Question

I would just like someone to take a look at my work and tell me where I am going wrong.

In another version of the "Giant Swing", the seat is connected to two cables as shown in the figure one of which is horizontal. The seat swings in a horizontal circle at a rate of 33.7rev/min .

If the seat weighs 214N and a 867-N person is sitting in it, find the tension in the horizontal cable.

Answer 1

Fc=TaSin(40)+Tb
Wtotal=1081N
m=110.306kg

33.7rpm x 2pir/rev x 1min/60s = 8.35pi m/s

\[Ta=\frac{ 1081 }{ \cos(40) }\]

\[Tb=Fc-TaSin(40)\]
\[Tb=ma _{c}-T _{a}\sin(40)\]
\[Tb=110.306kg(\frac{ v^2 }{ r})-Tasin(40)\]
\[Tb=110.306kg(\frac{ (8.35\pi)^2 }{ 7.5 }-1411.14N*\sin(40)\]
\[Tb=74998.19N\]

Answer 2

I tried to draw the diagram, but it wont let me post it for some reason, any help is appreciated!

Answer 3

The tension will vary with the location of the swing. It will be maximum at the bottom of the circle (centripetal force PLUS gravity) and minimum at the top of the circle (centripetal force MINUS gravity).

At a horizontal orientation, the tension will be pure centripetal force.

Let's derive the equation for all angles to see why!

A FBD reveals

The force in the radial direction, towards the pivot points being positive, \[F_r = mr \omega^2 + mg \cos(\theta)\]

You've converted rev/min to rad/sec properly (I didn't check the number, but the formula is right).

I assume you know the radius of the swing.