(Easy) Gravitational Force at an angle?

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Discussion Overview

The discussion revolves around the calculation of gravitational force required to balance a cable system used in gym equipment. Participants explore the effects of angles on weight distribution and the implications of cable sagging under load. The scope includes theoretical considerations and mathematical reasoning related to physics principles.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that to keep the cable from moving, a weight of 5 kg is needed based on a 30-degree angle calculation.
  • Another participant raises concerns that the cable will sag under load, affecting the angle and making it difficult to determine a single weight value that works.
  • There is a query about whether a 15-degree angle would require only 2.5 kg, with a calculation provided (10 sin 15° = 2.59 kg).
  • Participants discuss the need to clarify which angle should be used for calculations, with references to both 30 degrees and 60 degrees being mentioned.
  • One participant notes that using the sine of the side angle or the cosine of the top angle will yield the same result, indicating a potential method for resolving the angle issue.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate angle to use for calculations and the impact of cable sagging on weight requirements. There is no consensus on a definitive weight value that would work under all conditions.

Contextual Notes

Participants acknowledge the limitations of their calculations, particularly regarding the assumptions about cable strength and tension, as well as the effects of sagging on angle measurements.

JinglePeter
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Hi Guys,

I'm working on the theory for a piece of gym equipment, but I am afraid my physics-capabilities are somewhat limited.

In order to illustrate my question, I have made a drawing, please look at the attachment.

Excuse my english as well, it is not my first language.

So on the image you see a long cable pulled around 3 wheels. For the sake of theory let's assume there is no friction caused by the wheels.

How much weight do you need to put on the cable (Side C) in order to even out the weight distribution?

At first I thought 10 KGs, but I guess the angles make a lower requirement on side C than side A?

Is there someone who can help me confirm this? Preferrably with some kind of math behind it.

Thanks in advance!

- Peter
 

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Hi jinglepeter. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

To keep that cable from moving, you must oppose the weight acting in line with the cable, viz., 10. sin 30°, and this equals 5kg

So you could hang a 5kg weight from the vertical part.

Goog luck with your creation!
 
Last edited by a moderator:
If you actually build this device, you will find that the cable sags under the load so that the angle the cable makes with the top pulley is not the same as the angle between the pulleys, and changes with the position of the 10kg weight. Unless the cable is unrealistically strong and maintained under an unrealistic tension, I don't think you'll be able to find any single weight value that works.
 
Hi,

Thank you so much for your answer - Would that also mean that if the angle was 15%, then I would only need 2,5 kg?
 
Nugatory said:
If you actually build this device, you will find that the cable sags under the load so that the angle the cable makes with the top pulley is not the same as the angle between the pulleys, and changes with the position of the 10kg weight. Unless the cable is unrealistically strong and maintained under an unrealistic tension, I don't think you'll be able to find any single weight value that works.

The cable would sag under a bit yes. There should only be strong tension on line A and C.

While before it was calculated with 30 degrees - arent we supposed to calculate with 60 degrees, as that's the actual angle from the top piece?
 
JinglePeter said:
Hi,

Thank you so much for your answer - Would that also mean that if the angle was 15%, then I would only need 2,5 kg?

10 sin 15° = 2.59
 
NascentOxygen said:
10 sin 15° = 2.59

Thanks again,

Can you look over the calculation for the first one again? Aren't we supposed to use 60 degrees - as that's the angle for the top pulley?
 
JinglePeter said:
Thanks again,

Can you look over the calculation for the first one again? Aren't we supposed to use 60 degrees - as that's the angle for the top pulley?

You can use the sine of the side angle or the cosine of the top angle - you'll end up with the same value either way.
 

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