A uniform chain of length 7.00 m initiallylies stretched out on a horizontal tab

Question

A uniform chain of length 7.00 m initiallylies stretched out on a horizontal table. (a) If the coefficient of static frictionbetween chain and table is 0.600, show that the chain will begin toslide off the table if at least 2.63 m ofit hangs over the edge of the table.
(b) Determine the speed of the chain as its last link leaves thetable, given that the coefficient of kinetic friction between thechain and the table is 0.400. (a) If the coefficient of static frictionbetween chain and table is 0.600, show that the chain will begin toslide off the table if at least 2.63 m ofit hangs over the edge of the table.
(b) Determine the speed of the chain as its last link leaves thetable, given that the coefficient of kinetic friction between thechain and the table is 0.400.

Explanation / Answer

I hope this isn't a test problem: ------------------------------- a. Let m = mass/meter of the chain.
Let x = length of chain on table and 7-x be the length hanging offthe table Then frictional force on table = 0.600*(chain mass on table) *g = 0.6 * x*m*g Force of gravity on chain hanging over the table =(7-x)*m*g Set the two equations equal to get: 0.6 * x*m*g = (7-x)*m*g or 0.6 x = (7-x) or 1.6 x = 7 orx = 7/1.6 = 4.375 So, 7 - x = 2.625 ------------------------------- b. Balance point for kinectic friction = 0.4 0.4x = 7-x or 1.4 x= 7 or x = 5 and 7 - x = 2, so 2 m of chain hangs off table ------ ------------------------------- Note: There may be an easier way of solving this problem, buthere's what I have: Since m can be any value, assume 1 kg/meter Let x be the length of chain hanging over the table and 'a' bethe acceleration of the chain. Then force on chain = mass of chain * a = 7*a = mass ontable * g * coef of kinectic friction - mass off table *g or 7x'' =(7-x)*g*0.4 + x*g This is a differential equation problem as x is dynamicallychanging. The initial conditions for this problem are: x(0) = 2.625 and x' (0) = 0 and 2.625 <= x <= 7 The solution using an off-line solver is: x(t) = (175/48)*exp((1/5)*sqrt(21)*t)+(175/48)*exp(-(1/5)*sqrt(21)*t)-14/3 which can be solved for t when x(t) = 7 plug that value into x'(t) to determine the velocity of thechain x'(t) =(35/48)*sqrt(21)*exp((1/5)*sqrt(21)*t)-(35/48)*sqrt(21)*exp(-(1/5)*sqrt(21)*t) According to my calcs: x'(chain leaves table) = 8.35 m/s According to my calcs: x'(chain leaves table) = 8.35 m/s

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