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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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