- #1

SonOfOle

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## 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).

## Homework Equations

## 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?