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A swing ride at a carnival consists of chairs that are swung in a horizontal circle by 12.0 m cables that are attached to a vertical rotating pole, as the drawing shows. Suppose that the total mass of the chair and it occupant is 220 kg. i need just the free body diagram

Physics
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|dw:1350335809666:dw|
|dw:1350336122902:dw|
where is fn then

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Other answers:

What is fn
normal force
and how did u get t cos 65 because that is y coordinate right so it would be t sin 65 ?
can u plz help me with this
The chair doesn't touch any surface right? Then there's no normal force. To decide choosing cos or sin, you need to know what angle is given first.
I don't think you need a normal force for this one. (yeah, what @imron07 said!)
what do u mean by what angle
For the cos and sin relations, don't think of the y-coordinate... just look at the diagram and remember that cos = adj/hyp and sin = opp/hyp . You should see it as @imron07 drew it...
yes i know that cos =... and sin=.. but ..
?
can u draw it @JakeV8
I can't tell what you are asking... @imron07 has the right diagram. The vertical component of the force exerted by the 12m line is T cos(65) because T is the hyp and the vertical component is the adjacent side.
|dw:1350337034801:dw|
X = T cos(65), right?
yuppppp
thx
i was doing other way but same
glad to help. I'm pretty out of practice on these, but the diagram helped me remember a little.
|dw:1350337344014:dw|
yes, that's what I get. I have to take off, just signed on for a minute... good luck!
so how to find the tension of the rope @imron07
Tcos 65=mg
Yes, you can find T then.
then how to find the velocity
-T sin 65= mv^2/r
?
Correct :)
got it thx u
hold on what is the answer for tension
5102
but book gave 5.1*10^3N
\[T=\frac{mg}{\cos65}=5101N=5.1\times10^3N\]
and the last one velocity |dw:1350338528028:dw|
so what did u get @imron07 this is last one
Actually, you can omit the minus sign:\[v=\frac{Tr\sin65}{m}\]where r equals:\[r=l\sin65\]|dw:1350338783066:dw|
sorry,\[v=\sqrt{\frac{Tr\sin65}{m}}\]
how can i omit the -
the centripetal force is radially inward. This force itsel is minus, so: \[-\frac{mv^2}{r}=-T\sin65 \]

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