Question

Another problem on Mechanics?

Answer 1

`W_x = mu(3x+2h)g`

Explanation:

First, some asumptions.

i) The chain falls on the floor, resting at the floor level.
ii) The rope remaining above the table, keeps straigth.

For an excursion `x` the weigth supported on the floor is given by

`W_x=m_x g + I`

where `m_x = mu x` is the mass landed and `I` is the impulse exerted by the `dm` arriving at that instant.

`I = ((dm)/(dt)) v = (mu v dt)/(dt)v = mu v^2`

but we know that for any `dm` after the excursion `x`

`1/2v^2dm=dm (x+h)g` or

`v^2=2(x+h)g` Now, substituting into `I` we have

`I=2mu(x+h)g` and finally

`W_x =mu x g + 2mu(x+h)g = mu(3x+2h)g`