## 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

1. ksaimouli Group Title

|dw:1350335809666:dw|

2. imron07 Group Title

|dw:1350336122902:dw|

3. ksaimouli Group Title

where is fn then

4. imron07 Group Title

What is fn

5. ksaimouli Group Title

normal force

6. ksaimouli Group Title

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

7. ksaimouli Group Title

@imron07

8. ksaimouli Group Title

can u plz help me with this

9. imron07 Group Title

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.

10. JakeV8 Group Title

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

11. ksaimouli Group Title

what do u mean by what angle

12. JakeV8 Group Title

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...

13. ksaimouli Group Title

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

14. ksaimouli Group Title

?

15. ksaimouli Group Title

can u draw it @JakeV8

16. JakeV8 Group Title

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.

17. ksaimouli Group Title

|dw:1350337034801:dw|

18. JakeV8 Group Title

X = T cos(65), right?

19. ksaimouli Group Title

yuppppp

20. ksaimouli Group Title

thx

21. ksaimouli Group Title

i was doing other way but same

22. JakeV8 Group Title

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

23. ksaimouli Group Title

|dw:1350337344014:dw|

24. JakeV8 Group Title

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

25. ksaimouli Group Title

so how to find the tension of the rope @imron07

26. ksaimouli Group Title

Tcos 65=mg

27. imron07 Group Title

Yes, you can find T then.

28. ksaimouli Group Title

then how to find the velocity

29. ksaimouli Group Title

-T sin 65= mv^2/r

30. ksaimouli Group Title

?

31. imron07 Group Title

Correct :)

32. ksaimouli Group Title

got it thx u

33. ksaimouli Group Title

hold on what is the answer for tension

34. ksaimouli Group Title

@imron07

35. ksaimouli Group Title

5102

36. ksaimouli Group Title

but book gave 5.1*10^3N

37. imron07 Group Title

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

38. ksaimouli Group Title

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

39. ksaimouli Group Title

so what did u get @imron07 this is last one

40. imron07 Group Title

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

41. imron07 Group Title

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

42. ksaimouli Group Title

how can i omit the -

43. imron07 Group Title

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