A rope of length 30cm is on a horizontal table with maximum length hanging from edge A of the table. the coefficient of friction between the rope and table is 0.5.the distance of centre of mass of the rope from A is?
Answers
Answer:
Explanation:
=> According to the question, a rope of length 30cm is on a horizontal table with maximum length hanging from edge A of the table.
Total length of rope, L = 30 cm
the coefficient of friction, μ = 0.5
Length of hanging rope = l
Length of rope lying on a table = L - l
Suppose, mass density = λ
force of friction, Ff = (λ*l)g
=> The force of friction:
Ff = μmg
(λ*l)g = μ(λ*(L-l))g
l = μL - μl
μL = (1 + μ)*l
l / L = μ/ 1 + μ
l/L = 0.5 / 1 + 0.5 = 1/3
l/30 = 1/3
l = 10 cm
=> Suppose 'ρ' is the mass per unit length and the coordinates of 20ρ and 10ρ are (10,0) and (0,5) respectively from A.
Xcm = (20λ) (-10) + (10λ)(0) / 30 λ
= -200λ/30λ
= - 20/3
Ycm = (-5)(10λ) + 0(20λ) / 30λ
= -50λ / 30λ
= -5/3
=> The distance of centre of mass from A :
rcm=√x²cm+y²cm
=√ (-20/3)² - (-5/3)²
= √ 400/9 + 25/9
= √ 425/9
= √ 17 * 25 / 9
= 5√17 / 3
Thus, the distance of centre of mass of the rope from A is 5√17 / 3.