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Question-177214




Question Number 177214 by mr W last updated on 02/Oct/22
Commented by mr W last updated on 02/Oct/22
the friction coefficient between the  thin rod and the ground as well as the  wall is μ. the length of the rod is L.   find the condition which a and b  must fulfill such that the rod can  rest in equilibrium and is about to  slip as shown.
$${the}\:{friction}\:{coefficient}\:{between}\:{the} \\ $$$${thin}\:{rod}\:{and}\:{the}\:{ground}\:{as}\:{well}\:{as}\:{the} \\ $$$${wall}\:{is}\:\mu.\:{the}\:{length}\:{of}\:{the}\:{rod}\:{is}\:{L}.\: \\ $$$${find}\:{the}\:{condition}\:{which}\:{a}\:{and}\:{b} \\ $$$${must}\:{fulfill}\:{such}\:{that}\:{the}\:{rod}\:{can} \\ $$$${rest}\:{in}\:{equilibrium}\:{and}\:{is}\:{about}\:{to} \\ $$$${slip}\:{as}\:{shown}. \\ $$
Commented by mahdipoor last updated on 02/Oct/22
AB=(−a,b,c)     (√(c^2 +a^2 +b^2 ))=L^2   F_A =(p,q,s)       s=N_B ≥0  F_B =(u,v,w)     u=N_A ≥0  W=(0,0,−mg)  ΣF=ma=0=F_B +F_A +W  ⇒F_A =(−u,−v,mg−w)  ΣM_A =0=AA×F_A +AB×F_B +(1/2)AB×W  =(bw−((bmg)/2)−vc,cu+aw−((amg)/2),−av−bu)  ⇒ { ((F_A =(−u,(b/a)u,((mg)/2)+(c/a)u))),((F_B =(u,−(b/a)u,((mg)/2)−(c/a)u))) :}   { ((μ∣N_A ∣≥∣T_A ∣)),((μ∣N_B ∣≥∣T_B ∣)) :} ⇒  A: { ((0≥(((b^2 +c^2 )/a^2 )−μ^2 )u^2 −(((cmg)/a))u+(((mg)/2))^2 )),((0≤((c^2 /a^2 )−((b^2 +a^2 )/(μ^2 a^2 )))u^2 +(((cmg)/a))u+(((mg)/2))^2 )) :}  for ∀a,b  if ∃u≥0 for A, rod can rest.  and if at least one of the inequalities in  A equal zero , rod is about slip
$${AB}=\left(−{a},{b},{c}\right)\:\:\:\:\:\sqrt{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={L}^{\mathrm{2}} \\ $$$${F}_{{A}} =\left({p},{q},{s}\right)\:\:\:\:\:\:\:{s}={N}_{{B}} \geqslant\mathrm{0} \\ $$$${F}_{{B}} =\left({u},{v},{w}\right)\:\:\:\:\:{u}={N}_{{A}} \geqslant\mathrm{0} \\ $$$${W}=\left(\mathrm{0},\mathrm{0},−{mg}\right) \\ $$$$\Sigma{F}={ma}=\mathrm{0}={F}_{{B}} +{F}_{{A}} +{W} \\ $$$$\Rightarrow{F}_{{A}} =\left(−{u},−{v},{mg}−{w}\right) \\ $$$$\Sigma{M}_{{A}} =\mathrm{0}={AA}×{F}_{{A}} +{AB}×{F}_{{B}} +\frac{\mathrm{1}}{\mathrm{2}}{AB}×{W} \\ $$$$=\left({bw}−\frac{{bmg}}{\mathrm{2}}−{vc},{cu}+{aw}−\frac{{amg}}{\mathrm{2}},−{av}−{bu}\right) \\ $$$$\Rightarrow\begin{cases}{{F}_{{A}} =\left(−{u},\frac{{b}}{{a}}{u},\frac{{mg}}{\mathrm{2}}+\frac{{c}}{{a}}{u}\right)}\\{{F}_{{B}} =\left({u},−\frac{{b}}{{a}}{u},\frac{{mg}}{\mathrm{2}}−\frac{{c}}{{a}}{u}\right)}\end{cases} \\ $$$$\begin{cases}{\mu\mid{N}_{{A}} \mid\geqslant\mid{T}_{{A}} \mid}\\{\mu\mid{N}_{{B}} \mid\geqslant\mid{T}_{{B}} \mid}\end{cases}\:\Rightarrow \\ $$$${A}:\begin{cases}{\mathrm{0}\geqslant\left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\mu^{\mathrm{2}} \right){u}^{\mathrm{2}} −\left(\frac{{cmg}}{{a}}\right){u}+\left(\frac{{mg}}{\mathrm{2}}\right)^{\mathrm{2}} }\\{\mathrm{0}\leqslant\left(\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} }{\mu^{\mathrm{2}} {a}^{\mathrm{2}} }\right){u}^{\mathrm{2}} +\left(\frac{{cmg}}{{a}}\right){u}+\left(\frac{{mg}}{\mathrm{2}}\right)^{\mathrm{2}} }\end{cases} \\ $$$${for}\:\forall{a},{b}\:\:{if}\:\exists{u}\geqslant\mathrm{0}\:{for}\:{A},\:{rod}\:{can}\:{rest}. \\ $$$${and}\:{if}\:{at}\:{least}\:{one}\:{of}\:{the}\:{inequalities}\:{in} \\ $$$${A}\:{equal}\:{zero}\:,\:{rod}\:{is}\:{about}\:{slip} \\ $$
Commented by mr W last updated on 02/Oct/22
thanks for your solution sir!  i need some time to follow it.
$${thanks}\:{for}\:{your}\:{solution}\:{sir}! \\ $$$${i}\:{need}\:{some}\:{time}\:{to}\:{follow}\:{it}. \\ $$
Answered by mr W last updated on 03/Oct/22
Commented by mr W last updated on 03/Oct/22
background knowledge 1  an object doesn′t slip, when the angle  θ between the resultant contact force  and the normal of the contact surface  is less than θ^∗  which is tan^(−1) μ.   the object is about to slip,   when θ=θ^∗ =tan^(−1) μ.
$$\boldsymbol{{background}}\:\boldsymbol{{knowledge}}\:\mathrm{1} \\ $$$${an}\:{object}\:{doesn}'{t}\:{slip},\:{when}\:{the}\:{angle} \\ $$$$\theta\:{between}\:{the}\:{resultant}\:{contact}\:{force} \\ $$$${and}\:{the}\:{normal}\:{of}\:{the}\:{contact}\:{surface} \\ $$$${is}\:{less}\:{than}\:\theta^{\ast} \:{which}\:{is}\:\mathrm{tan}^{−\mathrm{1}} \mu.\: \\ $$$${the}\:{object}\:{is}\:{about}\:{to}\:{slip},\: \\ $$$${when}\:\theta=\theta^{\ast} =\mathrm{tan}^{−\mathrm{1}} \mu. \\ $$
Commented by mr W last updated on 02/Oct/22
Commented by mr W last updated on 03/Oct/22
let ξ=(a/L), η=(b/L)  c=(√(a^2 +b^2 ))=L(√(ξ^2 +η^2 ))  h=(√(L^2 −a^2 −b^2 ))=L(√(1−ξ^2 −η^2 ))  sin φ=(a/c)=(ξ/( (√(ξ^2 +η^2 ))))    tan θ_1 ≤μ  tan θ_2 ≤μ   ⇒cos θ_2 ≥(1/( (√(1+μ^2 ))))    tan β=(c/h)=((√(ξ^2 +η^2 ))/( (√(1−ξ^2 −η^2 ))))  sin ϕ×sin φ=cos θ_2   ⇒sin ϕ=((cos θ_2 )/(sin φ))≥(1/ξ)(√((ξ^2 +η^2 )/(1+μ^2 )))  ⇒tan ϕ≥(√((ξ^2 +η^2 )/(μ^2 ξ^2 −η^2 )))    ((CD)/(CB))=((sin (β+ϕ))/(sin ϕ))=((sin β)/(tan ϕ))+cos β  ((CD)/(AC))=((sin (β−θ_1 ))/(sin θ_1 ))=((sin β)/(tan θ_1 ))−cos β  ((sin β)/(tan θ_1 ))−cos β=((sin β)/(tan ϕ))+cos β  (1/(tan θ_1 ))−(1/(tan ϕ))=(2/(tan β))  (1/μ)−((√(μ^2 ξ^2 −η^2 ))/( (√(ξ^2 +η^2 ))))≤((2(√(1−ξ^2 −η^2 )))/( (√(ξ^2 +η^2 ))))   determinant (((2(√(1−ξ^2 −η^2 ))+(√(μ^2 ξ^2 −η^2 ))−((√(ξ^2 +η^2 ))/μ)≥0)))
$${let}\:\xi=\frac{{a}}{{L}},\:\eta=\frac{{b}}{{L}} \\ $$$${c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={L}\sqrt{\xi^{\mathrm{2}} +\eta^{\mathrm{2}} } \\ $$$${h}=\sqrt{{L}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }={L}\sqrt{\mathrm{1}−\xi^{\mathrm{2}} −\eta^{\mathrm{2}} } \\ $$$$\mathrm{sin}\:\phi=\frac{{a}}{{c}}=\frac{\xi}{\:\sqrt{\xi^{\mathrm{2}} +\eta^{\mathrm{2}} }} \\ $$$$ \\ $$$$\mathrm{tan}\:\theta_{\mathrm{1}} \leqslant\mu \\ $$$$\mathrm{tan}\:\theta_{\mathrm{2}} \leqslant\mu\: \\ $$$$\Rightarrow\mathrm{cos}\:\theta_{\mathrm{2}} \geqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }} \\ $$$$ \\ $$$$\mathrm{tan}\:\beta=\frac{{c}}{{h}}=\frac{\sqrt{\xi^{\mathrm{2}} +\eta^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−\xi^{\mathrm{2}} −\eta^{\mathrm{2}} }} \\ $$$$\mathrm{sin}\:\varphi×\mathrm{sin}\:\phi=\mathrm{cos}\:\theta_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\varphi=\frac{\mathrm{cos}\:\theta_{\mathrm{2}} }{\mathrm{sin}\:\phi}\geqslant\frac{\mathrm{1}}{\xi}\sqrt{\frac{\xi^{\mathrm{2}} +\eta^{\mathrm{2}} }{\mathrm{1}+\mu^{\mathrm{2}} }} \\ $$$$\Rightarrow\mathrm{tan}\:\varphi\geqslant\sqrt{\frac{\xi^{\mathrm{2}} +\eta^{\mathrm{2}} }{\mu^{\mathrm{2}} \xi^{\mathrm{2}} −\eta^{\mathrm{2}} }} \\ $$$$ \\ $$$$\frac{{CD}}{{CB}}=\frac{\mathrm{sin}\:\left(\beta+\varphi\right)}{\mathrm{sin}\:\varphi}=\frac{\mathrm{sin}\:\beta}{\mathrm{tan}\:\varphi}+\mathrm{cos}\:\beta \\ $$$$\frac{{CD}}{{AC}}=\frac{\mathrm{sin}\:\left(\beta−\theta_{\mathrm{1}} \right)}{\mathrm{sin}\:\theta_{\mathrm{1}} }=\frac{\mathrm{sin}\:\beta}{\mathrm{tan}\:\theta_{\mathrm{1}} }−\mathrm{cos}\:\beta \\ $$$$\frac{\mathrm{sin}\:\beta}{\mathrm{tan}\:\theta_{\mathrm{1}} }−\mathrm{cos}\:\beta=\frac{\mathrm{sin}\:\beta}{\mathrm{tan}\:\varphi}+\mathrm{cos}\:\beta \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\theta_{\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}=\frac{\mathrm{2}}{\mathrm{tan}\:\beta} \\ $$$$\frac{\mathrm{1}}{\mu}−\frac{\sqrt{\mu^{\mathrm{2}} \xi^{\mathrm{2}} −\eta^{\mathrm{2}} }}{\:\sqrt{\xi^{\mathrm{2}} +\eta^{\mathrm{2}} }}\leqslant\frac{\mathrm{2}\sqrt{\mathrm{1}−\xi^{\mathrm{2}} −\eta^{\mathrm{2}} }}{\:\sqrt{\xi^{\mathrm{2}} +\eta^{\mathrm{2}} }} \\ $$$$\begin{array}{|c|}{\mathrm{2}\sqrt{\mathrm{1}−\xi^{\mathrm{2}} −\eta^{\mathrm{2}} }+\sqrt{\mu^{\mathrm{2}} \xi^{\mathrm{2}} −\eta^{\mathrm{2}} }−\frac{\sqrt{\xi^{\mathrm{2}} +\eta^{\mathrm{2}} }}{\mu}\geqslant\mathrm{0}}\\\hline\end{array} \\ $$
Commented by mr W last updated on 03/Oct/22
background knowledge 2  when an object is in equilibrium under  the action of three forces, then these  three forces must intersect at the  same point and their sum (vector) is  zero.
$$\boldsymbol{{background}}\:\boldsymbol{{knowledge}}\:\mathrm{2} \\ $$$${when}\:{an}\:{object}\:{is}\:{in}\:{equilibrium}\:{under} \\ $$$${the}\:{action}\:{of}\:{three}\:{forces},\:{then}\:{these} \\ $$$${three}\:{forces}\:{must}\:{intersect}\:{at}\:{the} \\ $$$${same}\:{point}\:{and}\:{their}\:{sum}\:\left({vector}\right)\:{is} \\ $$$${zero}. \\ $$
Commented by mr W last updated on 03/Oct/22
Commented by mr W last updated on 02/Oct/22
Commented by mr W last updated on 02/Oct/22
Commented by mr W last updated on 02/Oct/22
= END =
$$=\:{END}\:= \\ $$
Commented by mr W last updated on 03/Oct/22
following diagram shows the range  for a and b such that the rod is in  equilibrium.  example μ=0.5.
$${following}\:{diagram}\:{shows}\:{the}\:{range} \\ $$$${for}\:{a}\:{and}\:{b}\:{such}\:{that}\:{the}\:{rod}\:{is}\:{in} \\ $$$${equilibrium}. \\ $$$${example}\:\mu=\mathrm{0}.\mathrm{5}. \\ $$
Commented by mr W last updated on 02/Oct/22
Commented by mahdipoor last updated on 02/Oct/22
very nice method for answer!  but ,   in lemma 1 only correctness of μN≤T  and  in lemma 2 only Σ Moment is zero  are guaranteed .  how about ΣF=0 ?  if ΣF=0 then lemma 2 is true.  i think balance of forces is not in solution  and that′s why (mg) is ineffective  in equilibrium conditions.
