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


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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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