Problem 3.9 from Kleppner and Kolenkow's text An Introduction to Mechanics involves a uniform rope of length $L$ and mass $m$ that is connected at one end (its "bottom" end) to a block of mass $M$ and is being pulled with a force $F$ at its other end (its "top" end). The problem asks to find the tension $T(x)$ in the rope a distance $x$ away from the end being pulled, while neglecting gravity. I am very unsure of the answer I came up with.

First, I assumed that the acceleration of the block and each segment of rope would be equal and that the force acting the block would be upwards with magnitude $T(L) = F$. Each small segment of rope (length $\Delta x$, mass $\Delta m = \frac{m}{L} \Delta x$), undergoes an upward force $T(x)$ and a downward force $T(x+\Delta x)$ such that

$$\frac{-F}{M} = \frac{T(x+\Delta x) - T(x)}{\Delta m}$$

Isolating the variables related to the variable tension and taking the limit $\Delta x \rightarrow 0$ yields

$$\frac{dT}{dx} = \frac{m}{ML}F$$

After integrating using the assumed boundary condition $T(L) = F$, I arrived at the answer

$$T(x) = (1 - \frac{m}{M})F + \frac{mF}{ML}x$$

Is there anything wrong with my thought process? I thought that the net force on any segment of rope would always be zero because of Newton's Third Law, yet my use of $T(x + \Delta x) - T(x)$ (which should be zero) provided an answer that seems reasonable. Does my answer contradict Newton's law, or is my understanding of it flawed?


Your thought process looks good to me. The net force on any part of the rope with nonzero mass can't be zero, if the mass and rope system is accelerating (which it sounds like it is). You can picture the rope as being made of a bunch of discrete masses--a beaded string. The beads will have a net force on them if the bead + mass system is accelerating.

To see that the third law is not violated more explicitly, imagine three adjacent beads, call them A, B, and C. Bead A exerts a force fa onto B, and B exerts -fa onto B, so the third law is satisfied between A and B. Bead C exerts force fc into B, which exerts force -fc onto C, so the third law is satisfied there too. But there is no requirement that fa+fc (the net force on B) should vanish. The third law says nothing about this combination.

  • $\begingroup$ Oh, I might have been mistaken about the vanishing force then (your first point revealed that). Thanks for clarifying! $\endgroup$ – trifork Jul 9 '15 at 0:53

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