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ksaimouli
Group Title
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
 2 years ago
 2 years ago
ksaimouli Group Title
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
 2 years ago
 2 years ago

This Question is Closed

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
dw:1350335809666:dw
 2 years ago

imron07 Group TitleBest ResponseYou've already chosen the best response.1
dw:1350336122902:dw
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
where is fn then
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
normal force
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
and how did u get t cos 65 because that is y coordinate right so it would be t sin 65 ?
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
can u plz help me with this
 2 years ago

imron07 Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
I don't think you need a normal force for this one. (yeah, what @imron07 said!)
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
what do u mean by what angle
 2 years ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
For the cos and sin relations, don't think of the ycoordinate... just look at the diagram and remember that cos = adj/hyp and sin = opp/hyp . You should see it as @imron07 drew it...
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
yes i know that cos =... and sin=.. but ..
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
can u draw it @JakeV8
 2 years ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
dw:1350337034801:dw
 2 years ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
X = T cos(65), right?
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
i was doing other way but same
 2 years ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
glad to help. I'm pretty out of practice on these, but the diagram helped me remember a little.
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
dw:1350337344014:dw
 2 years ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
yes, that's what I get. I have to take off, just signed on for a minute... good luck!
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
so how to find the tension of the rope @imron07
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
Tcos 65=mg
 2 years ago

imron07 Group TitleBest ResponseYou've already chosen the best response.1
Yes, you can find T then.
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
then how to find the velocity
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
T sin 65= mv^2/r
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
got it thx u
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
hold on what is the answer for tension
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
but book gave 5.1*10^3N
 2 years ago

imron07 Group TitleBest ResponseYou've already chosen the best response.1
\[T=\frac{mg}{\cos65}=5101N=5.1\times10^3N\]
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
and the last one velocity dw:1350338528028:dw
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
so what did u get @imron07 this is last one
 2 years ago

imron07 Group TitleBest ResponseYou've already chosen the best response.1
Actually, you can omit the minus sign:\[v=\frac{Tr\sin65}{m}\]where r equals:\[r=l\sin65\]dw:1350338783066:dw
 2 years ago

imron07 Group TitleBest ResponseYou've already chosen the best response.1
sorry,\[v=\sqrt{\frac{Tr\sin65}{m}}\]
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
how can i omit the 
 2 years ago

imron07 Group TitleBest ResponseYou've already chosen the best response.1
the centripetal force is radially inward. This force itsel is minus, so: \[\frac{mv^2}{r}=T\sin65 \]
 2 years ago
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