Which Cable Has Greater Tension in a Spinning Chair Problem?

  • Thread starter Thread starter harmeet_angel
  • Start date Start date
  • Tags Tags
    Forces
AI Thread Summary
In the spinning chair problem, the upper cable experiences a tension of 3500N while the chair's mass is 134kg. The discussion concludes that the tension in the upper cable (T1) is greater than that in the lower cable (T2). The tension in the lower cable is calculated by considering the vertical component of the upper cable's tension and the weight of the chair. By analyzing the forces acting on the chair, it is determined that the total tension in the upper cable includes both vertical and horizontal components, while the lower cable experiences a reduced tension due to the chair's weight. The approach taken in the calculations is confirmed to be correct.
harmeet_angel
Messages
26
Reaction score
0

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?
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top