Confused on how tensions can differ with angle 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?
 A: In the simple problem below:

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.
A: 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.
A: 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.
A: Imagine 2 ten foot ropes attached to a 20 foot high ceiling. The attachments on the ceiling are 1 foot apart. You are hanging in midair with one arm around each rope. Now imagine the attachments move 5 feet apart, beginning to create an angle between your rope/arms and the vertical. Next the attachments move 20 feet apart. Can you sense that it would now be much harder for you to hold yourself off the ground than it was when the ropes hung straight down.
