# Tension of a massless string

Suppose two ends of a massless string are attached at two different points along the same horizontal line such that the distance between the two points is less than the length of the string and the string is passed through a ring in such a way that the ring can move freely (downwards) along the string. Will the ring finally end up being suspended at the middle of the string (because of contant tension)? If it remains suspended at the middle why is it that the tension of the string at the two parts of the rope (on the two sides of the ring) are the same?

Again if we thing if another condition where we applied a horizontal force on the ring before it reaches to the midpoint and end up being static at any other point other than the mid point, will the tension remain still the same along both the parts? From the image, will the tension remain the same in the AB and AC part? If yes,then why? Edit: Added this image afterwards

• I don't understand. If the ends of the string are attached to two different points along the same horizontal line then it means that the string is a horizontal string. Then how would a ring be sliding along the string as well as downwards?
– ACat
Mar 20, 2021 at 18:46
• The distance between the two points where the ends are attached is less than the length of the string
– MSKB
Mar 20, 2021 at 18:50

this is Dr. Kaz. The tension in the string will be the same all over. It is understood that the string is longer than the distance between the points it is attached to. With the ring on the string, we will have a triangle whose base is the horizontal line between the attachment points. It is assumed that the ring has a weight. Without external force, the ring will stabilize in the middle of the string and forming a triangle with equal sides. The string is not stretchable. It’s length will determine the two angle on each side of the triangle. In this position, the horizontal components of the tension in each section will cancel each other and the weight of the ring will be twice the vertical component of the tension in one section. If an external horizontal force is applied to the ring, then the ring has to move to different point so that the sum of the horizontal components is zero including the external force. The sum of the vertical components also have to be zero including the weight of the ring. The problem will have to resolved as trigonometry problem in terms of the wight or the ring and the length of the string.

• Why is the tension same in both the parts of the string as stated in the last part of your answer?
– MSKB
Mar 20, 2021 at 20:56

If we suppose a real situation, with some air drag and/or friction between the disk and the string, it will rest in the middle, with or without a initial force. That is the point of smaller gravitational potential.

But while the disk is moving to that final point, the angles of each side of the string with the horizontal are not equal. As the tension is the same(*), the horizontal components are different. So there is a net horizontal force directed to the middle.

(*) It could be argued that the string is not at rest in this case, each point is accelerating as the disk moves. But the points of the string doesn't move in the direction of its length, because it is supposed not to stretch. So, the difference of tension at left and right of any small region is zero.

• Why are the tensions of both the part of the string same? Shouldn't those be different since the ring is bent at the point where the disk/ring is hanging/resting?
– MSKB
Mar 21, 2021 at 5:31
• @MohammadSakibShahriar Think of a small portion of the string. The force from the left must be equal to the force to the right. Otherwise the string would be accelerating and it is at rest. As it happens to all portions, the tension must be the same. Mar 21, 2021 at 12:48
• I agree this for the ring being suspended exactly at the midpoint but if it isn't at the middle why would the tensions be same at the both parts?
– MSKB
Mar 21, 2021 at 14:29