# Falling Rope on a scale

SonOfOle

## Homework Statement

A uniform flexible rope is suspended above a scale, with the bottom of the rope just touching the scale (gravity points downward). The rope has a length L and a total mass of M. The mass is uniformly distributed along it's length.

The rope is released. After a length x<L has fallen onto the scale, what does the scale read? Assume the scale can measure the force applied to it instantaneously. (Hint: the force exerted by the rope on the scale has two components).

## The Attempt at a Solution

F=dp/dt
p=m*t
m(t)=(M/L)*v(t)
so p(t)=(M/L)*(v(t)^2)
=(M/L)*(g^2)*(t^4)/4
thus dp/dt=(M/L)*(g^2)*t^3

also, x(t)=g(t^2)/2--> t=sqrt(2xg). plug that into the eqn for dp/dt, and get
F(x)=dp/dt=(M/L)*(g^2)*(2xg)^(3/2).

However, that solution doesn't treat the problem with a force of two components, nor do the units seem to work out. Any ideas?

konthelion
Let M=mass, L=length of the rope, v=velocity

You have to express in component form,
$$F_{net} + \frac{dm}{dt}v=m\frac{dv}{dt}$$
$$= F_{n} - mg+\frac{dm}{dt}v=0$$ (Eq 1), then

$$\frac{dm}{dt}=\frac{-M}{L}v$$ (*), where dm=mass of the section of the rope that falls on the scale.

Substitute (*) back into (Eq 1) and find v using $$v^2=v_{0}^2+2a\Delta y$$
Then find $$F_n$$

SonOfOle
thanks