Question

A chain of length $L=10\text{m}$ and mass $m=20\text{kg}$ is allowed to fall on a table such that the part falling on the table comes to rest immediately. The force acting on the table when $5\text{m}$ of it has fallen on the table is [Take $g=10{\text{m/s}}^2$]

• A. $300\text{N}$
• B. $600\text{N}$
• C. $200\text{N}$
• D. $100\text{N}$

Answer 1

The correct option is A $300\text{N}$

Mass of the chain, $m=20\text{kg}$
and length, $l=10\text{m}$
Total force on the table = Thrust force + weight of chain lying on table.

Mass per unit length of chain $\lambda =\frac{m}{l}$
Taking $dx$ element of the hanging part of the chain situated at distance $x$ from ground, whose mass is $dm$
$dm=\lambda dx$

Let velocity of this element when it hits the surface be $v$ :
From $W-E$ theorem
$W_{gravity}=△K.E$
$\left(dmg\right)x=\frac{1}{2}dm{v}^2$
or $v=\sqrt{2gx}$ ...(1)

As we know, thrust $F_T=v\cdot \frac{dm}{dt}$
$=v\cdot \lambda \cdot \left(\frac{dx}{dt}\right)$
$\Rightarrow F_T=v^2\cdot \lambda$
Putting the value of $v$ from (1)
$F_T=(\sqrt{2gx}{)}^2\times \frac{m}{l}=2gx\frac{m}{l}=\frac{2\times 10\times 5\times 20}{10}$
$F_T=200\text{N}$

Weight of chain lying on the table $W_l=\frac{M}{L}\times lg=\frac{20}{10}\times 5\times 10=100\text{N}$
Therefore, total force on the table $=200+100=300\text{N}$