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`        A rope of length 30cm is on a 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 Ais`
one year ago

Arun
23822 Points
```							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 cmthe coefficient of friction, μ = 0.5Length of hanging rope = lLength of rope lying on a table = L - lSuppose, mass density = λforce of friction, Ff = (λ*l)g=> The force of friction:Ff = μmg(λ*l)g = μ(λ*(L-l))gl = μL - μlμL = (1 + μ)*ll / L = μ/ 1 + μl/L = 0.5 / 1 + 0.5 = 1/3l/30 = 1/3l = 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= - 20/3Ycm = (-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 Hope it helps Regards
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6 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions