Cable Angle
January 13th, 2011 | 
Author:
How would you find the tension of two cables connected at an angle?
Here's the image: http://session.masteringphysics.com/problemAsset/1038593/5/YF-05-58.jpg
The seat weighs 233 N and the person weights 820 N
The seat swings in a horizontal circle at a rate of 33.3 rev/min.
What is the tension in each cable?
Wow, this is a monster ride--the numbers indicate that the poor girl is accelerating at over 9 g's !!
pateoh made a couple of errors in his analysis:
1) it is not true that T2 = mrω^2. He forgot to add the sideways component contributed by T1
2) It is not true that mrω^2 = (820/9.81)(7.5)(33.3/60)^2. He forgot to include the weight of the seat; and he expresses ω in rev/sec rather than radians/sec.
Let T1 be the tension in the diagonal cable, and T2 the tension in the horizontal cable.
Vertical forces on "object" (girl+seat):
-------------------------------
Weight (1053N) [down]
T1(cos40) [up]
(no vertical contribution from T2)
Since girl is not accelerating vertically, the vertical forces cancel out. So this implies:
upward force = downward force
T1(cos40) = 1053N
T1 = 1053N/cos40
Horizontal forces on object:
--------------------------------------
T1(sin40)
T2
object is accelerating sideways (toward post) at this rate:
a = ω²r
= (33.3rev/min)²(7.5m)
= (33.3(2π rad)/(60s))²(7.5m)
= (66.6π rad/(60s))²(7.5m)
By F_net = ma, we have:
T1(sin40) + T2 = m[(66.6π rad/(60s))²(7.5meters)]
= (1053N/g)(66.6π rad/(60s))²(7.5meters)
We previously showed that T1 = 1053N/cos40, so this means T1(sin40) = 1053N(tan40). So, substituting:
1053N(tan40) + T2 = (1053N/g)(66.6π rad/(60s))²(7.5meters)
or:
T2 = [(66.6π rad/(60s))²(7.5meters)/g – tan40] • 1053N
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