Tension And Definition of elasticity

Oct 2019
3
0


The formula of the tension on 2 ropes T giving θ is T=Fg/2*Sin(θ) which means T get closer to infinite when θ is small. But since I can hang on my horizontal bar at home without a problem probably there is no infinite tension on it. I'm guessing there is a fundamental difference between a steel bar and elastic rope also i'm not sure why it so according to newton law?
Appreciate your suggestions.
 

topsquark

Forum Staff
Apr 2008
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On the dance floor, baby!


The formula of the tension on 2 ropes T giving θ is T=Fg/2*Sin(θ) which means T get closer to infinite when θ is small. But since I can hang on my horizontal bar at home without a problem probably there is no infinite tension on it. I'm guessing there is a fundamental difference between a steel bar and elastic rope also i'm not sure why it so according to newton law?
Appreciate your suggestions.
Strings (wire, lace, etc.) used for introductory level students is "ideal": They have no mass and can bear, as you suggested, an infinite load. You don't run into more realistic strings until you have covered a bit of material science where expansion coefficients and the like are under your belt. Ideal strings are good because they teach the basic pattern for writing out Newton's 2nd without any extra complications. Ideal springs fall under the same category.

-Dan
 
Jun 2016
1,142
514
England
Are you sure the equation is correct?

should it not be:
T={Fg*sin(θ)}/2
 
Oct 2017
530
250
Glasgow
The formula of the tension on 2 ropes T giving θ is T=Fg/2*Sin(θ) which means T get closer to infinite when θ is small. But since I can hang on my horizontal bar at home without a problem probably there is no infinite tension on it. I'm guessing there is a fundamental difference between a steel bar and elastic rope also i'm not sure why it so according to newton law?
Appreciate your suggestions.
This is a case where the model you have in front of you is just not very representative of real cases.

Typically a horizontal "chinning" bar is mounted on brackets that connect to two vertical walls, so the weight of someone hanging from the bar is met by vertical forces at the brackets.

If a curved bar is instead hung from the ceiling, but with a very small curvature (and hence a very small angle with the horizontal ceiling), indeed the tension force is much larger and the bracket must be much stronger. You can consider this a poor design choice. In these cases, a vertical bracket (e.g. attached to a ceiling) that has a horizontal bolt that passes through the bar is a better design because the bolt will take the vertical load.
 
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Oct 2019
3
0


The curve on this bar is very close to 0, it really tolerates so much tension?
 
Oct 2017
530
250
Glasgow


The curve on this bar is very close to 0, it really tolerates so much tension?
Short answer: Yes, but you need to realise that in reality materials are never 100% rigid and have a finite structural integrity, so infinite tensions just don't exist. The mathematics you're investigating are only an approximation of reality.

Longer answer:

If you draw a free-body diagram, there are two forces vertically downwards due to the pull of the man on the bar (one at each hand) and two forces vertically upwards at each bracket (at each end of the bar) pushing up on the bar. Assuming that the bar is stationary, the sum of the two bracket forces will equal the sum of the two forces caused by the man pulling on the bar.

In terms of forces, that's enough for most people and you can solve most problems involving a person doing a chin-up just with that description. However, if you consider only the region near to the man's hands as he grips the bar, the vertical pulling force needs to be balanced by tension in the bar. This tension is at right-angles to the man's pulling force, so it would have to be infinite. I assume it's this infinite tension that you're curious about.

In reality, however, the bar will not be 100% rigid. It should bend very, very slightly as the man exerts the pulling force. As a consequence, there will be a lot of tension in the bar, but it won't affect the bar in an adverse way unless the force is very large. The bar will also have a finite size, so the line of transmittance is not completely horizontal. Consider, for example, a massive pulling force, much, much bigger than the weight of a person (for example, exerted by some sort of machine, digger or crusher). The force would become so high that the tension cannot transmit and the bar will bend and become distorted.

If you actually want to compute tension, you need to improve your model by including a finite size for the bar (e.g. by assuming it to be a cylinder with a radius and length) and a substance (like steel or carbon fiber). The equations for the stress and strain can probably help here.
 
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