Let's say we have a meter stick with a single rope attached to it. One end of the rope is attached to one end of the meter stick, the other end of the rope is attached to the opposite end of the meter stick. By hanging the rope from a beam in the center of the rope, the meter stick (which is attached to the rope) balances perfectly flat in equilibrium as it hangs from the rope. Now, say the rope is not hung perfectly in its center, and as a result the meter stick accelerates from its perfectly horizontal position to an almost vertical position. (Does this make sense so far?) At this point where the meter stick is now at rest at an angle which is at a diagonal (the actual angle is irrelevant, we could say perhaps 5 degrees off the vertical, just to clarify to the reader), is the tension in the rope still constant among the two halves of the rope? How can it be, when it appears that so much more of the meter stick's weight is being held up by one end of the rope?
I'd like to put forth an answer which directly addresses the title of your post, but not the particular situation in which you put forth with the meter stick and rope.
Consider instead a massive rope hanging vertically from a ceiling.
Give the rope a total mass of, say, $M$. Then use Newton's second law on the lower half of the rope to find the tension at the midpoint. Compare this value to the tension at the top of the rope by using Newton's second law for the entire rope. This should let you answer your question.
I believe the situation you are describing is this:
The ruler is the big blue line; the rope is the green line, and the forces on the ruler due to the rope are shown as colored vectors.
Now there can be no net horizontal force on the ruler, so the two red vectors must be equal and opposite. The tension has to point along the rope, so the direction of the total force is given (black dashed vector). This means that the tension in the part of the rope that is more vertical must be greater.
If the ruler has uniform mass(mass acts in the centre), and the rope is light and inextensible, then yes, the tension is equal throughout.
Take moments about A, to find the tension of the rope at B, or take moments about B, to find the tension of the rope at A. Either way you should get the same result, since the ruler is in equilibrium, so no resultant torque about any point. The tension should be half mg.
The angle that the ruler is balanced at is irrelevant, even when the ruler is vertical. Although for this to happen the rope has to pass through the ruler, which it won't, so at near vertical, things change as the rope bends around the ruler's edges.
Let's suppose the tension is not constant through out . Let it be T1 at point A , and T2 at point B , so then the net force on element AB is F=(T1-T2) , this causes a net acceleration F/m ( let m be the mass of the element AB). But since AB is massless in a massless string, a=F/m becomes infinitely large. Which is neither observed , nor is it intuitive . So our assumption is wrong and T1 =T2 as a=0. Also if it is an accelerating system T1 approaches T2 ( practically same). I hope my answer quantitatively deals with query.
The rope has mass, so tension varies, as one bit has to carry the other bits. The space elevator is a good example. No tension at the surface of the earth, and LOTS 5000 km up as that rope has to carry the weight of all 5000 km below it.
protected by AccidentalFourierTransform Jul 27 '18 at 16:39
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