Weight of a Falling Chain

Classical Mechanics Level 3

A 100 m rope of line density $\lambda=0.1$ kg/m is suspended vertically so that the bottom of the rope is just touching a scale. It is then released and falls onto the scale, which perceives a time dependent weight.

What is the reading on the scale in Newtons at the moment half the rope lies on the scale?

Assumptions and Details

$g=9.8$ m/s $^{2}$ .

The answer is 147.

1 solution

Beakal Tiliksew
Feb 23, 2014

The scale measure the weight of the object on the scale, plus the rate of change of momentum of the falling link.

Assuming constant weight distribution, the weight of the object at any instant of time is $\frac{Mg \times x}{L}$ , where x the length of the chain on the scale.

If we account for the change of momentum at the instant of it hitting the ground $F_{p} = \frac{dp}{dt} = \frac{M}{L}\frac{dx}{dt}V= \frac{M}{L}V ^{2}$

where V is the velocity of the element before hitting the ground, and we know form kinematics that $V^{2}$ = $2gx$ ,

Thus $F_{p} = 2\frac{M}{L}gx$ and $F_\text{total}= F_{p} + F_\text{weight} = 3\frac{M}{L}gx$

At half the distance, $F_\text{total} = \frac{3}{2}Mg$

cool

jinay patel - 7 years, 3 months ago

How can you get dp/dt into M/L dx/dt V?

William Nathanael Supriadi - 4 years, 6 months ago

Consider that $dP=dM \cdot V = \lambda dx \cdot V$

So $\dfrac{dP}{dt}=V \lambda \dfrac{dx}{dt}$ We know $M=\lambda L$ and hence $\dfrac{dP}{dt}=V \lambda \dfrac{dx}{dt}=\dfrac{M}{L}\cdot V \dfrac{dx}{dt}$

Harsh Poonia - 2 years, 3 months ago

why not consider dv/dt term when we differentiated dp/dt as the moving element also comes to rest after striking the ground
If we consider that f=2mg=196 N could also be correct

sggfhd rfdsgherg - 4 years, 5 months ago

You could do that as well. $\dfrac{dP}{dt}=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}$ Now since our $m$ is mass of a tiny element, the $m\dfrac{dv}{dt}$ term is negligible ( $dm\dfrac{dv}{dt}=0$ ). Rather, more easily, $\Delta \textbf{P}= \textbf{P}_f-\textbf{P}_i=(0)-(dm \textbf{v})=-\textbf{v}dm.$

Harsh Poonia - 1 year, 9 months ago

A new concept of calculation to me! Awesome

Kelvin Hong - 3 years, 9 months ago

Weight of a Falling Chain

The answer is 147.

1 solution

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