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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.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{compression}\:\mathrm{in}\:\mathrm{the}\:\mathrm{spring}\:\mathrm{if} \\ $$$$\mathrm{the}\:\mathrm{system}\:\mathrm{shown}\:\mathrm{below}\:\mathrm{is}\:\mathrm{in} \\ $$$$\mathrm{equilibrium}. \\ $$
Commented by Tinkutara last updated on 14/Sep/17
Commented by mrW1 last updated on 14/Sep/17
force in spring=0
$$\mathrm{force}\:\mathrm{in}\:\mathrm{spring}=\mathrm{0} \\ $$
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.
$$\mathrm{since}\:\mathrm{the}\:\mathrm{system}\:\mathrm{is}\:\mathrm{in}\:\mathrm{equilibrium}, \\ $$$$\mathrm{the}\:\mathrm{force}\:\mathrm{in}\:\mathrm{spring}\:\mathrm{must}\:\mathrm{be}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{friction}\:\mathrm{force}\:\mathrm{between}\:\mathrm{block}\:\mathrm{M}\: \\ $$$$\mathrm{and}\:\mathrm{ground}.\:\mathrm{since}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{of} \\ $$$$\mathrm{ground}\:\mathrm{is}\:\mathrm{smooth},\:\mathrm{the}\:\mathrm{friction}\:\mathrm{force} \\ $$$$\mathrm{is}\:\mathrm{zero}.\: \\ $$
Commented by ajfour last updated on 15/Sep/17
N=mgcos θ  Nsin θ=kx  ⇒  x=((mgcos θsin θ)/k) .
$${N}={mg}\mathrm{cos}\:\theta \\ $$$${N}\mathrm{sin}\:\theta={kx} \\ $$$$\Rightarrow\:\:{x}=\frac{{mg}\mathrm{cos}\:\theta\mathrm{sin}\:\theta}{{k}}\:. \\ $$
Commented by Tinkutara last updated on 15/Sep/17
My doubt is why it would not be this  as calculated by ajfour Sir?
$$\mathrm{My}\:\mathrm{doubt}\:\mathrm{is}\:\mathrm{why}\:\mathrm{it}\:\mathrm{would}\:\mathrm{not}\:\mathrm{be}\:\mathrm{this} \\ $$$$\mathrm{as}\:\mathrm{calculated}\:\mathrm{by}\:\mathrm{ajfour}\:\mathrm{Sir}? \\ $$
Commented by ajfour last updated on 15/Sep/17
this is the compression, in case  the wedge is in equilibrium,  i think.
$${this}\:{is}\:{the}\:{compression},\:{in}\:{case} \\ $$$${the}\:{wedge}\:{is}\:{in}\:{equilibrium}, \\ $$$${i}\:{think}. \\ $$
Commented by alex041103 last updated on 15/Sep/17
Because of the gravitation the small  body acts on the bigger with some  force mg cosθ so  F_(spring) =mgcosθsinθ=(1/2)mgsin(2θ)  But F_(spring) ^→ =−kx^→ ⇒F_(spring) =kx  ⇒x=((mgsin(2θ))/(2k))
$${Because}\:{of}\:{the}\:{gravitation}\:{the}\:{small} \\ $$$${body}\:{acts}\:{on}\:{the}\:{bigger}\:{with}\:{some} \\ $$$${force}\:{mg}\:{cos}\theta\:{so}\:\:{F}_{{spring}} ={mgcos}\theta{sin}\theta=\frac{\mathrm{1}}{\mathrm{2}}{mgsin}\left(\mathrm{2}\theta\right) \\ $$$${But}\:\overset{\rightarrow} {{F}}_{{spring}} =−{k}\overset{\rightarrow} {{x}}\Rightarrow{F}_{{spring}} ={kx} \\ $$$$\Rightarrow{x}=\frac{{mgsin}\left(\mathrm{2}\theta\right)}{\mathrm{2}{k}} \\ $$
Commented by Tinkutara last updated on 16/Sep/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by alex041103 last updated on 15/Sep/17
Note: The sinθ  in F_(spring) =mgcosθsinθ is there  because part of the force is applied on the  surface ⇒ there is some force(normal reaction)  which compensates for mgcosθcosθ  where mgcos(θ)sin(θ) and mgcos^2 (θ)  are the magnitudes of the vectors  one normal and another parallel to the  surface, and the sum of those vectors  is the vector of the force with which  the smaller body acts on the bigger one.
