In another version of the 'Giant Swing' (see Exercise $5.50$ ), the seat is connected to two cables, one of which is horizontal (Fig. E $5.51$ ). The seat swings in a horizontal circle at a rate of $28.0$ rpm (rev/min). If the seat weighs $255$ N and an $825$ -N person is sitting in it, find the tension in each cable.

Question

In another version of the 'Giant Swing' (see Exercise $5.50$ ), the seat is connected to two cables, one of which is horizontal (Fig. E $5.51$ ). The seat swings in a horizontal circle at a rate of $28.0$ rpm (rev/min). If the seat weighs $255$ N and an $825$ -N person is sitting in it, find the tension in each cable.

Illustration of a seat on a giant swing, showing angles and distances for calculating centripetal forces.

Accepted Answer

Hey everyone today, we're dealing with the problem about centripetal acceleration and tension. So being told that in an amusement park a design of the carousel has two wires, one of which is horizontal and a seat is attached to them, Horizontal circle of seven or 7.5 m in radius is formed by the seat as it swings at a speed of 20 rpm. With this for being asked to determine the tension in each cable. If the seat weighs 200 newtons and a person weighing 850 newtons is seated in it. So the object is moving in a horizontal circuit, right? It's moving in a horizontal circle, let's say the positive X. Direction, which means that it has an acceleration. A centripetal acceleration directed towards the center of the circle. Now let positive X. P. To the right in the direction of the acceleration and the positive Y. The upwards this is positive. Why? So we're told a few things. The radius here is 7.5 m Radius of the circular path. The total mass will be 200 newtons Plus 815 Newtons. However, mass needs to be in kilograms. So we need to convert uh huh Newton's two kg. That we can do this simply by dividing by the force of gravity or the force of acceleration due to gravity. Meters per second square. This is because the weight that would be experienced here Would be MG. The total mass times the acceleration due to gravity. So M will simply be the total weight divided by the force of gravity, which will give us an answer of 107 kg. Now, since the rotation rate is 20 rpm. So revolution rate, Which is the period is 20 rev per minute, Which is 0.30. revolutions per second. The period that's a really step from here will be equal to one over 0.33 revs per second, which is simply equal to three seconds. With this in mind. And we can recall that the tension that will be acting here we have two tensions tension can never push it can only be pulled away. Right? So let's just call this tension A and tension B. Now applying Newton's second law and projecting along the vertical axis. We can use our Socotra rules and we can find that the Tension horses are diagonally up can be given as Kassian which is simply MG. Excuse me? Which means solving for the tension in that direction. It'll give us 1371 Newtons because we'll have MG which is simply 1107 kg times 9.8 divided by cosine With this. And using Newton's second law protecting along the horizontal axis, gives us the tension and the horizontal axis. Plus ah tension cosign will give us the mass into the centripetal acceleration. But we know we can recall that centripetal acceleration is simply V squared over R. So we can also rewrite that as for pie squared R over t squared oops. So substituting that in, we get to solve for T V, which is simply kilograms into four pi squared into 7.5 Because that is the radius divided by nine -1371 Newtons into sign 40, Which gives us a final answer of 2, newtons. So the tension in rope, A or in one cable in the horizontal cable is 2639 newtons. And the tension in the diagonal cable is 1371 newtons or answer choice B. I hope this helps. And I look forward to seeing you all in the next one.

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Key Concepts

Centripetal Force

Tension in Cables

Equilibrium of Forces