A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and tab

Question

A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is:

Accepted Answer

1. **System Analysis:** Let the total length of the chain be $L$ and its total mass be $M$. Let the length of the chain hanging over the edge be $l$. The length of the chain remaining on the table is $L - l$.
2. **Forces Involved:**
   - **Driving Force:** The weight of the hanging portion pulls the chain down. Mass of hanging part = $(l/L)M$. Force $F_g = (l/L)Mg$.
   - **Resisting Force:** The limiting static friction acting on the portion on the table balances the pulling force. Mass on table = $((L-l)/L)M$. Normal reaction $N = ((L-l)/L)Mg$. Friction $f = \mu N = \mu ((L-l)/L)Mg$ [Source 74].
3. **Equilibrium Condition:** For the maximum length to hang without sliding, the driving force must equal the limiting friction:
   $$ \frac{l}{L}Mg = \mu \left(\frac{L-l}{L}ight)Mg $$
   Canceling common terms ($Mg/L$):
   $$ l = \mu(L - l) $$
   $$ l = \mu L - \mu l $$
   $$ l(1 + \mu) = \mu L $$
   $$ \frac{l}{L} = \frac{\mu}{1 + \mu} $$
4. **Calculation:** Given $\mu = 0.25$:
   $$ 	ext{Fraction} = \frac{0.25}{1 + 0.25} = \frac{0.25}{1.25} = \frac{1}{5} = 0.2 $$
   $$ 	ext{Percentage} = 0.2 	imes 100\% = 20\% $$

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