## ksaimouli 2 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

1. ksaimouli

|dw:1350335809666:dw|

2. imron07

|dw:1350336122902:dw|

3. ksaimouli

where is fn then

4. imron07

What is fn

5. ksaimouli

normal force

6. ksaimouli

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

7. ksaimouli

@imron07

8. ksaimouli

can u plz help me with this

9. imron07

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

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

11. ksaimouli

what do u mean by what angle

12. JakeV8

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

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

14. ksaimouli

?

15. ksaimouli

can u draw it @JakeV8

16. JakeV8

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

|dw:1350337034801:dw|

18. JakeV8

X = T cos(65), right?

19. ksaimouli

yuppppp

20. ksaimouli

thx

21. ksaimouli

i was doing other way but same

22. JakeV8

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

23. ksaimouli

|dw:1350337344014:dw|

24. JakeV8

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

25. ksaimouli

so how to find the tension of the rope @imron07

26. ksaimouli

Tcos 65=mg

27. imron07

Yes, you can find T then.

28. ksaimouli

then how to find the velocity

29. ksaimouli

-T sin 65= mv^2/r

30. ksaimouli

?

31. imron07

Correct :)

32. ksaimouli

got it thx u

33. ksaimouli

hold on what is the answer for tension

34. ksaimouli

@imron07

35. ksaimouli

5102

36. ksaimouli

but book gave 5.1*10^3N

37. imron07

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

38. ksaimouli

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

39. ksaimouli

so what did u get @imron07 this is last one

40. imron07

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

41. imron07

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

42. ksaimouli

how can i omit the -

43. imron07

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