$${very}\:{nice}\:{method}\:{for}\:{answer}! \\ $$$${but}\:,\: \\ $$$${in}\:{lemma}\:\mathrm{1}\:{only}\:{correctness}\:{of}\:\mu{N}\leqslant{T} \\ $$$${and} \\ $$$${in}\:{lemma}\:\mathrm{2}\:{only}\:\Sigma\:{Moment}\:{is}\:{zero} \\ $$$${are}\:{guaranteed}\:. \\ $$$${how}\:{about}\:\Sigma{F}=\mathrm{0}\:? \\ $$$${if}\:\Sigma{F}=\mathrm{0}\:{then}\:{lemma}\:\mathrm{2}\:{is}\:{true}. \\ $$$${i}\:{think}\:{balance}\:{of}\:{forces}\:{is}\:{not}\:{in}\:{solution} \\ $$$${and}\:{that}'{s}\:{why}\:\left({mg}\right)\:{is}\:{ineffective} \\ $$$${in}\:{equilibrium}\:{conditions}. \\ $$
Commented by mr W last updated on 03/Oct/22
1) μN≤T means θ≤tan^(−1) μ  2) here i just want to say:         if equilibrium, then 3 forces        intersect at same point.        it is not said: if 3 forces intersect        at same point, then equilbrium.        certainly i know Σ_(1,2,3) F_i  should be        zero for equilibrium. but for solving        this problem, i only need to know        “if equilibrium, then 3 forces        intersect at same point”. that        ΣF=0 is also true, is clear. but i        don′t need this for solving the        problem.  3)  for solving the problem, only        the directions of the forces are        interesting, not their sizes.         therefore the weight of the rod mg        doesn′t affect the result. in fact        it is more a question of geometry.
$$\left.\mathrm{1}\right)\:\mu{N}\leqslant{T}\:{means}\:\theta\leqslant\mathrm{tan}^{−\mathrm{1}} \mu \\ $$$$\left.\mathrm{2}\right)\:{here}\:{i}\:{just}\:{want}\:{to}\:{say}:\: \\ $$$$\:\:\:\:\:\:{if}\:{equilibrium},\:{then}\:\mathrm{3}\:{forces} \\ $$$$\:\:\:\:\:\:{intersect}\:{at}\:{same}\:{point}. \\ $$$$\:\:\:\:\:\:{it}\:{is}\:{not}\:{said}:\:{if}\:\mathrm{3}\:{forces}\:{intersect} \\ $$$$\:\:\:\:\:\:{at}\:{same}\:{point},\:{then}\:{equilbrium}. \\ $$$$\:\:\:\:\:\:{certainly}\:{i}\:{know}\:\underset{\mathrm{1},\mathrm{2},\mathrm{3}} {\sum}{F}_{{i}} \:{should}\:{be} \\ $$$$\:\:\:\:\:\:{zero}\:{for}\:{equilibrium}.\:{but}\:{for}\:{solving} \\ $$$$\:\:\:\:\:\:{this}\:{problem},\:{i}\:{only}\:{need}\:{to}\:{know} \\ $$$$\:\:\:\:\:\:“{if}\:{equilibrium},\:{then}\:\mathrm{3}\:{forces} \\ $$$$\:\:\:\:\:\:{intersect}\:{at}\:{same}\:{point}''.\:{that} \\ $$$$\:\:\:\:\:\:\Sigma{F}=\mathrm{0}\:{is}\:{also}\:{true},\:{is}\:{clear}.\:{but}\:{i} \\ $$$$\:\:\:\:\:\:{don}'{t}\:{need}\:{this}\:{for}\:{solving}\:{the} \\ $$$$\:\:\:\:\:\:{problem}. \\ $$$$\left.\mathrm{3}\right)\:\:{for}\:{solving}\:{the}\:{problem},\:{only} \\ $$$$\:\:\:\:\:\:{the}\:{directions}\:{of}\:{the}\:{forces}\:{are} \\ $$$$\:\:\:\:\:\:{interesting},\:{not}\:{their}\:{sizes}.\: \\ $$$$\:\:\:\:\:\:{therefore}\:{the}\:{weight}\:{of}\:{the}\:{rod}\:{mg} \\ $$$$\:\:\:\:\:\:{doesn}'{t}\:{affect}\:{the}\:{result}.\:{in}\:{fact} \\ $$$$\:\:\:\:\:\:{it}\:{is}\:{more}\:{a}\:{question}\:{of}\:{geometry}. \\ $$
Commented by mahdipoor last updated on 02/Oct/22
thank you , you are right ! ♥
$${thank}\:{you}\:,\:{you}\:{are}\:{right}\:!\:\heartsuit \\ $$
Commented by mahdipoor last updated on 03/Oct/22
and one more question:  in special case: let a^2 +b^2 =L^2  (⇒ h=0   and ξ^2 +η^2 =1)     its means , rod is horizontal  its seems that in this case,0≤ξ≤1  and μ is ineffective  from your answer :  0+(√(μ^2 ξ^2 −η^2 ))−(1/μ)≥0 ⇒ ξ≥(1/μ)    what is wrong?
