In one of the versions of the \"Giant Swing\", the seat is connected to two cabl

Question

In one of the versions of the "Giant Swing", the seat is connected to two cables, one of which is horizontal (Figure 1) .

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The seat swings in a horizontal circle at a rate of 31.0 rev/min . Part A If the seat weighs 255 N and a 839-N person is sitting in it, find the tension in the horizontal cable. Express your answer with the appropriate units. Part B Find the tension in the inclined cable. Express your answer with the appropriate units.

Explanation / Answer

the tension in the inclined cable is,

T2 = mg/cos@ = [255+839] / cos40 = 1428.1 N

the time t is,

t = 2*pi/w = 2*pi /31 rpm = 2*pi / 31(2*pi/60) = 1.935 s

thus, the tension in the horizontal cable:

T = m[4*pi^2*r / t^2] = ([255+839]/9.8)[4*pi^2*7.5 / 1.935* 1.935] = 8814.4 N

Let T1 be a tenson in horizontal one and T2 be the tension in the inclined cable.

T2cos(40) = (278+869)

T2 = 1497.3 N

v= (2*3.14*36.1*r)/60 =3.77r = 3.77*7.5 =28.33

1497.3*sin(40)+T1 = (278+869)*28.33^2/7.5*9.8

T1=11596.8

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