Question-15534

Question Number 15534 by ajfour last updated on 11/Jun/17

Commented by ajfour last updated on 11/Jun/17

F i n d t h e e x i s t i n g s t a t i c f r i c t i o n

c o e f f i c i e n t b e t w e e n r o p e a n d

p u l l e y i f t h e t e n s i o n i n r o p e a t

t h e h o o k t o w h i c h {i t}^{'} s o n e e n d

i s t i e d i s e q u a l t o \frac{M g}{2} .

A s s u m e m a s s o f t h r e a d i s

n e g l i g i b l e c o m p a r e d t o t h e

m a s s o f t h e b l o c k (M) .

Answered by mrW1 last updated on 11/Jun/17

T = tension in rope

r = radius of pulley

dT = - μ \frac{T}{r} rd θ = - μ Td θ

\frac{dT}{T} = - μ d θ

\ln T = - μ θ + C

at θ = 0 : T = Mg

\Rightarrow C = \ln Mg

\Rightarrow \ln \frac{T}{Mg} = - μ θ

\Rightarrow T = Mg \times e^{- μ θ} = \frac{Mg}{e^{μ θ}}

at θ = \frac{3 π}{2} : T = \frac{Mg}{2}

\frac{Mg}{2} = \frac{Mg}{e^{\frac{3 π μ}{2}}}

\Rightarrow e^{\frac{3 π μ}{2}} = 2

\Rightarrow μ = \frac{2 \ln 2}{3 π} \approx 0.147

Commented by ajfour last updated on 11/Jun/17

t h a n k y o u s i r . a n s w e r i s c o r r e c t .

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