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-

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 \dots 1 (a)

N \sin θ = F + f \cos θ \dots 2 (a)

a n d f = μ N, s o

(μ + 1) N = m g \sqrt{2} \dots 1 (b)

(1 - μ) N = F \sqrt{2} \dots 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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