Question

A chain hangs over a nail with 2.0 m on one side and 6.0 m on the other side. If the force of friction is equal to the weight of 1.0 m of the chain, calculate the time required for the chain to slide off the nail.

Answer 1

Answer on Question #70087, Physics / Other

Question A chain hangs over a nail with 2.0 m on one side and 6.0 m on the other side. If the force of friction is equal to the weight of 1.0 m of the chain, calculate the time required for the chain to slide off the nail.

Solution Let us use following notation. The forces that acts on the rope are:

$F_{g}=\rho gL,\quad F_{l}=\rho gl,\quad F_{f}=\rho g\cdot f$

where $F_{g}$ is force from longer part ($L=6$), $F_{l}$ is force from shorter ($l=2$) and $F_{f}$ is force from friction, proportional to $f=1$ m of chain. Let us denote total length $L_{0}=L+l$.

Now we can write down balance of forces:

$F_{tot}=F_{g}-F_{l}-F_{f}=ma=\rho L_{0}a$

$\rho gL-\rho gl-\rho g\cdot f=\rho L_{0}a$

$g(L-l-f)=L_{0}a$

We also remember that resulting acceleration describes the change of L:

$a=\frac{d^{2}L}{dt^{2}}$

So now we have differential equation for $L(t)$:

$\frac{d^{2}L}{dt^{2}}=g\frac{L-l-f}{L_{0}}=g\frac{L-(L_{0}-L)-f)}{L_{0}}=g\frac{2L-(L_{0}+f)}{L_{0}}$

The general solution of this equation is

$L(t)=C_{1}+C_{2}e^{\sqrt{\frac{2a}{L_{0}}}}$

Using initial condition $L(0)=6$ and $dL/dt(0)=0$ we obtain solution:

$L=\frac{L_{0}+f}{2}+\frac{L+0/2-f}{2}e^{\sqrt{\frac{2a}{L_{0}}}}$

From this we can find time of falling is ($L=8$):

$t=\sqrt{\frac{8}{2\cdot 9.8}}\ln(7/3)\approx 0.54\,s$