Question

a rope lies on a table such that a part of it hangs down the table when the length of hanging part is 1/3 of entire length the rope just begins to slide the coefficient of friction between the Rope and the table is

Answer 1

Given: Let the total length of the rope be 'L' and m be the mass per unit length.

Length of hanging rope 'L₂' = (1/3)L

Mass of hanging rope 'M₂' = (1/3)L× m

Length of the rope on the table 'L₁' = L - (1/3) L = (2/3)L

Mass of the rope on the table 'M₁' = (2/3)L× m

To find: Coefficient of friction between the rope and table.

Let 'μ' be the coefficient of friction between the rope and table.

Apply the motion of equation,

The weight of rope on the table = The weight of hanging rope

or, μ (2/3)L× mg = (1/3) L × mg

or, μ = 1/2 = 0.5

Hence, the required coefficient of friction between the rope and table will be 0.5

Answer 2

Given:  

Length of hanging rope = (1/3) of entire length the rope

Mass of hanging rope = (1/3) of the length of the rope * mass of the rope

Length of the rope on the table = (2/3) Length of hanging rope

Mass of the rope on the table = (2/3) Length of hanging rope * mass of the rope

To find:  

The coefficient of friction between the rope and table.

Solution:

Consider,

x as the coefficient of friction

By law,

The weight of rope on the table = The weight of hanging rope

x * (2/3) Length of hanging rope × mass of the rope * gravity  = (1/3)  Length of hanging rope × mass of the rope * gravity

x = 1/2

We get,

x = 0.5

Hence, the required coefficient of friction between the rope and table is 0.5.

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