Which Cable Has Greater Tension in a Spinning Chair Problem?

  • Thread starter Thread starter harmeet_angel
  • Start date Start date
  • Tags Tags
    Forces
Click For Summary
SUMMARY

The discussion centers on the tension forces in a spinning chair problem, specifically comparing the tension in the upper cable (T1) and the lower cable (T2). The tension in the upper cable is established at 3500N, while the mass of the chair is 134kg. The analysis concludes that the upper cable experiences greater tension due to the combined effects of vertical and horizontal forces acting on the chair during rotation. The vertical component of the upper cable's tension is calculated to determine the tension in the lower cable.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of tension forces in static and dynamic systems
  • Familiarity with rotational motion concepts
  • Basic principles of force decomposition
NEXT STEPS
  • Study the principles of tension in rotating systems
  • Learn about force decomposition in physics problems
  • Explore the effects of mass and speed on tension in cables
  • Investigate the dynamics of circular motion and its applications
USEFUL FOR

Physics students, educators, and anyone interested in understanding the mechanics of tension in rotating systems.

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.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
2K
Replies
5
Views
2K
  • · Replies 26 ·
Replies
26
Views
4K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 31 ·
2
Replies
31
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
7K
Replies
3
Views
2K
Replies
4
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K