Find-the-compression-in-the-spring-if-the-system-shown-below-is-in-equilibrium-

Question Number 21148 by Tinkutara last updated on 14/Sep/17

Find the compression in the spring if

the system shown below is in

equilibrium .

Commented by Tinkutara last updated on 14/Sep/17

Commented by mrW1 last updated on 14/Sep/17

force in spring = 0

Commented by mrW1 last updated on 15/Sep/17

since the system is in equilibrium,

the force in spring must be equal to

the friction force between block M

and ground . since the surface of

ground is smooth, the friction force

is zero .

Commented by ajfour last updated on 15/Sep/17

N = m g \cos θ

N \sin θ = k x

\Rightarrow x = \frac{m g \cos θ \sin θ}{k} .

Commented by Tinkutara last updated on 15/Sep/17

My doubt is why it would not be this

as calculated by ajfour Sir ?

Commented by ajfour last updated on 15/Sep/17

t h i s i s t h e c o m p r e s s i o n, i n c a s e

t h e w e d g e i s i n e q u i l i b r i u m,

i t h i n k .

Commented by alex041103 last updated on 15/Sep/17

B e c a u s e o f t h e g r a v i t a t i o n t h e s m a l l

b o d y a c t s o n t h e b i g g e r w i t h s o m e

f o r c e m g c o s θ s o F_{s p r i n g} = m g c o s θ s i n θ = \frac{1}{2} m g s i n (2 θ)

B u t {\vec{F}}_{s p r i n g} = - k \vec{x} \Rightarrow F_{s p r i n g} = k x

\Rightarrow x = \frac{m g s i n (2 θ)}{2 k}

Commented by Tinkutara last updated on 16/Sep/17

Thank you very much Sir!

Commented by alex041103 last updated on 15/Sep/17

N o t e : T h e s i n θ i n F_{s p r i n g} = m g c o s θ s i n θ i s t h e r e

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

s u r f a c e \Rightarrow t h e r e i s s o m e f o r c e (n o r m a l r e a c t i o n)

w h i c h c o m p e n s a t e s f o r m g c o s θ c o s θ

w h e r e m g c o s (θ) s i n (θ) a n d {m g c o s}^{2} (θ)

a r e t h e m a g n i t u d e s o f t h e v e c t o r s

o n e n o r m a l a n d a n o t h e r p a r a l l e l t o t h e

s u r f a c e, a n d t h e s u m o f t h o s e v e c t o r s

i s t h e v e c t o r o f t h e f o r c e w i t h w h i c h

t h e s m a l l e r b o d y a c t s o n t h e b i g g e r o n e .

Commented by mrW1 last updated on 16/Sep/17

so that the system remains in

equilibrium, m should not slide alone

M, i . e . friction must exist between both

blocks . they act as if they were a single

block which is connected on the

spring . since no other horizontal force

is acting on this block, the force in the

spring also must be zero .

if there is no friction between m and

M, block m will move with an

acceleration alone block M . but the

question says that the system is in

equilibrium, it means no part of it

is in motion .

Commented by alex041103 last updated on 16/Sep/17

y e s o k

Commented by ajfour last updated on 16/Sep/17

y e s s i r, p e r f e c t l o g i c .

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