$${Note}:\:{The}\:{sin}\theta\:\:{in}\:{F}_{{spring}} ={mgcos}\theta{sin}\theta\:{is}\:{there} \\ $$$${because}\:{part}\:{of}\:{the}\:{force}\:{is}\:{applied}\:{on}\:{the} \\ $$$${surface}\:\Rightarrow\:{there}\:{is}\:{some}\:{force}\left({normal}\:{reaction}\right) \\ $$$${which}\:{compensates}\:{for}\:{mgcos}\theta{cos}\theta \\ $$$${where}\:{mgcos}\left(\theta\right){sin}\left(\theta\right)\:{and}\:{mgcos}^{\mathrm{2}} \left(\theta\right) \\ $$$${are}\:{the}\:{magnitudes}\:{of}\:{the}\:{vectors} \\ $$$${one}\:{normal}\:{and}\:{another}\:{parallel}\:{to}\:{the} \\ $$$${surface},\:{and}\:{the}\:{sum}\:{of}\:{those}\:{vectors} \\ $$$${is}\:{the}\:{vector}\:{of}\:{the}\:{force}\:{with}\:{which} \\ $$$${the}\:{smaller}\:{body}\:{acts}\:{on}\:{the}\:{bigger}\:{one}. \\ $$
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.
$$\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{system}\:\mathrm{remains}\:\mathrm{in} \\ $$$$\mathrm{equilibrium},\:\mathrm{m}\:\mathrm{should}\:\mathrm{not}\:\mathrm{slide}\:\mathrm{alone} \\ $$$$\mathrm{M},\:\mathrm{i}.\mathrm{e}.\:\mathrm{friction}\:\mathrm{must}\:\mathrm{exist}\:\mathrm{between}\:\mathrm{both} \\ $$$$\mathrm{blocks}.\:\mathrm{they}\:\mathrm{act}\:\mathrm{as}\:\mathrm{if}\:\mathrm{they}\:\mathrm{were}\:\mathrm{a}\:\mathrm{single} \\ $$$$\mathrm{block}\:\mathrm{which}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{spring}.\:\mathrm{since}\:\mathrm{no}\:\mathrm{other}\:\mathrm{horizontal}\:\mathrm{force} \\ $$$$\mathrm{is}\:\mathrm{acting}\:\mathrm{on}\:\mathrm{this}\:\mathrm{block},\:\mathrm{the}\:\mathrm{force}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{spring}\:\mathrm{also}\:\mathrm{must}\:\mathrm{be}\:\mathrm{zero}. \\ $$$$\mathrm{if}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{m}\:\mathrm{and} \\ $$$$\mathrm{M},\:\mathrm{block}\:\mathrm{m}\:\mathrm{will}\:\mathrm{move}\:\mathrm{with}\:\mathrm{an} \\ $$$$\mathrm{acceleration}\:\mathrm{alone}\:\mathrm{block}\:\mathrm{M}.\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{question}\:\mathrm{says}\:\mathrm{that}\:\mathrm{the}\:\mathrm{system}\:\mathrm{is}\:\mathrm{in} \\ $$$$\mathrm{equilibrium},\:\mathrm{it}\:\mathrm{means}\:\mathrm{no}\:\mathrm{part}\:\mathrm{of}\:\mathrm{it} \\ $$$$\mathrm{is}\:\mathrm{in}\:\mathrm{motion}. \\ $$
Commented by alex041103 last updated on 16/Sep/17
yes ok
$${yes}\:{ok} \\ $$
Commented by ajfour last updated on 16/Sep/17
yes sir, perfect logic.
$${yes}\:{sir},\:{perfect}\:{logic}. \\ $$

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