Another problem on Mechanics?

Nov 13, 2016

${W}_{x} = \mu \left(3 x + 2 h\right) g$

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 $\mathrm{dm}$ arriving at that instant.

$I = \left(\frac{\mathrm{dm}}{\mathrm{dt}}\right) v = \frac{\mu v \mathrm{dt}}{\mathrm{dt}} v = \mu {v}^{2}$

but we know that for any $\mathrm{dm}$ after the excursion $x$

$\frac{1}{2} {v}^{2} \mathrm{dm} = \mathrm{dm} \left(x + h\right) g$ or

${v}^{2} = 2 \left(x + h\right) g$ Now, substituting into $I$ we have

$I = 2 \mu \left(x + h\right) g$ and finally

${W}_{x} = \mu x g + 2 \mu \left(x + h\right) g = \mu \left(3 x + 2 h\right) g$