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
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ksaimouli
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|dw:1350335809666:dw|
imron07
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|dw:1350336122902:dw|
ksaimouli
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where is fn then
imron07
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What is fn
ksaimouli
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normal force
ksaimouli
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and how did u get t cos 65 because that is y coordinate right so it would be t sin 65 ?
ksaimouli
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@imron07
ksaimouli
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can u plz help me with this
imron07
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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.
JakeV8
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I don't think you need a normal force for this one. (yeah, what @imron07 said!)
ksaimouli
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what do u mean by what angle
JakeV8
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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
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yes i know that cos =... and sin=.. but ..
ksaimouli
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?
ksaimouli
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can u draw it @JakeV8
JakeV8
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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
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|dw:1350337034801:dw|
JakeV8
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X = T cos(65), right?
ksaimouli
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yuppppp
ksaimouli
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thx
ksaimouli
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i was doing other way but same
JakeV8
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glad to help. I'm pretty out of practice on these, but the diagram helped me remember a little.
ksaimouli
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|dw:1350337344014:dw|
JakeV8
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yes, that's what I get. I have to take off, just signed on for a minute... good luck!
ksaimouli
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so how to find the tension of the rope @imron07
ksaimouli
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Tcos 65=mg
imron07
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Yes, you can find T then.
ksaimouli
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then how to find the velocity
ksaimouli
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-T sin 65= mv^2/r
ksaimouli
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?
imron07
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Correct :)
ksaimouli
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got it thx u
ksaimouli
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hold on what is the answer for tension
ksaimouli
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@imron07
ksaimouli
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5102
ksaimouli
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but book gave 5.1*10^3N
imron07
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\[T=\frac{mg}{\cos65}=5101N=5.1\times10^3N\]
ksaimouli
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and the last one velocity |dw:1350338528028:dw|
ksaimouli
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so what did u get @imron07 this is last one
imron07
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Actually, you can omit the minus sign:\[v=\frac{Tr\sin65}{m}\]where r equals:\[r=l\sin65\]|dw:1350338783066:dw|
imron07
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sorry,\[v=\sqrt{\frac{Tr\sin65}{m}}\]
ksaimouli
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how can i omit the -
imron07
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the centripetal force is radially inward. This force itsel is minus, so: \[-\frac{mv^2}{r}=-T\sin65 \]