Which Cable Has Greater Tension in a Spinning Chair Problem?

[harmeet_angel]

1.A chair attached to a vertical rotating pole by two cables, is spun in a horizontal circle at a constant speed. The speed is sufficient to create tension in both the upper and lower cables. The tension in the upper cable is 3500N, L=10 m, and the mass of the chair is 134kg.


c. Predict which tension force is greater, T1(tension in the upper cable) or T2 (tension in the lower cable).

d. What is the tension in the lower cable?
http://www.physics247.com/physics-homework-help/rotatingpole.jpg




2. Homework Equations
Mathematical equations for equilateral triangle
like height =(((3)^0.5)/2)x side
and F=ma equation




3. The Attempt at a Solution
for c, I think upper cable has more tension
and for d, I just found the vertical component of the upper cable, and also the weight of the chair, and assumed that the vertical component of the lower cable is vertical comp of the upper cable minus the Weight

wondering if i am right in my assumption?

 

[AlephZero]

That looks right.

For part (c), the forces on the chair are its weight (downwards) and a horizontal force because it is rotating.

You can consider the force in the cables caused by each of those separately and add up the result.

By symmetry the horizontal force will produce an equal tension F1 in both cables.

The vertical force will produce a tension F2 in the top cable and a compression -F2 in the bottom cable.

So the total force in the top cable is F1+F2 and in the bottom cable is F1-F2.

(d) That's a good way to do it.

 

[m2003]

I would first clarify a few things about the problem. Is the chair attached to the pole at the center or at one end? Is the rotation of the pole and chair in a horizontal plane or is there an incline? These details can affect the solution.

Assuming that the chair is attached to the pole at the center and the rotation is in a horizontal plane, we can use the equation F=ma to determine the tension in the lower cable. Since the chair is moving in a circular motion, there must be a centripetal force acting towards the center of the circle. This force is provided by the tension in the cables.

For part c, the tension in the upper cable will be greater because it not only has to provide the centripetal force, but also counteract the weight of the chair.

For part d, we can use the equation F=ma to solve for the tension in the lower cable. The mass of the chair and the acceleration due to gravity will provide the force of weight acting downward. The vertical component of the upper cable's tension will provide the upward force, and the remaining force will be provided by the tension in the lower cable. The equation would look like this:

Tlower - (mg + Tupper*sinθ) = ma

Where Tlower is the tension in the lower cable, mg is the weight of the chair, Tupper is the tension in the upper cable, and θ is the angle between the upper cable and the vertical. We can rearrange this equation to solve for Tlower:

Tlower = ma + mg + Tupper*sinθ

Substituting in the given values, we get:

Tlower = (134kg)(10m/s^2) + (134kg)(9.8m/s^2) + (3500N)*sin(90°)

Tlower = 1340N + 1313N + 3500N = 6153N

Therefore, the tension in the lower cable is 6153N. It is important to note that this solution is based on the assumptions mentioned earlier. If those assumptions are different, the solution may also be different.