Question

A chain of length L, and mass M is suspended vertically by one end with the bottom end just above a table. The chain is released and falls, and the links do not rebound off the table, but they spread out so that the top link falls very nearly the full distance L. Just before the instant when the entire chain has fallen onto the table, how much force does the table exert on the chain? Assume that the chain links have negligible interaction with each other as the chain drops, and make the approximation that there is very large number of links. Hint: Consider the instantaneous rate of change momentum of the chain as the last link hits the table.

Short answer

The force exerted by the table on the chain is 3 mg

Step-by-step solution

Identification of the given data

The chain length is L

The mass is M

Concept of the force exerted by the chain

The total force exerted by the chain is calculatedby adding the gravitationalforce and impulse force acting on the chain.

Determination of the force exerted by the chain

The force exerted by the chain,

$F=F_1=F_2.........\left(1\right)$

Where,

$$\begin{gathered} F_1=\frac{dp}{dt} \\ =\frac{d\left(mv\right)}{dt} \\ =m\frac{dv}{dt}+v\frac{dm}{dt}.....\left(2\right) \end{gathered}$$

The first term is 0 because only the rate of change of the amount of chain hitting the table per unit is considered. So a differential mass element of the chain will be,

$dm=\frac{M}{L}dx.....\left(3\right)$

Substitute Equation (3) in Equation (2),

$$\begin{gathered} F_1=v\frac{Mdx}{Ldt} \\ =\frac{M}{L}{v}^2 \end{gathered}$$

F₁ = impulse Force acting on the chain $\Rightarrow \frac{M}{L}{v}^2$

After period of all , the velocity of the chain $v^2=2gx$

So F₁ = $2mg\frac{x}{L}$

x = The distance con ered

$$\begin{gathered} F_1=TheGravitationalforceisgivensimplybytheweightonthechain \\ {F}_2=mg=\frac{x}{L} \end{gathered}$$

The total force exerted by the chain is addition of impulse force and the gravitational force,

From Equation (1),

$$\begin{gathered} F=2mg\frac{x}{L}+mg\frac{x}{L} \\ =3mg\frac{x}{L}....\left(4\right) \end{gathered}$$

If we consider the instantaneous rate of change momentum of the chain as the last link hits the table, x will be L

So, put x = L in Equation (4),

F = 3mg

Hence, the force exerted by the table on the chain is