What is the ratio between tensions in different positions for a swinging ball?

AI Thread Summary
The discussion revolves around calculating the tension ratios in a swinging ball held by two cords before and after one cord is cut. At position A, the ball is at rest, and the tension can be expressed as T1cosx = mg, while at position B, the ball accelerates, leading to T2 = mgcosx. The key point is that the weight of the ball causes it to accelerate in the direction of the arc at position B, affecting the tension calculations. Participants clarify that different approaches to resolving forces can yield the same results, emphasizing the importance of understanding the forces acting on the ball in both positions. The conversation highlights the differences in force resolution methods and their implications for tension ratios.
konichiwa2x
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Hi, Can someone please explain how to do this problem? I really don't understand.

A ball is held at rest in position A as shown in the figure by 2 light cords. The horizontal cord is cut and the ball swings as a pendulum. What is the ratio of the tensions in the supporting cord, in position A, to that in position B?

http://img445.imageshack.us/img445/3927/lomprobtensionvx9.png

thanks for your time.
 
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konichiwa2x said:
Hi, Can someone please explain how to do this problem? I really don't understand.

A ball is held at rest in position A as shown in the figure by 2 light cords. The horizontal cord is cut and the ball swings as a pendulum. What is the ratio of the tensions in the supporting cord, in position A, to that in position B?

thanks for your time.
I assume they want you to compare the tension at A before the horizontal cord is cut to the tension at B. Otherwise, it is not much of a problem. At A, the net force is zero. At B, the ball is accelerating in the direction of the arc. Compare the forces in the two cases.
 
In case-1, T1cosx=mg, while in case-2, T2=mgcosx because in case-2, T cannot be resolved while mg can be resolved as there is resultant acceleration. So T1/T2=secx*secx. Its a situation of temporary rest.
 
thanks

I had earlier done this for case 2: Tcosx=mg. why is that wrong? can't tension be resolved along the vertical?
T cannot be resolved while mg can be resolved as there is resultant acceleration.
can you please explain that?


I assume they want you to compare the tension at A before the horizontal cord is cut to the tension at B. Otherwise, it is not much of a problem
yup, tension before the cord is cut. Otherwise, the tensions are same right! so T1/T2 = 1..

At B, the ball is accelerating in the direction of the arc. Compare the forces in the two cases.
but what is causing it to accelerate in the direction of the arc? its own weight?
 
konichiwa2x said:
thanks

but what is causing it to accelerate in the direction of the arc? its own weight?
Yes. At B, the weight has a component in the direction of the arc, and a component perpendicular to the arc (in the direction of the string). The ball is not accelerating perpendicular to the arc, so what does that tell you about the sum of the forces in that direction?

I'm not sure exactly how the interpret shramana's words, but the idea expressed by the equations is the same. I think he is just saying that in one case you are going to resolve the vectors in one set of directions, and in the other case you are going to resolve them in a different set. In fact you could resolve them in the same set of directions in both cases and get the same result. It is just easier to do the two cases with different directions, keeping one force "unresolved" in both cases.
 
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