Solving Giant Swing Horizontal Cable Tension

AI Thread Summary
The discussion focuses on calculating the tension in a horizontal cable for a Giant Swing scenario, where a seat swings at 33.0 revolutions per minute with a combined weight of 1164 N (seat plus person). The user attempts to derive the tension using free body diagrams and centripetal force equations but struggles with unit conversion from revolutions per minute to meters per second. It is suggested that treating the person and chair as a single object simplifies the calculations, emphasizing that the vertical tension must balance the total weight while the horizontal components provide centripetal acceleration. The user is advised to check the units in their calculations to ensure they yield the correct velocity in meters per second. Accurate unit conversion is crucial for determining the correct tension in the cable.
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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