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ksaimouli
 3 years ago
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
ksaimouli
 3 years ago
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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350335809666:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350336122902:dw

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0I don't think you need a normal force for this one. (yeah, what @imron07 said!)

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0what do u mean by what angle

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For 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...

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

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0can u draw it @JakeV8

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350337034801:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0X = T cos(65), right?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0glad to help. I'm pretty out of practice on these, but the diagram helped me remember a little.

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350337344014:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, that's what I get. I have to take off, just signed on for a minute... good luck!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, you can find T then.

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0then how to find the velocity

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[T=\frac{mg}{\cos65}=5101N=5.1\times10^3N\]

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, you can omit the minus sign:\[v=\frac{Tr\sin65}{m}\]where r equals:\[r=l\sin65\]dw:1350338783066:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry,\[v=\sqrt{\frac{Tr\sin65}{m}}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the centripetal force is radially inward. This force itsel is minus, so: \[\frac{mv^2}{r}=T\sin65 \]
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