In another version of the "Giant Swing", the seat is connected to two cables as shown in the figure , one of which is horizontal. The seat swings in a horizontal circle at a rate of 33.3 .

(A) If the seat weighs 281 and a 855- person is sitting in it, find the tension in the horizontal cable.

(B)If the seat weighs 281 and a 855- person is sitting in it, find the tension in the inclined cable.

I will assume that 33.3 is rpm (revolutions per minute) and that the units of the weights are N.

1) Circular motion equations

angular velocity: 33.3 rev/min * 1min /60s * 2pi rad/rev = 3.4872 rad / s

centripetal acceleration = (angular velocity)^2 * r = (3.4872rad/s)^2 * 7.50 m =

centripetal acceleration = 91.2 m/s^2

2) Part A

Horizontal forces

Second Law of Newton: F = m*a

=> Centripetal force = m * centripetal acceleration =

= [(281N + 855N) / 9.81m/s^2] * 91.2 m/s^2 = 10,561 N

This is the horizontal component of the tension of the inclined cable + the horizontal force of the horizontal cable.

To find the force of the horizontal cable yet we need to find the tension of the inclined cable

3) Part B

free body diagram

Vertical forces: Tension up - weight = 0

Tension up = weight = 281N + 855N =1,136 N

cos (40°) = Tension up / Tension => Tension = Tension up / cos(40)

Tension = 1136 / cos(40) = 1482.9 N (this is the answer to part (B)).

Now we can find the horizontal component of tension of the inclined cable as

sin (40) = horizontal component / Tension =>

horizontal component = Tension*sin(40) = 1482.9 * sin(40) = 953.2 N

Then the tension of the horizontal cable is the centripetal forc minus the horizontal component of the tension of the inclined cable

Tension of the horizontal cable = 10561N - 953.2 N = 9607,8 N (this is the answer to part A)

batolisis batolisis · Answer 2 · 2025-03-31T12:08:47-04:00

A) The tension in the horizontal cable is : 9607.8 N

B) The tension in the inclined cable is : 1482.9 N

Given that :

Angular velocity = 3.4872 rad/s

Centripetal acceleration = 91.2 m/s²

Analysis of the solutions

A) The tension in the horizontal cable can be calculated by considering the horizontal forces and newton's second law

F = ma

Centripetal force = Mass * centripetal acceleration --- ( 1 )

= (( 281 + 855 ) / 9.81 m/s² ) * 91.2 m/s²

= 10561 N

Horizontal component = 1482.9 * sin 40⁰ = 953.2 N

Therefore the tension in the horizontal cable

= 10561 - horizontal component

= 10561 - 953.2 N = 9607.8 N

B ) Determine the tension in the inclined cabe

considering the vertical forces

vertical force = Tension - weight = 0

therefore the tension upwards = weight

= 281 + 855 = 1136 N

To determine the tension in the inclined cable we will apply the formula below

Cos 40° = Tension up / Tension in the inclined cable

Tension in the inclined cable = 1136 / cos 40⁰

= 1482.9 N

Hence we can conclude that A) The tension in the horizontal cable is : 9607.8 N and B) The tension in the inclined cable is : 1482.9 N

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