In another version of the \"Giant Swing\", the seat is connected to two cables a

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 35.5 rev/min .

Part A

If the seat weighs 223 N and a 818-N person is sitting in it, find the tension in the horizontal cable.

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Part B

If the seat weighs 223 N and a 818-N person is sitting in it, find the tension in the inclined cable.

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Figure 1 of 1

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 35.5 rev/min .

Part A

If the seat weighs 223 N and a 818-N person is sitting in it, find the tension in the horizontal cable.

T1 =   N  

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Incorrect; Try Again; 3 attempts remaining

Part B

If the seat weighs 223 N and a 818-N person is sitting in it, find the tension in the inclined cable.

T2 =   N  

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Figure 1 of 1

Explanation / Answer

part A:

angular speed=35.5 rev/min

=35.5*2*pi rad/(60 seconds)

=3.7176 rad/s

let tension in slanted cable be T2 and tension in horizontal cable be T1.

balancing vertical forces:

T2*cos(40)=223+818

==>T2=1358.9 N

balancing forces in horizontal direction:

T2*sin(40)+T1=((223+818)/9.8)*3.7176^2*7.5

==>T1=((223+818)/9.8)*3.7176^2*7.5-1358.9*sin(40)=10137.12 N

part B:

tension in the inclined cable is T2=1358.9 N

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