A horizontal force of 10N just prevents a mass of 2kg from sliding down a rough plane inclined at 45° to the horizontal. Find the coefficient of friction.


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Question Number 22892 by NECx last updated on 23/Oct/17

$A h o r i z o n t a l f o r c e o f 10 N j u s t$ $p r e v e n t s a m a s s o f 2 k g f r o m$ $s l i d i n g d o w n a r o u g h p l a n e$ $i n c l i n e d a t 45 ° t o t h e h o r i z o n t a l .$ $F i n d t h e c o e f f i c i e n t o f f r i c t i o n .$
Commented by ajfour last updated on 23/Oct/17

Commented by ajfour last updated on 23/Oct/17

$f \sin θ + N \cos θ = m g . . . 1 (a)$ $N \sin θ = F + f \cos θ . . . 2 (a)$ $a n d f = μ N, s o$ $(μ + 1) N = m g \sqrt{2} . . . 1 (b)$ $(1 - μ) N = F \sqrt{2} . . . 2 (b)$ $d i v i d i n g w e o b t a i n$ $\frac{μ + 1}{1 - μ} = \frac{m g}{F}$ $\Rightarrow \frac{μ + 1}{2} = \frac{m g}{m g + F}$ $\Rightarrow μ = \frac{m g - F}{m g + F} = \frac{20 - 10}{20 + 10} = \frac{1}{3} .$
Commented by NECx last updated on 24/Oct/17

$y o u r a n s w e r i s v e r y a c c u r a t e s i r .$ $T h a n k s f o r t h e h e l p$
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