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 #67151 – Math – Differential Equations

Question

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

Solution

Let $l = 8$ be the length of the chain, $x(t)$ be the length of the longest side of the chain, $m$ be the mass of one meter of the chain.

Mass of the chain is $M = ml$, the friction force is $m \cdot 1 \cdot g = mg$.

$$F = M a.$$ $$m x (t) g - m (l - x (t)) g - m g = M a.$$ $$2 m g x (t) g - m l g - m g = m l \frac {d ^ {2} x}{d t ^ {2}}.$$ $$\frac {d ^ {2} x}{d t ^ {2}} - 2 \frac {g}{l} x (t) = - \frac {(l + 1) g}{l}. \quad (1)$$

Let $x(t) = C$ be a particular solution of nonhomogeneous differential equation (1), where $C$ is a real constant.

Then

$$\begin{array}{l} \frac {d ^ {2} C}{d t ^ {2}} - 2 \frac {g}{l} C = - \frac {(l + 1) g}{l}, \\ - 2 \frac {g}{l} C = - \frac {(l + 1) g}{l}, \\ \end{array}$$

hence $C = \frac{l + 1}{2}$

The general solution of the homogeneous equation $\frac{d^2x}{dt^2} - 2\frac{g}{l} x(t) = 0$ is

$$x _ {*} (t) = A e ^ {\sqrt {\frac {2 g}{l}} t} + B e ^ {- \sqrt {\frac {2 g}{l}} t}.$$

The general solution of nonhomogeneous differential equation (1) is

$$\begin{array}{l} x (t) = A e ^ {\sqrt {\frac {2 g}{l}} t} + B e ^ {- \sqrt {\frac {2 g}{l}} t} + \frac {l + 1}{2}; \\ x (t) = A e ^ {1.566t} + B e ^ {- 1.566t} + 4.5. \\ \end{array}$$

Initial conditions: $x(0) = 6, \frac{dx}{dt} = 0$.

$$\text{So} \left\{ \begin{array}{c} A + B + 4.5 = 6 \\ A - B = 0 \end{array} \right. \to A = B = 0.75.$$ $$x(t) = 0.75e^{1.566t} + 0.75e^{-1.566t} + 4.5.$$

When $x(t) = 8$:

$$\begin{array}{l} 0.75e^{1.566t} + 0.75e^{-1.566t} + 4.5 = 8 \rightarrow e^{1.566t} + e^{-1.566t} - 4.667 = 0 \rightarrow \\ e^{2*1.566t} - 4.667e^{1.566t} + 1 = 0 \rightarrow e^{1.566t} = 0.225 \text{ or } e^{1.566t} = 4.442. \end{array}$$

If $e^{1.566t} = 0.225$ then $t = -0.953 < 0$ is impossible.

If $e^{1.566t} = 4.442$ then $t = 0.953$ sec.

Answer: $t = 0.953$ sec.

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