Question-22481

Question Number 22481 by ajfour last updated on 19/Oct/17

Commented by ajfour last updated on 19/Oct/17

Q . 22479 (S o l u t i o n)

Answered by ajfour last updated on 19/Oct/17

W h e n t h e l a d d e r i s a b o u t t o s l i p :

I f μ_{1} \neq 0, μ_{2} \neq 0

f_{2} = μ_{2} N_{2} = N_{1} \dots . (i)

f_{1} + N_{2} = μ_{1} N_{1} + N_{2} = m g

\Rightarrow (μ_{1} μ_{2} + 1) N_{2} = m g

N_{2} = \frac{m g}{1 + μ_{1} μ_{2}}

c o n s i d e r i n g τ o r q u e a b o u t c e n t r e :

N_{2} (\frac{l}{2}) \cos θ = f_{2} (\frac{l}{2}) \sin θ +

f_{1} (\frac{l}{2}) \cos θ + N_{1} (\frac{l}{2}) \sin θ

\Rightarrow N_{2} (1 - μ_{2} \tan θ) = N_{1} (μ_{1} + \tan θ)

\dots . . (i i i)

\Rightarrow I f μ_{1} = 0, μ_{2} \neq 0 t h e n

f r o m (i i) w e g e t N_{2} = m g, a n d

N_{1} = μ_{2} m g . T h e n f r o m (i i i) w e g e t

m g (1 - μ_{2} \tan θ) = N_{1} \tan θ

\Rightarrow m g - N_{1} \tan θ = N_{1} \tan θ

o r N_{1} \tan θ = \frac{m g}{2} .

I f μ_{1} \neq 0, μ_{2} = 0

f o r t h e a b o u t t o s l i p c a s e

f_{2} = 0 \Rightarrow N_{1} = 0, θ = \frac{π}{2}

N_{2} = m g .

H e n c e (3), (4) a r e c o r r e c t .

Commented by Tinkutara last updated on 19/Oct/17

Thank you very much Sir!