]*cosΘ

To get the velocity, we use period = circumference /velocity
33 rev/min * 1min/60 sec = 0.55 rev/sec
0.55 = 2π(7.5)/v
v=85.68m/s
m= (Wp + Wc)/9.8
So (Wp+Wc/9.8)(81.68)2/7.5 - (295+869)/tan40

Basically I end up getting a huge number and I'm fairly certain that it's wrong but can't see why
 
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jasonchiang97 said:

I drew two FBD(free body diagrams). One for the person and one for the chair
I haven't checked your work in detail (sorry) but you should be able to treat the person and chair as a single object. The vertical component of the tension in the slanted cable must balance the weight of the person+chair, and the horizontal components of the two tensions must provide the centripetal acceleration.
 
Nathanael said:
I haven't checked your work in detail (sorry) but you should be able to treat the person and chair as a single object. The vertical component of the tension in the slanted cable must balance the weight of the person+chair, and the horizontal components of the two tensions must provide the centripetal acceleration.

yea that's basically what I did. I guess my main problem is I don't know how to convert rev/min to m/s so I just took the rev/min and divided it by 60 to get rev/sec which gives 0.55rev/sec

I then took 2π(7.5)/0.55 to get velocity. Wondering if that step is correct
 
##2\pi\frac{\text{radians}}{\text{revolution}}\cdot 7.5\frac{\text{meters}}{\text{radian}}\cdot\frac{1}{0.55\frac{\text{revolutions}}{\text{second}}}##

Check the units in the above equation. If it is not meters/second then try to fix it so it is.
 

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