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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
 one year ago
 one year 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
 one year ago
 one year ago

This Question is Closed

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

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

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

imron07 Group TitleBest ResponseYou've already chosen the best response.1
What is fn
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
normal force
 one year 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 ?
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
@imron07
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
can u plz help me with this
 one year 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.
 one year 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!)
 one year ago

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

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

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

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

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

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

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
dw:1350337344014:dw
 one year 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!
 one year ago

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

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

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

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

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

imron07 Group TitleBest ResponseYou've already chosen the best response.1
Correct :)
 one year ago

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

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

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
@imron07
 one year ago

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

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

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

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

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

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

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