Classical-Mechanics · Unit 8 · Video 5 · Interactive Practice

Two Roads to a Rope's Tension: Force Balance and a Differential Equation

IKey Formulas

Formula	Meaning	Where it comes from
$T(0) = Mg$	Tension at the top	The top holds the whole rope's weight
$T(y) = Mg\left(1 - \dfrac{y}{L}\right)$	Tension at depth $y$	Force balance on the segment above the cut
$\dfrac{dT}{dy} = -\dfrac{M}{L}\,g$	Rate of change of tension	Newton's law on an infinitesimal element

Key Insight: Slice the rope, apply Newton's second law to the piece, then take the limit — the finite force balance and the differential equation give the same $T(y) = Mg\left(1 - \tfrac{y}{L}\right)$ .

IIThe Cut — Tension vs. Depth

The tension at depth $y$ holds up only the rope below it, so it shrinks toward the free end.

Cut depth y 2

💡 The tension also equals the weight of rope hanging below the cut: $\frac{M}{L}(L-y)\,g = Mg\left(1-\tfrac{y}{L}\right)$ .

IIIThe Infinitesimal Element

Zoom in on an infinitesimal slice: Newton's second law on it becomes a differential equation for $T(y)$ .

Step 1

Step 1 — Isolate the element

\Delta m = \frac{M}{L}\,\Delta y

Step 2 — Three forces (down positive)

T(y)\ \uparrow,\qquad T(y+\Delta y) = T(y) + \Delta T\ \downarrow,\qquad \Delta m\,g\ \downarrow

Step 3 — Newton's second law, $a = 0$

\Delta m\,g - T(y) + \big(T(y) + \Delta T\big) = 0

\Rightarrow\quad \Delta m\,g + \Delta T = 0

Step 4 — Divide by $\Delta y$

\Delta T = -\Delta m\,g = -\frac{M}{L}\,g\,\Delta y

\frac{\Delta T}{\Delta y} = -\frac{M}{L}\,g

Step 5 — Take the limit $\Delta y \to 0$

\frac{dT}{dy} = \lim_{\Delta y \to 0}\frac{\Delta T}{\Delta y} = -\frac{M}{L}\,g

IVPredict the Tension

By a given depth, how much of its tension has the rope already shed?

Depth y (m) 2.5 Your guess T (N) 20

VQuiz Questions

Problem 1 · Tension at the Top

Given: A uniform rope of mass $M$ and length $L$ hangs straight down from a ceiling; gravity has magnitude $g$ . Find the tension $T(0)$ where the rope meets the ceiling.

✅ Correct! Taking the whole rope as the system, the ceiling force balances gravity:

Mg - T(0) = 0

, so the top carries the entire weight

Mg

❌ Not quite. The top supports all of the rope below it, not half — the full weight is

Mg

❌ Not quite. With down positive, equilibrium of the whole rope gives

Mg - T(0) = 0

Show solution

Take the entire rope as the system. Two forces act: the ceiling pulls up with $T(0)$ , and gravity pulls down with $Mg$ . The rope hangs at rest, so $a = 0$ .

With down taken as positive, Newton's second law reads:

Mg - T(0) = 0 \quad\Longrightarrow\quad T(0) = Mg

The top of the rope carries the rope's entire weight.

Problem 2 · How the Tension Changes

Given: $T(y) = Mg\left(1 - \dfrac{y}{L}\right)$ . Find the rate of change $\dfrac{dT}{dy}$ . ⚠️ Watch the sign.

✅ Correct! The tension drops at the constant rate

-\tfrac{M}{L}g

— a straight line from

Mg

at the top to

0

at the bottom.

❌ Close, but check the sign. Tension decreases as

y

grows, so the slope must be negative.

❌ Not quite. Differentiate

Mg - \tfrac{Mg}{L}y

term by term: the constant drops out, leaving

-\tfrac{M}{L}g

Show solution

Write the tension as $T(y) = Mg - \dfrac{Mg}{L}\,y$ . The term $Mg$ is constant, and $\dfrac{d}{dy}\!\left(1 - \dfrac{y}{L}\right) = -\dfrac{1}{L}$ :

\frac{dT}{dy} = Mg\cdot\left(-\frac{1}{L}\right) = -\frac{M}{L}\,g

The rate is constant — it does not depend on $y$ — so the tension falls off steadily all the way down.

Problem 3 · Plug in the Numbers

Given: A rope with $M = 6\ \text{kg}$ , $L = 3\ \text{m}$ , and $g = 10\ \text{m/s}^2$ . Find the tension at depth $y = 1\ \text{m}$ .

✅ Correct!

T = Mg\left(1 - \tfrac{y}{L}\right) = 60\left(1 - \tfrac{1}{3}\right) = 40\ \text{N}

❌ Not quite. That is

Mg\cdot\tfrac{y}{L}

— but the formula uses the factor

\left(1 - \tfrac{y}{L}\right)

, not

\tfrac{y}{L}

❌ Not quite. First find

Mg = 60\ \text{N}

, then multiply by

\left(1 - \tfrac{1}{3}\right) = \tfrac{2}{3}

Show solution

The weight of the whole rope is:

Mg = 6 \times 10 = 60\ \text{N}

Substitute into $T(y) = Mg\left(1 - \dfrac{y}{L}\right)$ with $y = 1\ \text{m}$ and $L = 3\ \text{m}$ :

T(1) = 60\left(1 - \frac{1}{3}\right) = 60 \times \frac{2}{3} = 40\ \text{N}

Check the ends: $T(0) = 60\ \text{N}$ (full weight) and $T(3) = 0$ (free end), as expected.

Problem 4 · Where Is the Tension Halved?

Given: $T(y) = Mg\left(1 - \dfrac{y}{L}\right)$ . Find the depth $y$ at which the tension is exactly half its maximum value $Mg$ .

✅ Correct! Because the tension falls linearly, half its value sits exactly at the midpoint

y = \tfrac{L}{2}

❌ Not quite. Set

Mg\left(1 - \tfrac{y}{L}\right) = \tfrac{1}{2}Mg

and solve for

y

Show solution

Set the tension equal to half of $Mg$ and cancel the common $Mg$ :

Mg\left(1 - \frac{y}{L}\right) = \frac{1}{2}Mg \quad\Longrightarrow\quad 1 - \frac{y}{L} = \frac{1}{2}

Solve for $y$ :

\frac{y}{L} = \frac{1}{2} \quad\Longrightarrow\quad y = \frac{L}{2}

Because $T$ is a straight line from $Mg$ to $0$ , the halfway tension lands at the halfway depth.

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