- ksaimouli

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

- schrodinger

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- ksaimouli

|dw:1350335809666:dw|

- anonymous

|dw:1350336122902:dw|

- ksaimouli

where is fn then

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## More answers

- anonymous

What is fn

- ksaimouli

normal force

- ksaimouli

and how did u get t cos 65 because that is y coordinate right so it would be t sin 65 ?

- ksaimouli

- ksaimouli

can u plz help me with this

- anonymous

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.

- anonymous

I don't think you need a normal force for this one. (yeah, what @imron07 said!)

- ksaimouli

what do u mean by what angle

- anonymous

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...

- ksaimouli

yes i know that cos =... and sin=.. but ..

- ksaimouli

?

- ksaimouli

can u draw it @JakeV8

- anonymous

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.

- ksaimouli

|dw:1350337034801:dw|

- anonymous

X = T cos(65), right?

- ksaimouli

yuppppp

- ksaimouli

thx

- ksaimouli

i was doing other way but same

- anonymous

glad to help. I'm pretty out of practice on these, but the diagram helped me remember a little.

- ksaimouli

|dw:1350337344014:dw|

- anonymous

yes, that's what I get. I have to take off, just signed on for a minute... good luck!

- ksaimouli

so how to find the tension of the rope @imron07

- ksaimouli

Tcos 65=mg

- anonymous

Yes, you can find T then.

- ksaimouli

then how to find the velocity

- ksaimouli

-T sin 65= mv^2/r

- ksaimouli

?

- anonymous

Correct :)

- ksaimouli

got it thx u

- ksaimouli

hold on what is the answer for tension

- ksaimouli

- ksaimouli

5102

- ksaimouli

but book gave 5.1*10^3N

- anonymous

\[T=\frac{mg}{\cos65}=5101N=5.1\times10^3N\]

- ksaimouli

and the last one velocity |dw:1350338528028:dw|

- ksaimouli

so what did u get @imron07 this is last one

- anonymous

Actually, you can omit the minus sign:\[v=\frac{Tr\sin65}{m}\]where r equals:\[r=l\sin65\]|dw:1350338783066:dw|

- anonymous

sorry,\[v=\sqrt{\frac{Tr\sin65}{m}}\]

- ksaimouli

how can i omit the -

- anonymous

the centripetal force is radially inward. This force itsel is minus, so: \[-\frac{mv^2}{r}=-T\sin65 \]

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