A heavy uniform chain lies on a horizontal ta | vedclass

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(A) Let the total length of the chain be $L$ and its total mass be $M$. The mass per unit length is $\lambda = M/L$.Let $x$ be the length of the chain

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) Let the total length of the chain be $L$ and its total mass be $M$. The mass per unit length is $\lambda = M/L$.Let $x$ be the length of the chain hanging over the edge. Then the length of the chain on the table is $(L - x)$.The mass of the hanging part is $m_h = \lambda x$ and the mass of the part on the table is $m_t = \lambda (L - x)$.The force pulling the chain down is the weight of the hanging part: $F_g = m_h g = \lambda x g$.The maximum static frictional force acting on the part on the table is $f_{max} = \mu N = \mu m_t g = \mu \lambda (L - x) g$.For the chain to be on the verge of sliding,the pulling force must equal the maximum frictional force: $\lambda x g = \mu \lambda (L - x) g$.Dividing both sides by $\lambda g$,we get $x = \mu (L - x)$.Substituting $\mu = 0.25$: $x = 0.25(L - x) \implies x = 0.25L - 0.25x \implies 1.25x = 0.25L$.Thus,the fraction $x/L = 0.25 / 1.25 = 1/5 = 0.20$.Converting to percentage,the fraction is $20\%$.

$A$ 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 length of the chain that can hang over one edge of the table is ...... $\%$

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