Could somebody explain to me intuitively how tension is not the same in a rope with mass? My physics teacher (when regarding a massless string) told me that the tension is always equal because if you pull at one side more then the other side has to pull just as much to keep it in equilibrium resulting in the tensions being equal. This fits really well intuitively with me, but if the string has a non-negligible mass how can I adapt this idea? Or replace it if necessary?

# [Physics] Non-constant tension in rope

forcesfree-body-diagrammassnewtonian-mechanicsstring

#### Related Solutions

Say you have a weight tied to each side a a rope which is strung over a pulley with friction. Here's a really easy way to see why the tensions on each side of the rope can't be equal.

Imagine a really stiff pulley - in other words, ${\bf F}_\text{friction}$ is high. If that's the case, it'll be possible to balance unequal loads on this pulley system - i.e. a heavy weight on the right side and a lighter weight on th left - without the system moving. If the weights don't move, then we can say that the forces acting on each weight add up to zero:

For the heavy weight, there's the weight downward, ${\bf w}_\text{heavy}$ and there's the tension of the right side of the rope upward, ${\bf T}_\text{right}$. The tension pulls up and the weight down, and the system doesn't move, so

$$ {\bf T}_\text{right} - {\bf w}_\text{heavy} = 0 $$

or

$$ {\bf T}_\text{right} = {\bf w}_\text{heavy} $$

Similarly for the left (light) side,

$$ {\bf T}_\text{left} - {\bf w}_\text{light} = 0 \quad \Rightarrow \quad{\bf T}_\text{left} = {\bf w}_\text{light} $$

As you can see, the tension on the right, ${\bf T}_\text{right}$ is equal in magnitude to the heavy weight, while the tension on the left, ${\bf T}_\text{left}$ is equal to that of the lighter weight. The friction is introducing an extra force which changes the tensions on each side.

As far as your question about rope stretching goes, if you anchor a rope on one side and pull, the rope will pull back, creating a tension. This is indeed because of stretching in the rope. This is not really what Newton's 3rd law is referring to. Newton's third law, in this case, tells us that the force that we feel from the rope, tension, is exactly the force the rope feels from us pulling. The two are equal and opposite. You can change the tension by changing the stiffness of the rope, but whatever the tension, Newton's 3rd law will still be true - the rope will feel us pulling it as much as we feel it pulling us.

For the second question- COnsider the string to be made up two parts separated by a vertical line passing through the lowest point.

Now,
**consider the point where the string meets the wall**.The string exerts a force on the wall(Normal force,tangential to the curve at that point) and in trun experiences a force in the opposite direction.

Now ** resolve** these normal force on the string into its two components.
The

**is balanced by the**

*horizontal component***which the string experiences on the lowest point due to the pull of the other segment of the string.**

*tension force*Also use the fact that the **vertical component balances the weight of the half-segment of the string.**

Solve for tension.

As for your first question, the tension at a pint 1m away from the end is the force that pulls on the remaining string(the mass of which you can calculate by **-** ***linear mass density times length**) to move it with the common acceleration, which would be given by external force force divided by total mass.Use this.

## Best Answer

Imagine a load hanging in a vertical rope:

Clearly, the top particle carries the most whereas the bottom particle carries the least.

Tension increases up through the rope since the particles gradually carry more total weight. Only in the special case of massless particles - a massless rope - is this not the case, since more particles don't add extra weight.