Question

In another version of the "Giant Swing," the seat is connected to two cables as shown in the figure (Figure 1), one of which is horizontal. The seat swings in a horizontal circle at a rate of 36.1 ${\rm rev/min}$ .

A) If the seat weighs 278 ${\rm N}$ and a 869- ${\rm N}$ person is sitting in it, find the tension $T_1$ in the horizontal cable.

B)If the seat weighs 278 ${\rm N}$ and a 869- ${\rm N}$ person is sitting in it, find the tension $T_2$ in the inclined cable.

Answer 1

Let T1 be a tenson in horizontal one anD T2 in other one

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

Answer 2

T1 = (278+869)cos 40 = 878.65 N

T2 = 278+869) sin 40 = 738.56 N

Answer 3

the angular velocity is w = 36.1 rev/min = 36.1 x (2pi/60) rad/s

the weight of seat is W = 278 N

the weight of person is W1 = 869 N

the total weight of seat and person is

F = W + W1

the tension T1 in the horizontal cable is

T1 = F x cosA - m x r x w^2

where A = 40.0o,m = (W +W1/g),g = 9.8 m/s^2 and r = 7.50 m

the tension T2 in the vertical cable is

T2 = F x sinA - m x r x w^2

Answer 4

A) W=36.1 rev/min = 36.1*2*3.14/60 = 3.778rad/s.........

for horizontal T1=r*W^2==========>T1= 107.07N..........

B)for inclined rope

T2 sin 50 = mg =========>T2= mg / sin 50 =(278+869)/sin50=1497.3 N..............

Answer 5

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