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So I understand the mechanics of resolving forces via Trig, what confuses me is how the tension can change.

Imagine you have a 100N weight suspended from a rope tied to 2 points. It seems to me (but clearly I'm wrong), that all you have in the system in that 100N of downward force and yet as the angle decreases to 0 (perfectly straight rope) the tension goes to infinity. It seems bizarre to me. Help?

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    $\begingroup$ Could I suggest that you include the actual trig. formula that you have given to use, it might help you get an answer. Best of luck with it. $\endgroup$ – user81619 Sep 20 '15 at 2:31
  • $\begingroup$ Sure. Its just the normal vector resolution stuff. Given a vector (say 15N at 30 degrees): X component = 15Ncos(30) Y component = 15Nsin(30) $\endgroup$ – mikeboensel Sep 20 '15 at 2:38
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We consider the string to be unbreakable, thus any amount of force in it cannot break it. In order to support a block, we need that the upward force due to the combination of strings be equal to the downward force exerted by gravity on the block. As we increase the angle, more of the force exerted by the string is directed in the horizontal direction. Thus, there is less force exerted on the block in the vertical direction. Therefore, to compensate for that decrease, the string exerts a larger overall force on the block.

To understand how the force actually increases:

The string is made up of a certain material. The material follows the following relation:

$$Y = \frac{FL}{A\triangle L}$$

Not sure if you have learnt of this yet. $Y$ represents the Young's modulus, a constant for a given material. $F$ represents the force exerted on the string, $L$ represents the initial length of the string, $A$ the given cross-section, and $\triangle L$ the change in length. Rearranging: $$F = \frac{AY\triangle L}{L}$$ Thus: $F$ is inversely proportional to $L$. The vertical component of $L$ decreases with an increase in angle. (Sorry, but can't make diagram at the moment, hope you understand). Thus, the force increases, as everything else is constant.

Hope that answers your question.

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  • $\begingroup$ Ah, thanks. Got it. Don't know why but I didn't get that the horizontal component is increasing (but nets to zero), while the vertical component remains the same. $\endgroup$ – mikeboensel Sep 20 '15 at 20:30
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In the simple problem below:

Tension problem.

The tension $T$ in the rope, needed to balance the vertical forces, is given by:

$2T\sin\alpha=mg$.

So $T=\frac{mg}{2\sin\alpha}$.

As $\alpha$ becomes smaller $T$ rises and at $\alpha =0$, then $\sin\alpha=0$ and $T \to \infty$.

In reality this will never occur: even for small $mg$ a weight suspended from the middle of a (horizontal) rope, $\alpha$ will never be really $0$ because the rope is always a little extensible. But to approximate a horizontal rope, tension is higher than for a 'sagging' one, that's an unavoidable consequence of the trigonometry of the system.

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    $\begingroup$ Great graphic, yes. I see the math. I still don't understand though at a fundamental level how tension can change. The weight on the rope doesn't. That seems like the only input into the system. It seems strange that a rope could support a given weight in a given configuration, but not another. The weight's force doesn't change. Sorry if I'm dense. $\endgroup$ – mikeboensel Sep 20 '15 at 2:57
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    $\begingroup$ @mikeboensel Try to imagine yourself holding the ropes that are attached to the box. Can you imagine whether it will be easier to hold the box up when the ropes are straight up and down, versus if you try to pull the ropes into a horizontal position? If not, maybe you can rig up some sort of contraption yourself to test it. $\endgroup$ – wgrenard Sep 20 '15 at 7:50
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It may help if you think of yourself as the weight. You don't have to actually do the following, except in your imagination.

Hang from a chin-up bar with your hands shoulder-width apart. Each of your arms is supporting half your weight.

Now hang from a chin-up bar with your hands far apart and your arms in a wide V. Don't try to use any strength from your shoulders, just let the stretch of your arms hold your weight.

Your left arm is pulling you up and to the left; your right arm is pulling you up and to the right. The 'up' part of the pull on each arm is what keeps you from falling to the floor. The pull to your right and the pull to your left (which should be equal) counter-act each other, but don't help you stay up. Therefore they are additional forces beyond what is needed when your arms are vertical.

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  • $\begingroup$ Of all the answers, this one best helped me to visualize the forces involved. (By comparison, imagining myself and a helper on each end of the rope at different angles was not as clear to me). $\endgroup$ – mm2001 Oct 7 '15 at 3:36

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