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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.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{compression}\:\mathrm{in}\:\mathrm{the}\:\mathrm{spring}\:\mathrm{if} \\ $$$$\mathrm{the}\:\mathrm{system}\:\mathrm{shown}\:\mathrm{below}\:\mathrm{is}\:\mathrm{in} \\ $$$$\mathrm{equilibrium}. \\ $$
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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