The problem I was trying to solve was this one.

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Using the conservation of energy, I obtained that the velocity $v$ equals $$\sqrt{\frac{(2lx-x^2)g}{l-x}}$$

In the left segment, the momentum doesn't change, so the net force is zero. The reaction force on hinge A can be equated to the combined effect of the gravitational force and the thrust force(force due to changing mass) on the left segment of the chain . The thrust force is given by $$ \textbf{v}\frac{dm}{dt}$$

Here, the thrust force equals $k\frac{v^2}{2}$, where $k$ is the linear mass density. The $2$ in the denominator is because $$\frac{dm}{dt} = k\frac{d(x/2)}{dt}=k\frac{v}{2}$$

However, when we solve the problem this way, we arrive at a different answer than we would if approached using energy conservation. (Using energy conservation and $F=\frac{dp}{dt}$ for the system, we would get $R(x)=\frac{kg(2l^2+2lx-3x^2)}{4(l-x)}$)

In many such chain problems, using thrust force and energy conservation gives two different answers. Here's an example of another one.

A chain of length $L$ and mass per unit length $\rho$ is pulled on a horizontal surface. One end of the chain is lifted vertically with constant velocity $v$ by a force $P$. Find $P$ as a function of height $x$, and the work done by the force

My general doubt is: Why is there a discrepancy in answers? Which answer is more correct?

  • $\begingroup$ Experiments seem to indicate that at least some (but probably not all), of the energy of the links being captured by the stationary section of chain, is transferred to the falling section of chain. The problem is to explain the mechanism by which this occurs. $\endgroup$
    – R.W. Bird
    May 18, 2021 at 14:33

1 Answer 1


Momentum conservation states that: $$ \frac{dp}{dt} = F. $$

You have two forces here and two components of $\dot p$, since both $m$ and $v$ (of the moving part) changes: $$ m\frac{dv}{dt} + \frac{dm}{dt}v = Mg - R. $$

So when you write force balance in your head, you forget that the system has acceleration, so the forces (including thrust) are not balanced.

Finally, we get $$ R=klg-k\frac{l-x}2\frac{dv}{dt}-\frac{d\left(k\frac{l-x}2\right)}{dt}v. $$

To simplify it, we use $\frac{dx}{dt} = v$ and $\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=\frac12\frac{d(v^2)}{dx}$. I believe, you can do the algebra and find out that the answer is the same.

  • $\begingroup$ Thank you for your answer. This is the same method used to arrive at one of the answers, using Newton's 2nd Law for the whole system. My doubt revolves around a different method; equating the reaction force on the hinge to gravitational force and force of the lower links of the chain (thrust force) on the left segment. The left segment is stationary. $\endgroup$
    – Aspirant
    May 18, 2021 at 10:46
  • $\begingroup$ Also view the second question, for reference. Solving via energy conservation and thrust force yields different answers. $\endgroup$
    – Aspirant
    May 18, 2021 at 10:49
  • $\begingroup$ equating the reaction force on the hinge to gravitational force and force of the lower links of the chain (thrust force) on the left segment If you consider partial chain, you need to take chain tension in the equation. Chain tension includes thrust force but isn't equal to it. $\endgroup$ May 18, 2021 at 10:59
  • $\begingroup$ And for the second example I quoted? Isn't tension at the bottom due to other factors negligible there? $\endgroup$
    – Aspirant
    May 18, 2021 at 11:55
  • $\begingroup$ What part of the chain are you considering in the 2nd example? Consider a small chunk $dm$ of the chain just before the vertical part. At moment $t$ it had velocity $0$, at moment $t+dt$ it has velocity $v$. What could increase its velocity except for tension in the chain? In a real chain, there is a transitional segment of the chain where velocity drops from $v$ to $0$ and tension drops from $T$ to $0$, so the motionless part of the chain doesn't experience any forces. In the textbook chain this segment has zero size, so the plots $T(x)$ and $v(x)$ have a discontinuity. $\endgroup$ May 18, 2021 at 12:19

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