The tension in an ideal massless rope can be uniform, but a rope with nonzero mass generally has different tension at different positions because each section must accelerate or support different amounts of rope.

A useful way to analyze a massive rope is to apply Newton's second law to a short differential element.

Vertical rope at rest

Consider a uniform rope of length $L$ and linear mass density $\lambda$, hanging vertically at rest from its top. Let $y$ measure distance upward from the free lower end.

The portion of rope below position $y$ has length $y$ and mass

$$m(y)=\lambda y.$$

The tension at that position must support the weight of all rope below it:

$$T(y)=m(y)g.$$

Therefore

$$\boxed{T(y)=\lambda gy}.$$

The tension is zero at the free lower end and increases linearly toward the support.

At the top,

$$T(L)=\lambda gL=Mg,$$

where $M=\lambda L$ is the total rope mass.

Differential force balance

The same result can be obtained locally. Consider a small rope element of length $dy$ and mass

$$dm=\lambda,dy.$$

Let the tension at its lower end be $T(y)$ and at its upper end be $T(y+dy)$. In static equilibrium,

$$T(y+dy)-T(y)-\lambda g,dy=0.$$

Dividing by $dy$ and taking the limit gives

$$\boxed{\frac{dT}{dy}=\lambda g}.$$

Integrating with the free-end boundary condition $T(0)=0$ gives

$$T(y)=\lambda gy.$$

This differential-element method generalizes to continuous ropes, cables, fluids, beams, and other systems whose forces vary through space.

Rope accelerating vertically

Suppose the same vertical rope accelerates upward with acceleration $a$. For the part of the rope below position $y$, Newton's second law gives

$$T(y)-\lambda yg=\lambda y a.$$

Thus

$$\boxed{T(y)=\lambda y(g+a)}.$$

If the rope accelerates upward, every section must both support the material below and provide its upward acceleration, so the tension is larger than in the static case.

For downward acceleration of magnitude smaller than $g$, replace $a$ by a negative value. In ideal free fall with $a=-g$,

$$T(y)=0$$

throughout the rope: no section needs to support another because all parts fall together with the same gravitational acceleration.

Worked example

A uniform rope has

$$L=8.0,\mathrm m,\qquad \lambda=0.50,\mathrm{kg/m}.$$

When it hangs at rest, the tension $3.0,\mathrm m$ above the free end is

$$T=(0.50)(9.81)(3.0)\approx14.7,\mathrm N.$$

At the top,

$$T=(0.50)(9.81)(8.0)\approx39.2,\mathrm N.$$

The top must support the entire rope, while the lower point supports only the mass beneath it.

The key distinction is therefore structural: uniform tension is an idealization of a massless rope. Once the rope's own mass matters, tension becomes a spatially varying field determined by local force balance and boundary conditions.