$${and}\:{one}\:{more}\:{question}: \\ $$$${in}\:{special}\:{case}:\:{let}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={L}^{\mathrm{2}} \:\left(\Rightarrow\:{h}=\mathrm{0}\right. \\ $$$$\left.\:{and}\:\xi^{\mathrm{2}} +\eta^{\mathrm{2}} =\mathrm{1}\right)\:\:\: \\ $$$${its}\:{means}\:,\:{rod}\:{is}\:{horizontal} \\ $$$${its}\:{seems}\:{that}\:{in}\:{this}\:{case},\mathrm{0}\leqslant\xi\leqslant\mathrm{1} \\ $$$${and}\:\mu\:{is}\:{ineffective} \\ $$$${from}\:{your}\:{answer}\:: \\ $$$$\mathrm{0}+\sqrt{\mu^{\mathrm{2}} \xi^{\mathrm{2}} −\eta^{\mathrm{2}} }−\frac{\mathrm{1}}{\mu}\geqslant\mathrm{0}\:\Rightarrow\:\xi\geqslant\frac{\mathrm{1}}{\mu}\:\: \\ $$$${what}\:{is}\:{wrong}? \\ $$
Commented by Tawa11 last updated on 03/Oct/22
Great sirs
$$\mathrm{Great}\:\mathrm{sirs} \\ $$
Commented by mr W last updated on 03/Oct/22
to mahdipoor sir:  thanks for your feedback!   i appreciate such substantial feedback  very much.  your question is interesting. i also  noticed this issue as i saw the graph  for the case μ=2, which i didn′t post.  i considered to study it and try to   explain it later, when i have more time.  since you have asked the same   question about this issue, i have spent   some time to take a closer look at it.  i think it is not a fault in the  solution. it can be explained as  following.
$$\boldsymbol{{to}}\:\boldsymbol{{mahdipoor}}\:\boldsymbol{{sir}}: \\ $$$${thanks}\:{for}\:{your}\:{feedback}!\: \\ $$$${i}\:{appreciate}\:{such}\:{substantial}\:{feedback} \\ $$$${very}\:{much}. \\ $$$${your}\:{question}\:{is}\:{interesting}.\:{i}\:{also} \\ $$$${noticed}\:{this}\:{issue}\:{as}\:{i}\:{saw}\:{the}\:{graph} \\ $$$${for}\:{the}\:{case}\:\mu=\mathrm{2},\:{which}\:{i}\:{didn}'{t}\:{post}. \\ $$$${i}\:{considered}\:{to}\:{study}\:{it}\:{and}\:{try}\:{to}\: \\ $$$${explain}\:{it}\:{later},\:{when}\:{i}\:{have}\:{more}\:{time}. \\ $$$${since}\:{you}\:{have}\:{asked}\:{the}\:{same}\: \\ $$$${question}\:{about}\:{this}\:{issue},\:{i}\:{have}\:{spent}\: \\ $$$${some}\:{time}\:{to}\:{take}\:{a}\:{closer}\:{look}\:{at}\:{it}. \\ $$$${i}\:{think}\:{it}\:{is}\:{not}\:{a}\:{fault}\:{in}\:{the} \\ $$$${solution}.\:{it}\:{can}\:{be}\:{explained}\:{as} \\ $$$${following}. \\ $$
Commented by mr W last updated on 03/Oct/22
Commented by mr W last updated on 03/Oct/22
in case of μ≥1, θ^∗ =tan^(−1) μ≥45°.  in this case the rod can lie in a very  flat position, close to horizontal or  even in horizontal position, and still  be kept in equilbrium. we can see in  the diagram above that within the  angles θ^∗  contact forces R_1  and R_2    can be found which meet the mg at  the same point, therefore a equilibrium  state is possible. certainly a absolutely  horizontal equilibrium position is  just theory. but mathematically it is   possible.
$${in}\:{case}\:{of}\:\mu\geqslant\mathrm{1},\:\theta^{\ast} =\mathrm{tan}^{−\mathrm{1}} \mu\geqslant\mathrm{45}°. \\ $$$${in}\:{this}\:{case}\:{the}\:{rod}\:{can}\:{lie}\:{in}\:{a}\:{very} \\ $$$${flat}\:{position},\:{close}\:{to}\:{horizontal}\:{or} \\ $$$${even}\:{in}\:{horizontal}\:{position},\:{and}\:{still} \\ $$$${be}\:{kept}\:{in}\:{equilbrium}.\:{we}\:{can}\:{see}\:{in} \\ $$$${the}\:{diagram}\:{above}\:{that}\:{within}\:{the} \\ $$$${angles}\:\theta^{\ast} \:{contact}\:{forces}\:{R}_{\mathrm{1}} \:{and}\:{R}_{\mathrm{2}} \: \\ $$$${can}\:{be}\:{found}\:{which}\:{meet}\:{the}\:{mg}\:{at} \\ $$$${the}\:{same}\:{point},\:{therefore}\:{a}\:{equilibrium} \\ $$$${state}\:{is}\:{possible}.\:{certainly}\:{a}\:{absolutely} \\ $$$${horizontal}\:{equilibrium}\:{position}\:{is} \\ $$$${just}\:{theory}.\:{but}\:{mathematically}\:{it}\:{is}\: \\ $$$${possible}. \\ $$
Commented by mr W last updated on 03/Oct/22
Commented by mr W last updated on 03/Oct/22
Commented by mr W last updated on 03/Oct/22
for μ<1, θ^∗ <45°, we can see that an  equilibrium state in such a flat   position is not possible, because we  can not find within the angles θ^∗  such  contact forces R_1  and R_2  which meet  the mg at the same point. in this case  c<L, i.e. ξ^2 +η^2 <1.
$${for}\:\mu<\mathrm{1},\:\theta^{\ast} <\mathrm{45}°,\:{we}\:{can}\:{see}\:{that}\:{an} \\ $$$${equilibrium}\:{state}\:{in}\:{such}\:{a}\:{flat}\: \\ $$$${position}\:{is}\:{not}\:{possible},\:{because}\:{we} \\ $$$${can}\:{not}\:{find}\:{within}\:{the}\:{angles}\:\theta^{\ast} \:{such} \\ $$$${contact}\:{forces}\:{R}_{\mathrm{1}} \:{and}\:{R}_{\mathrm{2}} \:{which}\:{meet} \\ $$$${the}\:{mg}\:{at}\:{the}\:{same}\:{point}.\:{in}\:{this}\:{case} \\ $$$${c}<{L},\:{i}.{e}.\:\xi^{\mathrm{2}} +\eta^{\mathrm{2}} <\mathrm{1}. \\ $$
Commented by mr W last updated on 03/Oct/22
Commented by mahdipoor last updated on 04/Oct/22
exactly   in this answer,it is important that the  three forces are concurrent  in horizontal positon,the forces are parallel  and are not concurrent .  that is why the answer for this case is fals
$${exactly}\: \\ $$$${in}\:{this}\:{answer},{it}\:{is}\:{important}\:{that}\:{the} \\ $$$${three}\:{forces}\:{are}\:{concurrent} \\ $$$${in}\:{horizontal}\:{positon},{the}\:{forces}\:{are}\:{parallel} \\ $$$${and}\:{are}\:{not}\:{concurrent}\:. \\ $$$${that}\:{is}\:{why}\:{the}\:{answer}\:{for}\:{this}\:{case}\:{is}\:{fals} \\ $$

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