In-the-figure-shown-mass-m-is-placed-on-the-inclined-surface-of-a-wedge-of-mass-M-All-the-surfaces-are-smooth-Find-the-acceleration-of-the-wedge-

Question Number 20777 by Tinkutara last updated on 02/Sep/17

I n t h e f i g u r e s h o w n, m a s s^{'} m^{'} i s

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

w e d g e o f m a s s M . A l l t h e s u r f a c e s

a r e s m o o t h . F i n d t h e a c c e l e r a t i o n o f

t h e w e d g e .

Commented by Tinkutara last updated on 02/Sep/17

Commented by ajfour last updated on 02/Sep/17

N \sin θ = M A . . (i)

m g \cos θ - N = m A \sin θ

o r u s i n g (i)

m g \cos θ \sin θ - M A = m A \sin^{2} θ

\Rightarrow \boldsymbol A = \frac{\boldsymbol m g \sin \boldsymbol θ \cos \boldsymbol θ}{\boldsymbol M + \boldsymbol m \sin^{2} \boldsymbol θ} .

Commented by Tinkutara last updated on 02/Sep/17

Thank you very much Sir!

Commented by mrW1 last updated on 03/Sep/17

what is the result if the friction coe . of all

surfaces is μ ?

Commented by ajfour last updated on 03/Sep/17

Commented by ajfour last updated on 03/Sep/17

m g \cos θ - N = m A \sin θ \dots (i)

N \sin θ - μ N \cos θ - μ R = M A . . (i i)

R = M g + N \cos θ + μ N \sin θ \dots (i i i)

s o l v i n g f o r A w e g e t,

\boldsymbol A = \frac{(1 - μ^{2}) m g \cos θ \sin θ - 2 μ m g \cos^{2} θ - μ M g}{M + (1 - μ^{2}) m \sin^{2} θ} .

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