What percentage of the total length of a chai | vedclass

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Question

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

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) Let $L$ be the total length of the chain and $M$ be its total mass. The mass per unit length is $\lambda = M/L$.Let $l$ be the length of the chain hanging off the table. The length of the chain on the table is $(L - l)$.The weight of the hanging part is $W_h = (\lambda l)g$,which acts as the pulling force.The normal force on the part of the chain on the table is $N = (\lambda(L - l))g$.The limiting friction force is $f_{max} = \mu N = \mu \lambda(L - l)g$.For the chain to be in equilibrium,the pulling force must equal the limiting friction force:$\lambda l g = \mu \lambda(L - l)g$$l = \mu(L - l)$$l = \mu L - \mu l$$l(1 + \mu) = \mu L$$l/L = \mu / (1 + \mu)$Given $\mu = 0.25$:$l/L = 0.25 / (1 + 0.25) = 0.25 / 1.25 = 1/5 = 0.20$.Thus,the percentage of the length is $0.20 	imes 100 = 20\%$.

What percentage of the total length of a chain can hang off the edge of a table without sliding,given that the coefficient of friction between the chain and the table is $0.25$?

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