A-block-is-placed-on-a-rough-horizontal-surface-The-minimum-force-required-to-slide-the-block-is-

Question Number 20412 by Tinkutara last updated on 26/Aug/17

A block is placed on a rough horizontal

surface . The minimum force required

to slide the block is

Commented by Tinkutara last updated on 26/Aug/17

Commented by ajfour last updated on 26/Aug/17

P \cos θ = f_{m a x} = μ N

N + P \sin θ = m g

S o P \cos θ = μ (m g - P \sin θ)

\Rightarrow P = \frac{μ m g}{\cos θ + μ \sin θ}

= \frac{μ}{\sqrt{1 + μ^{2}}} \times \frac{m g}{\sin [θ + \tan^{- 1} (\frac{1}{μ})]}

P i s l e a s t w h e n θ = \frac{π}{2} - \cot^{- 1} μ

o r w h e n θ = \tan^{- 1} μ

P_{m i n i m u m} = \frac{μ m g}{\sqrt{1 + μ^{2}}} .

Commented by Tinkutara last updated on 26/Aug/17

Thank you very much Sir!