Determine the tension in the cable attached to the chair

  • Thread starter abstrkt
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A "swing" ride at a carnival consists of chairs that are swung in a circle by 12.0 m cables attached to a vertical rotating pole, as the drawing shows. ( = 70.0°) Suppose the total mass of a chair and its occupant is 200 kg

(a) Determine the tension in the cable attached to the chair.

(b) Find the speed of the chair.



I'm really lost..can anyone help me?
 

Answers and Replies

  • #2
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i couldnt really help with this question without working through it. So,
try reading it a stage at a time and understanding it fully before moving on.
This will hopefully help you to understand it further :smile:

firstly, see attached diagram to go with this question.
First we must resolve the tension in the vertical direction (Tv)
in centripetal questions like this Tv = Mg (or the weight)
so
Tv = 200 x 9.81
Tv = 1962N

Since we know the angle we can now find T using trigonometry
so
T (or the hypotenuse H) = 1962/cos 70
T = 5737N


Now to find the velocity (v)
firstly we must find the tension in the radius or horizontal (Th). We do this the same way as we found the overall tension in the rope, using trig, so:

Th = 1962 tan 70
Th = 5391N but we know Th = (M.v^2)/r
so substituting between formulas we get
5391 (Th) = (200(M) x V^2)/12 sin 70 where 12 sin 70 is the radius(in metres)

re-arranging to find
(5391 x 11.3)/200 = V^2
V = 17.5 m/s^-1

hope this helps and is right, ive kinda rushed a bit though so check it
:smile:
cobhc

btw, what school year was this set for?
 

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